• MHB
Gold Member
MHB
Re: Integration lessons (continued)

4.7.2 Exercises

Exercise 1

$$\displaystyle \int^\infty_0 \text{erfc}^2(x)\,dx = \frac{2-\sqrt{2}}{\sqrt{\pi}}$$​

Proof

Integration by parts

$$I = x\text{erfc}^2(x)\left. \right]^\infty_0 -2\int^\infty_0 x\text{erfc}'(x)\text{erfc}(x)\,dx$$

The first integral goes to 0

$$I = -2\int^\infty_0 x\,\text{erfc}'(x)\text{erfc}(x)\,dx$$

The derivative of the complementary error function

$$\text{erfc}'(x) = (1-\text{erf}(x))'= -\frac{2}{\sqrt{\pi}}e^{-x^2}$$

$$I= \frac{4}{\sqrt{\pi}}\int^\infty_0xe^{-x^2}\text{erfc}(x)\,dx$$

Integration by parts again we have

$$I=\frac{2}{\sqrt{\pi}}e^{-x^2}\text{erfc}(x)\left. \right]^\infty_0-\frac{4}{\pi}\int^\infty_0e^{-2t^2}\,dt$$

At infinity the integral goes to 0 and at 0 we have

$$\frac{2}{\sqrt{\pi}}e^{-x^2}\text{erfc}(x)_{x=0}=\frac{2}{\sqrt{\pi}}(1-\text{erf}(0))=\frac{2}{\sqrt{\pi}}$$

$$\int^\infty_0e^{-2t^2}\,dt = \frac{\sqrt{\pi}}{2\sqrt{2}}$$

Collecting the results together we have

$$I = \frac{2}{\sqrt{\pi}}-\frac{\sqrt{\pi}}{2\sqrt{2}} \times \frac{4}{\pi}=\frac{2-\sqrt{2}}{\sqrt{\pi}}$$

Exercise 2

$$\displaystyle \int^\infty_0 \sin(x^2) \text{erfc}(x) \,dx =\frac{\pi-2 \coth^{-1}\sqrt{2}}{4 \sqrt{2\pi}}$$​

Proof

Using the substitution $x=\sqrt{t}$

$$\frac{1}{2}\int^\infty_0 \sin(t) t^{-\frac{1}{2}}\text{erfc}(\sqrt{t}) \,dx$$

Consider the function

$$I(a) = \frac{1}{2}\int^\infty_0 \sin(t) t^{-\frac{1}{2}}\text{erfc}(a\sqrt{t}) \,dt$$

Differentiating with respect to $a$ we have

$$I'(a) = \frac{-1}{\sqrt{\pi}}\int^\infty_0 \sin(t) e^{-a^2t}\,dt=\frac{-1}{\sqrt{\pi}}\cdot \frac{1}{a^4+1}$$

Now integrating with respect to $a$

$$I(a) = \frac{-1}{\sqrt{\pi}}\int^a_0 \frac{dx}{x^4+1}+C$$

To evaluate the constant we take $a \to \infty$

$$I(\infty) = \frac{-1}{\sqrt{\pi}}\int^\infty_0\frac{dx}{x^4+1}+C$$

The function has an anti-derivative and the value is

$$\frac{-1}{\sqrt{\pi}} \int^\infty_0\frac{dx}{x^4+1}=-\frac{\sqrt{\pi}}{2\sqrt{2}}$$

and knowing that

$$\text{erfc}(\infty)=0 \,\,\, \implies \,\,\,\, C =\frac{\sqrt{\pi}}{2\sqrt{2}}$$

Finally we get

$$I(a) = \frac{-1}{\sqrt{\pi}}\int^a_0 \frac{dx}{x^4+1}+\frac{\sqrt{\pi}}{2\sqrt{2}}$$

Plugging $a=1$ we have our integral

$$I(1) = \int^\infty_0 \sin(x^2) \text{erfc}(x) \,dx=\frac{\sqrt{\pi}}{2\sqrt{2}}-\frac{1}{\sqrt{\pi}}\int^1_0 \frac{dx}{x^4+1}$$

Also knowing that

$$\frac{1}{\sqrt{\pi}}\int^1_0 \frac{dx}{x^4+1}=\frac{\pi+2 \coth^{-1}\sqrt{2}}{4 \sqrt{2\pi}}$$

$$\int^\infty_0 \sin(x^2) \text{erfc}(x) \,dx=\frac{\pi-2 \coth^{-1}\sqrt{2}}{4 \sqrt{2\pi}}$$

[HW]

Can you find closed forms for

$$\displaystyle \int^\infty_0 \text{erfc}^3(x)\, dx = ?$$

$$\displaystyle \int^\infty_0 \text{erfc}^4(x)\, dx = ?$$

$$\displaystyle \int^\infty_0 \text{erfc}^n(x)\, dx = ?$$

Last edited:
Gold Member
MHB
Re: Integration lessons (continued)

4.Integration using special functions (continued)

4.8 Exponential integral function

Definition

$$\displaystyle E(x)=\int^\infty_x\frac{e^{-t}}{t}\,dt$$​

4.8.1 Exercises

Prove that

$$\lim_{x \to 0}\left[\log(x)+E(x) \right]=-\gamma$$

Integration by parts for $E(x)$

$$E(x)=e^{-t}\log(t) \left.\right]^\infty_x+\int^\infty_x\log(x)e^{-t}\,dt$$

$$E(x)=-e^{-x}\log(x)+\int^\infty_x\log(x)e^{-t}\,dt$$

$$\lim_{x \to 0}\left[\log(x)+E(x) \right]=\lim_{x \to 0}\left(\log(x)-e^{-x}\log(x)\right)+\int^\infty_0\log(x)e^{-t}\,dt$$

The first limit goes to 0

$$\lim_{x \to 0}\left[\log(x)+E(x) \right]=\int^\infty_0\log(x)e^{-t}\,dt=\psi(1)=-\gamma$$

Prove that

$$\int^\infty_0 x^{p-1} E(ax)\,dx =\frac{\Gamma(p)}{pa^{p}} \,\,\, p>0$$

Integrating by parts we have

$$\int^\infty_0 x^{p-1} E(ax)\,dx=\frac{1}{p}x^pE(ax) \left. \right]^\infty_0+\frac{1}{ap} \int^\infty_0 x^{p-1}e^{-ax}\,dx$$

The first limit goes to 0

$$I=\frac{1}{ap} \int^\infty_0 x^{p-1}e^{-ax}\,dx= \frac{1}{pa^p }\int^\infty_0 x^{p-1}e^{-x}\,dx=\frac{\Gamma(p)}{p \, a^p}$$

Prove the more general case

$$\int^\infty_0 x^{p-1}e^{ax}E(ax) \, dx = \frac{\pi}{\sin(a\pi)}\cdot \frac{\Gamma(p)}{a^p}$$

Switch to the integral representation

$$\int^\infty_0 x^{p-1}e^{ax}\int^\infty_{ax}\frac{e^{-t}}{t}\,dt \, dx$$

Use the substitution $t=ax \,y$

$$\int^\infty_0\int^\infty_{1} x^{p-1}e^{ax}\frac{e^{-ax\,y}}{y}\,dy \, dx$$

By switching the two integrals

$$\int^\infty_1\frac{1}{y}\int^\infty_{0} x^{p-1}e^{-ax(y-1)} dx\,dy$$

By the Laplace identities

$$\frac{\Gamma(p)}{a^p}\int^\infty_1 \frac{1}{y(y-1)^p} \, dy$$

Now let $y=1/x$

$$\frac{\Gamma(p)}{a^p}\int^1_0 x^{p-1}(1-x)^{-p} \, dx$$

Using the the reflection formula for the Gamma function

$$\frac{\Gamma(p)}{a^p}\int^1_0 x^{p-1}(1-x)^{-p} \, dx = \frac{\pi}{\sin(a\pi)}\cdot \frac{\Gamma(p)}{a^p}$$

Gold Member
MHB
Re: Integration lessons (continued)

7.8.1 Exercises (continued)

Prove that

$$\int^\infty_0 e^{z}E^2(z) \, dz = \frac{\pi^2}{6}$$

Using the integral representation

$$E^2(z) = \int^\infty_1\int^\infty_1\frac{e^{-x z}e^{-y z}}{x y}\,dx \,d y$$

$$\int^\infty_0 e^{z}E^2(z) \, dz = \int^\infty_0 \int^\infty_1 \,\int^\infty_1\frac{e^{-z(x+y-1)}}{x y}\,dx \,d y \, dz$$

Switching the integration

$$\int^\infty_1 \frac{1}{y} \int^\infty_1 \frac{1}{x}\,\int^\infty_0e^{-z(x+y-1)}\,dz\,dx \,d y$$

$$\int^\infty_1 \frac{1}{y} \int^\infty_1\frac{1}{x(x+y-1)}\,dx \, dy$$

The inner integral is an elmetnary integral

$$\int^\infty_1\frac{1}{x(x+y-1)}\,dx = -\frac{\log(y)}{1-y}$$

$$\int^\infty_1 \frac{\log(y)}{y(y-1)} \,dy$$

Now use the substitution $y=1/x$

$$-\int^1_0 \frac{\log(x)}{(1-x)} \,dx=-\int^1_0\frac{\log(1-x)}{x}\,dx = \text{Li}_2(1)=\frac{\pi^2}{6}$$

Prove that

$$\int^\infty_0 z^{p-1} E^2(z)\,dz=\frac{2\Gamma(p)}{p^2}\, _2F_1(p,p;p+1;-1)$$

Consider the general case

$$F(p)=\int^\infty_0 z^{p-1} E^2(z)\,dz$$

Integrating by parts

$$F(p)=\frac{2}{p} \int^\infty_0 z^{p-1}e^{-z}E(z)\,dz$$

Write the integral representation

$$\frac{2}{p} \int^\infty_0 z^{p-1}e^{-z}\int^\infty_1 \frac{e^{-zt}}{t},dtdz$$

$$\frac{2}{p} \int^\infty_1\frac{1}{t}\int^\infty_0 z^{p-1}e^{-z(1+t)} \,dz\,dt$$

Take the inner integral

$$\frac{2\Gamma(p)}{p} \int^\infty_1\frac{dt}{t(1+t)^p}$$

Use the substitution $t=1/x$

$$\frac{2\Gamma(p)}{p} \int^1_0\frac{x^{p-1}}{(1+x)^p}\,dx$$

Using the Hypergeomtirc identity

$$\beta(c-b,b) \, _2F_1(a,b;c;z)=\int_0^1 \frac{x^{b-1}(1-x)^{c-b-1}}{(1-xz)^a}\, dx$$

put $c=p+1;b=p;a=p;z=-1$

$$\beta(1,p) \, _2F_1(p,p;p+1;-1)=\int_0^1 \frac{x^{p-1}}{(1+x)^p}\, dx$$

Hence the result

$$\int^\infty_0 z^{p-1} E^2(z)\,dz=\frac{2\Gamma(p)}{p^2}\, _2F_1(p,p;p+1;-1)$$

[HW]

Find a general formula for

$$\int^\infty _0 x^n E^2(x) \, dx = ? \,\,\,\,\,\, n \in \mathbb{N}$$

Gold Member
MHB
Re: Integration lessons (continued)

4.Integration lessons (continued)

4.9.Complete elliptic integrals

Complete elliptic of first kind

$$\displaystyle K(k)= \int^{\frac{\pi}{2}}_0\frac{d\theta}{\sqrt{1-k^2\sin^2 \theta }}=\int^1_0\frac{dx}{\sqrt{1-x^2}\sqrt{1-k^2x^2}}$$​

Complete elliptic of second kind

$$\displaystyle E(k)= \int^{\frac{\pi}{2}}_0\sqrt{1-k^2\sin^2 \theta }\, d\theta =\int^1_0\frac{\sqrt{1-k^2x^2}}{\sqrt{1-x^2}}dx$$​

Some special values

$$K(i) = \frac{1}{4\sqrt{2\pi}}\Gamma^2\left( \frac{1}{4}\right)$$

By definition we have

$$K(i)=\int^1_0\frac{dx}{\sqrt{1-x^2}\sqrt{1+x^2}}=\int^1_0 \frac{dx}{\sqrt{1-x^4}}$$

Let $x=\sqrt[4]{t}$ we have $dx=\frac{1}{4}t^{\frac{-3}{4}}\,dt$

$$K(i)=\frac{1}{4}\int^1_0t^{\frac{-3}{4}}(1-t)^{\frac{-1}{2}}\,dt$$

By beta function

$$K(i) = \frac{\Gamma\left( \frac{1}{4}\right)\Gamma\left( \frac{1}{2}\right)}{4\Gamma\left( \frac{3}{4}\right)}$$

By reflection formula

$$\Gamma\left( \frac{1}{4}\right)\Gamma\left( \frac{3}{4}\right)=\pi \csc\left( \frac{\pi}{4}\right)=\pi \sqrt{2}$$

$$K(i) = \frac{\Gamma^2\left( \frac{1}{4}\right)\Gamma\left( \frac{1}{2}\right)}{4\pi \sqrt{2}}=\frac{\Gamma^2\left( \frac{1}{4}\right)}{4\sqrt{2\pi}}$$

$$E(i)=\frac{\Gamma^2\left( \frac{1}{4}\right)}{4\sqrt{2\pi}}+\frac{\Gamma^2\left(\frac{3}{4} \right)}{\sqrt{2\pi}}$$

By definition we have

$$E(i)=\int^1_0\frac{\sqrt{1+x^2}}{\sqrt{1-x^2}}dx$$

Separating the two integrals

$$E(i)=\int^1_0\frac{1+x^2}{\sqrt{1-x^4}}dx=\int^1_0\frac{1}{\sqrt{1-x^4}}dx+\int^1_0\frac{x^2}{\sqrt{1-x^4}}dx$$

The first integral is $K(i)$ for the second integral use $x=\sqrt[4]{t}$

$$\frac{1}{4}\int^1_0t^{\frac{3}{4}-1}(1-t)^{-\frac{1}{2}}\,dt=\frac{\Gamma\left( \frac{3}{4}\right)\Gamma\left( \frac{1}{2}\right)}{4\Gamma\left( \frac{5}{4}\right)}=\frac{\Gamma^2\left(\frac{3}{4} \right)}{\sqrt{2\pi}}$$

Hence we have

$$E(i) = K(i) +\frac{\Gamma^2\left(\frac{3}{4} \right)}{\sqrt{2\pi}} = \frac{\Gamma^2\left( \frac{1}{4}\right)}{4\sqrt{2\pi}}+\frac{\Gamma^2\left(\frac{3}{4} \right)}{\sqrt{2\pi}}$$

4.9.2.Some integrals involving elliptic integrals

$$\int^1_0 K(k) \, dk = 2G \,\,\,\,\, ;\,G=\text{Catalan's constant }$$

$$I=\int^1_0 \int^1_0\frac{1}{\sqrt{1-x^2}\sqrt{1-k^2x^2}} \,dx\, dk$$

Switching the two integrals

$$I=\int^1_0 \frac{1}{\sqrt{1-x^2}}\int^1_0\frac{1}{\sqrt{1-k^2x^2}} \,dk\, dx$$

$$I=\int^1_0 \frac{\arcsin x}{x\sqrt{1-x^2}} dx$$

Now let $\arcsin x = t$ hence we have $x=\sin t$

$$I=\int^{\frac{\pi}{2}}_0 \frac{t}{\sin \, t} dt$$

The previous integral is a representation of the constant

$$G=\frac{I}{2} \,\,\, \implies \,\,\, I=2G$$

[HW]

Find the Values

$$E(0) \,\,\, , \,\,\,\, K \left( \frac{1}{\sqrt{2}}\right)$$

Last edited:
Gold Member
MHB
Re: Integration lessons (continued)

4.9.3.Properties of elliptic integrals

$$\displaystyle \tag{1}K(\sqrt{k}) = \frac{1}{\sqrt{1-k}}K\left(\sqrt{\frac{k}{k-1} }\right)\,\,;\,k\notin [1,\infty)$$​

Starting by the integral representation

$$K(k)=\int^1_0 \frac{dx}{\sqrt{1-x^2}\sqrt{1-k^2x^2}}$$

By using that $x=\sqrt{1-y^2}$

$$\int^1_0\frac{y\,dy}{\sqrt{1-y^2}\sqrt{1-(1-y^2)}\sqrt{1-k^2(1-y^2)}}$$

$$\int^1_0\frac{ dy}{\sqrt{1-y^2}\sqrt{1-k^2+k^2y^2}}=\int^1_0\frac{ dy}{\sqrt{1-k^2}\sqrt{1-y^2}\sqrt{1-\frac{k^2}{k^2-1}y^2}}$$

$$K(k)=\frac{1}{\sqrt{1-k^2}}K\left( \sqrt{\frac{k^2}{k^2-1}}\right)$$

which is equivalent to the property by taking the root

$$\displaystyle \tag{2} E(\sqrt{k}) = \sqrt{1-k}\,E\left(\sqrt{\frac{k}{k-1}} \right)\,\,;\,k\notin [1,\infty)$$​

Staring by the integral representation

$$E(\sqrt{k})=\int^1_0 \frac{\sqrt{1-kx^2}}{\sqrt{1-x^2}}\,dx$$

By using that $x=\sqrt{1-y^2}$

$$E(\sqrt{k})=\sqrt{1-k}\int^1_0 \frac{\sqrt{1-\frac{k}{k-1}y}}{\sqrt{1-y^2}}\,dx=\sqrt{1-k}\,E\left(\sqrt{\frac{k}{k-1} }\right)$$

Using the above formulas we can have some values

For the value $k=-1$

$$K(i) = \frac{1}{\sqrt{2}}K\left(\frac{1}{\sqrt{2}} \right)$$

$$K\left(\frac{1}{\sqrt{2}} \right) = \sqrt{2}K(i)= \frac{1}{4\sqrt{\pi}}\Gamma^2\left( \frac{1}{4}\right)$$

Similarly we have

$$E\left ( \frac{1}{\sqrt{2}}\right)= \frac{\Gamma^2\left( \frac{1}{4}\right)}{8\sqrt{\pi}}+\frac{\Gamma^2\left(\frac{3}{4} \right)}{2\sqrt{\pi}}$$

Hence using $k\to -k$

$$K(\sqrt{k}\,i) = \frac{1}{\sqrt{1+k}}K\left(\frac{\sqrt{k}}{\sqrt{1+k}}\right)\,\,$$

$$E(\sqrt{k}\,i) = \sqrt{1+k}\,E\left(\frac{\sqrt{k}}{\sqrt{1+k}}\right)\,\,$$

Gold Member
MHB
Re: Integration lessons (continued)

4.9.4.Elliptic integrals as a Hypergeometric function

Definition

$$\displaystyle K(k)=\frac{\pi}{2}{}_2F_1\left(\frac{1}{2},\frac{1}{2},1,k^2 \right)$$​

Using the integral representation of the hypergeometric function

$$\beta(c-b,b) \, _2F_1(a,b,c,z)=\int_0^1 \frac{t^{b-1}(1-t)^{c-b-1}}{(1-tz)^a}\, dt$$

Now use the substitution $t=x^2$ and $z=k^2$

$$\beta(c-b,b) \, _2F_1(a,b,c,k^2)=2\int_0^1 \frac{x^{2b-1}(1-x^2)^{c-b-1}}{(1-k^2x^2)^a}\, dx$$

Put $a=\frac{1}{2}\,;$ $b=\frac{1}{2}$ and $c=1$

$$\int_0^1 \frac{dx}{\sqrt{1-x^2}\sqrt{1-k^2x^2}}\, =\frac{1}{2}\beta(1/2,1/2) \, _2F_1\left(\frac{1}{2},\frac{1}{2},1,k^2 \right)$$

By the beta function we have

$$\frac{1}{2}\beta(1/2,1/2)=\frac{1}{2}\Gamma^2\left(\frac{1}{2} \right)=\frac{\pi}{2}$$

Hence the result

$$K(k) =\frac{\pi}{2}\, _2F_1\left(\frac{1}{2},\frac{1}{2},1,k^2 \right)$$

By the same approach we have

$$\displaystyle E(k) =\frac{\pi}{2}\, _2F_1\left(\frac{1}{2},-\frac{1}{2},1,k^2 \right)$$​

A relation using the Quadratic transformation

$$_2F_1\left( a,b,2b,\frac{4z}{(1+z)^2} \right)=(1+z)^{2a}\,_2F_1\left(a,a-b+\frac{1}{2},b+\frac{1}{2},z^2 \right).$$

Hence we can deduce by putting $a=b=1/2$

$$K\left( \frac{2\sqrt{k}}{1+k} \right)=(1+k)K(k)$$

Or we have

$$K(k)=\frac{1}{k+1}K\left( \frac{2\sqrt{k}}{1+k} \right)$$

Hence we have for $k=\frac{1}{\sqrt{2}}$

$$K\left(2 \sqrt{-4+3 \sqrt{2}} \right)=\frac{1+\sqrt{2}}{\sqrt{2}}K\left(\frac{1}{\sqrt{2}} \right)= \frac{1+\sqrt{2}}{4\sqrt{2\pi}}\Gamma^2\left( \frac{1}{4}\right)$$

For the elliptic integral of second kind using the hypergeomtric representation with $a=\frac{-1}{2}$ and $b=\frac{1}{2}$

$$_2F_1\left( -1/2,1/2,1,\frac{4z}{(1+z)^2} \right)=(1+z)^{-1}{}_2F_1\left(-1/2,-1/2,1,z^2 \right)$$

The later hypergeometric series can be written in terms of elliptic integrals using some general contiguity relations

$${}_2F_1\left(-1/2,-1/2,1,z^2 \right)=\frac{2}{\pi}\left( 2 E(k)+(k^2-1)K(k)\right)$$

So we have

$$2 E(k)+(k^2-1)K(k)=(k+1)E\left(\frac{2\sqrt{k}}{1+k} \right)$$

For $k=\frac{1}{\sqrt{2}}$

$$E\left(2 \sqrt{-4+3 \sqrt{2}} \right)=\frac{\sqrt{2}}{1+\sqrt{2}}\left[\frac{\Gamma^2\left( \frac{1}{4}\right)}{8\sqrt{\pi}}+\frac{\Gamma^2\left(\frac{3}{4} \right)}{\sqrt{\pi}} \right]$$

Values for complex arguments

Start by the following

$$K\left(\frac{2\sqrt{k}}{1+k} \right)=\int^1_0 \frac{1}{\sqrt{1-x^2}\sqrt{1-\frac{4k}{(1+k)^2}x^2}}\,dx$$

By some simplifications we have

$$E\left(\frac{2\sqrt{k}}{1+k} \right)=(1+k)\int^1_0 \frac{1}{\sqrt{1-x^2}\sqrt{(1+k)^2-4k \, x^2}}\,dx$$

Use $x=\sqrt{1-y^2}$

$$\int^1_0 \frac{1+k}{\sqrt{1-y^2}\sqrt{(1+k)^2-4k\,(1-y^2)}}\,dy=\frac{1+k}{1-k}\int^1_0 \frac{1}{\sqrt{1-y^2}\sqrt{1+\frac{4k}{(1-k)^2}y^2}}\,dy$$

Hence we have

$$K\left(\frac{2\sqrt{k}}{1+k} \right)=\frac{1+k}{1-k}K\left(\frac{2\sqrt{-k}}{1-k} \right)$$

Similarly we have

$$E\left(\frac{2\sqrt{k}}{1+k} \right)=\frac{1-k}{1+k}E\left(\frac{2\sqrt{-k}}{1-k} \right)$$

Using these formulas and the results we got earlier we have for $x=1/\sqrt{2}$
$$K\left(2 \sqrt{-4-3 \sqrt{2}} \right)=\frac{\pi\,\sqrt{\pi}}{4} \cdot \frac{2-\sqrt{2}}{\Gamma^2\left( \frac{3}{4}\right)}$$

$$E\left(2 \sqrt{-4-3 \sqrt{2}} \right)=\frac{(2+\sqrt{2})\left(\pi^2+4 \Gamma^4\left( \frac{3}{4}\right)\right)}{4\sqrt{\pi}\Gamma^2\left( \frac{3}{4}\right)}$$

Last edited:
Gold Member
MHB
Re: Integration lessons (continued)

4.9.5.Differentiation of elliptic integrals

Note : We should remove the variable $k$ and denote elliptic integrals $E$ and $K$ once there is no confusion. It is assumed that the variable is $k$ when we use these symbols.

Differentiation:

Interestingly the derivative of elliptic integrals can be written in terms of elliptic integrals

Derivative of complete elliptic integral of second kind

$$\frac{d}{dk}E=\int^1_0 \frac{\frac{\partial\,}{\partial\,k}\sqrt{1-k^2 x^2}}{\sqrt{1-x^2}}\,dx$$

$$\frac{d}{dk}E=\int^1_0 \frac{-k\,x^2}{\sqrt{1-x^2}\sqrt{1-k^2\,x^2}}\,dx$$

Adding and subtracting 1 results in

$$\frac{1}{k}\int^1_0 \frac{\sqrt{1-k^2x^2}}{\sqrt{1-x^2}}\,dx-\frac{1}{k}\int^1_0 \frac{dx}{\sqrt{1-x^2}\sqrt{1-k^2\,x^2}}$$

Upon realizing the relation to elliptic integrals we conclude

$$\displaystyle \frac{d}{dk}E=\frac{E-K}{k}$$​

For the complete elliptic integral of first kind we need more work

Start by the following

$$\frac{d}{dk}K=\int^1_0 \frac{1}{\sqrt{1-x^2}}\,\frac{\partial\,}{\partial\,k}\left[\frac{1}{\sqrt{1-k^2 x^2}}\right]dx$$

$$\frac{d}{dk}K=\int^1_0 \frac{kx^2}{\sqrt{1-x^2}\sqrt{1-k^2x^2}(1-k^2x^2)}\,dx$$

Adding and subtracting 1 we have

$$\frac{-1}{k}\int^1_0 \frac{1-kx^2-1}{\sqrt{1-x^2}\sqrt{1-k^2x^2}(1-k^2x^2)}\,dx=\frac{1}{k}\int^1_0 \frac{dx}{\sqrt{1-x^2}\sqrt{1-k^2x^2}(1-k^2x^2)}-\frac{K}{k}$$

Let us focus on the first integral

$$\int^1_0 \frac{1}{\sqrt{1-x^2}(1-\, k^2 \, x^2)^{\frac{3}{2}}}\,dx$$

Let $x=\sqrt{t}$ and we have $dx= \frac{1}{2\sqrt{t}}\,dt$

$$\frac{1}{2}\int^1_0 \frac{t^{-\frac{1}{2}}}{\sqrt{1-t}(1-\, k^2 \, t)^{\frac{3}{2}}}\,dx$$

Using the hypergeometric integral representation

$$\frac{1}{2}\int^1_0 \frac{t^{-\frac{1}{2}}}{\sqrt{1-t}(1-\, k^2 \, t)^{\frac{3}{2}}}=\frac{\pi}{2}{}_2F_1\left(\frac{3}{2},\frac{1}{2},1,k^2 \right)$$

Using the linear transformation

$$_2F_1\left(a,b,c,z \right)=(1-z)^{c-a-b}\,{}_2F_1\left(c-a,c-b,c,z \right)$$

We get by putting $k'=\sqrt{1-k^2}$

$$\frac{\pi}{2}{}_2F_1\left(\frac{3}{2},\frac{1}{2},1,k^2 \right)=\frac{1}{1-k^2}\frac{\pi}{2}{}_2F_1\left(-\frac{1}{2},\frac{1}{2},1,k^2 \right)=\frac{E}{k'^2}$$

So finally we get

$$\displaystyle \frac{d}{dk}K=\frac{1}{k}\left( \frac{E}{ k'^2}-K \right)$$​

Note : $k'$ is called the complementary modulus and it should be of interest for us in the next set of lectures.

Gold Member
MHB
Integration lessons continued ...

4.Integration using special functions (continued)

4.10. Euler sums

General definition:

$$\displaystyle S_{p^{\,r},q} = \sum_{k\geq 1} \frac{(H_k^{(p)})^r}{k^q}$$​

Where we define the general harmonic number

$$\displaystyle H_k^{(p)} = \sum_{n=1}^k \frac{1}{n^p} \,\,\,;\,\, H^{(1)}_k \equiv H_k = \sum_{n=1}^k \frac{1}{n}$$​

Euler sums were greatly studied by Euler, hence the name.

Generating function:

$$\displaystyle \sum_{k\geq 1} H_k^{(p)}x^k = \frac{\mathrm{Li}_p(x)}{1-x}$$​

Proof

$$\sum_{k\geq 1} H_k^{(p)}x^k = \sum_{k\geq 1} \sum_{n=1}^k \frac{1}{n^p}x^k$$

By interchanging the two series we have

$$\sum_{n\geq 1} \sum_{k\geq n} \frac{x^k}{n^p} =\sum_{n\geq 1}\frac{1}{n^p} \sum_{k\geq n}x^k$$

The inner sum is a geometric series

$$\frac{1}{1-x} \sum_{n\geq 1} \frac{x^n}{n^p} =\frac{\mathrm{Li}_p(x)}{1-x}$$

We can use this to generate some more functions by integrating. Hence assume $p=1$

$$\sum_{k\geq 1} H_k x^k = -\frac{\log(1-x)}{1-x}$$

Divide by $x$ and integrate to get

$$\sum_{k\geq 1} \frac{H_k}{k} x^k =\mathrm{Li}_2(x)+\frac{1}{2}\log^2(1-x)$$

Now divide by $x$ and integrate again

$$\sum_{k\geq 1} \frac{H_k}{k^2} x^k =\mathrm{Li}_3(x)+\frac{1}{2}\int^x_0 \frac{\log^2(1-t)}{t}\,dt$$

Now let us look at the integral

$$\int^x_0 \frac{\log^2(1-t)}{t}\, dt$$

Integrating by parts we get the following

$$\int^x_0 \frac{\log^2(1-t)}{t}\, dt = - \log(1-x) \text{Li}_2(x) -\int^x_0\frac{\text{Li}_2 (t)}{1-t} \, dt$$

Now we are left with the following integral

$$\int^x_0\frac{\text{Li}_2 (t)}{1-t} \, dt = \int^{1}_{1-x} \frac{\text{Li}_2 (1-t)}{t} \, dt$$

$$\int^{1}_{1-x} \frac{\frac{\pi^2}{6} -\text{Li}_2(t) - \log(1-t) \log(t)}{t} \, dt$$

$$-\frac{\pi^2}{6}\log(1-x)- \int^{1}_{1-x} \frac{\text{Li}_2(t)}{t}\,dt -\int^{1}_{1-x}\frac{ \log(1-t) \log(t)}{t} \, dt$$

The first integral

• $$\int^{1}_{1-x} \frac{\text{Li}_2(t)}{t}\,dt =\text{Li}_3(1)-\text{Li}_3(1-x)$$

The second integral by parts we obtain

• $$\int^{1}_{1-x}\frac{ \log(1-t) \log(t)}{t} =\text{Li}_3(1) +\log(1-x)\text{Li}_2(1-x)-\text{Li}_3(1-x)$$

Collecting the results together we obtain

$$\int^x_0\frac{\text{Li}_2 (t)}{1-t} \, dt =-\frac{\pi^2}{6}\log(1-x)-\text{Li}_2(1-x) \log(1-x)+2\, \text{Li}_3(1-x)- 2 \zeta(3)$$

Hence we solved the integral

$$\int^x_0 \frac{\log^2(1-t)}{t}\, dt = - \log(1-x) \text{Li}_2(x) +\frac{\pi^2}{6}\log(1-x)+\text{Li}_2(1-x) \log(1-x)-2\, \text{Li}_3(1-x)+ 2 \zeta(3)$$

So we have got our Harmonic sum

$$\sum_{k\geq 1} \frac{H_k}{k^2} \, x^{k} = \text{Li}_3(x)+\frac{1}{2} \left( - \log(1-x) \text{Li}_2(x) +\frac{\pi^2}{6}\log(1-x)+\text{Li}_2(1-x) \log(1-x)-2\, \text{Li}_3(1-x)+ 2 \zeta(3) \right)$$

$$\sum_{k\geq 1} \frac{H_k}{k^2} \, x^{k} = \text{Li}_3(x)-\, \text{Li}_3(1-x)+\, \log(1-x) \text{Li}_2(1-x) +\frac{1}{2}\log(x) \log^2(1-x)+\zeta(3)$$

The expression can be further simplified but I will leave it for the reader.

To be continued ...

Gold Member
MHB
4.Integration using special functions (continued)

4.10. Euler sums (continued)

General definition:

$$\displaystyle \sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)\,\,;q\geq 2$$​

This can be proved using complex analysis or using basic techniques of series manipulations as in this answer.

Where by definition we have for $q=2$

$$\sum_{n=1}^\infty \frac{H_n}{n^2}= 2\zeta(3)$$

Let us look at a way of solving that sum using integration

First we need that

$$H_n = \int^1_0 \frac{1-x^n}{1-x}\,dx$$

proof

$$\int^1_0 \frac{1-x^n}{1-x}\,dx =\sum_{k\geq 0} \int^1_0 x^k-x^{n+k}\,dx$$

$$\sum_{k\geq 0} \frac{1}{k+1}-\frac{1}{n+k+1} = H_n$$

Plugging the result in the sum we get

$$\int^1_0 \frac{1}{1-x}\sum_{n = 1}^\infty\frac{1-x^n }{n^2}\,dx = \int^1_0 \frac{\zeta(2)-\mathrm{Li}_2(x)}{1-x}dx$$

Now use the duplication formula for dilogarithm

$$\zeta(2)-\mathrm{Li}_2(x) = \mathrm{Li}_2(1-x)+\log(x)\log(1-x)$$

Hence we have

$$\int^1_0 \frac{\mathrm{Li}_2(1-x)+\log(x)\log(1-x)}{1-x}dx = \int^1_0 \frac{\mathrm{Li}_2(x)+\log(1-x)\log(x)}{x}dx$$

First integral

$$\int^1_0 \frac{\mathrm{Li}_2(x)}{x} = \mathrm{Li}_3(1) =\zeta(3)$$

Second integral

$$\int^1_0 \frac{\log(1-x)\log(x)}{x}dx =\int^1_0 \frac{\mathrm{Li}_2(x)}{x} = \zeta(3)$$

Finally

$$\sum_{n=1}^\infty \frac{H_n}{n^2} = \zeta(3)+\zeta(3) = 2\zeta(3)$$

Examples

$$\int^1_0\frac{\log^2(1-x)\log(x)}{x} = -\frac{\pi^4}{180}$$

Using the generating function

$$\sum_{k\geq 1}H_k x^{k-1} = -\frac{\log(1-x)}{x(1-x)}$$

By integrating both sides

$$\sum_{k\geq 1}\frac{H_k}{k} x^{k} =\mathrm{Li}_2(x) +\frac{1}{2}\log^2(1-x)$$

Or

$$\log^2(1-x) =2\sum_{k\geq 1}\frac{H_k}{k} x^{k} -2\mathrm{Li}_2(x)$$

plugging this in our integral we have

$$2\int^1_0\left(\sum_{k\geq 1}\frac{H_k}{k} x^{k} -\mathrm{Li}_2(x)\right)\frac{\log(x)}{x}\,dx$$

First integral

$$-2\int^1_0\mathrm{Li}_2(x)\frac{\log(x)}{x}\,dx =2\int^1_0\frac{\mathrm{Li}_3(x)}{x}\,dx = 2\zeta(4)$$

Second integral

$$2\sum_{k\geq 1}\frac{H_k}{k}\int^1_0x^{k-1}\log(x)\,dx$$

Using integration by parts twice and the formula presented first

$$-2\sum_{k\geq 1}\frac{H_k}{k^3} =-5\zeta(4)+\zeta^2(2)$$

Finally we get

$$\int^1_0\frac{\log^2(1-x)\log(x)}{x} =-5\zeta(4)+\zeta^2(2)+2\zeta(4) =\zeta^2(2)-3\zeta(4)$$

Gold Member
MHB
4.Integration using special functions (continued)

4.10. Euler sums (continued)

Show that

$$\displaystyle \int_{0}^{\infty} \, e^{-at} \sin(bt) \frac{\ln t}{t}\, dt =- \left( \frac{ \log(a^2+b^2) }{2} +\gamma \right) \arctan \left( \frac{b}{a} \right)$$​

Proof

We can start by the following integral

$$I(s) = \int_{0}^{\infty} t^{s-1} \, e^{-at} \, \sin(bt) dt$$

By using the the expansion of the sine function

$$I(s)=\int_{0}^{\infty}t^{s-1} \, e^{-at}\sum_{n\geq 0} \frac{(-1)^n (bt)^ {2n+1}}{\Gamma(2n+2)}$$

By swapping the summation and integration

$$I(s)= \sum_{n\geq 0} \frac{(-1)^n (b)^{2n+1}}{\Gamma(2n+2)} \int_{0}^ {\infty}t^{s+2n} \, e^{-at} dt= \frac{1}{a^s} \sum_{n\geq 0} \frac{(-1)^n (b)^{2n+1} \Gamma (s+2n+1)}{\Gamma(2n+2) a^{2n+1}}$$

By differentiating and plugging $s=0$ we have
$$I'(0) = \sum_{n\geq 0} \frac{(-1)^n \psi_0(2n+1)}{2n+1} \left( \frac{b} {a} \right)^{2n+1} -\log(a) \sum_{n\geq 0} \frac{(-1)^n }{2n+1} \left( \frac {b}{a} \right)^{2n+1}$$

Now use that $\psi(n+1) +\gamma = H_n$

$$I'(0) = \sum_{n\geq 0} \frac{(-1)^n H_{2n}-\gamma}{2n+1} \left( \frac{b} {a} \right)^{2n+1} -\log(a) \arctan \left( \frac{b}{a} \right)$$
$$I'(0) = \sum_{n\geq 0} \frac{(-1)^n H_{2n}}{2n+1} \left( \frac{b}{a} \right)^{2n+1}-(\gamma +\log(a))\arctan \left( \frac{b}{a} \right)$$

Now we look at the harmonic sum

\begin{align} \sum_{k\geq 0}(-1)^k H_{2k} x^{2k}&= \sum_{k\geq 0}(-1)^k x^ {2k} \int^1_0 \frac{1-t^{2k}}{1-t} \, dt\\ &= \int^1_0 \frac{1}{1-t} \sum_{k\geq 0}(-1)^k x^{2k} \left(1-t^{2k}\right) \, dt\\ &= \int^1_0 \frac{1}{1-t} \sum_{k\geq 0}(-1)^k \left(x^{2k}-(xt)^{2k}\right) \, dt\\ &= \int^1_0 \frac{1}{1-t}\left(\frac{1}{1+x^2}-\frac{1}{1+t^2x^2}\right) \, dt\\ &=\frac{1}{1+x^2} \int^1_0 \frac{1+t^2x^2-1-x^2}{(1-t)(1+t^2x^2)} \, dt\\ &=\frac{-x^2}{1+x^2} \int^1_0 \frac{(1-t^2)}{(1-t)(1+t^2x^2)} \, dt\\ &=\frac{-x^2}{1+x^2} \int^1_0 \frac{1+t}{(1+t^2x^2)} \, dt\\ &=\frac{-x^2}{1+x^2} \left( \int^1_0 \frac{1}{1+t^2x^2}+\frac{t}{1+t^2x^2} \, dt \right)\\ &= \frac{-1}{2(1+x^2)} \left(2x \arctan (x) + \log(1+x^2) \right) \end{align}

Using this we conclude by integrating

$$\sum_{k\geq 0}\frac{(-1)^k H_{2k}}{2k+1} x^{2k}=-\frac{1}{2} \log(1+x^2) \arctan(x)$$

Hence the following

$$\sum_{k\geq 0}\frac{(-1)^k H_{2k}}{2k+1} \left(\frac{b}{a} \right)^{2k+1}=- \frac{1}{2} \log \left( \frac{a^2+b^2}{a^2} \right) \arctan \left(\frac{b}{a} \right)$$

Substituting that in our integral

\begin{align} \int_{0}^{\infty} \, e^{-at} \sin(bt) \frac{\ln t}{t}\, dt &= -\left( \frac{1}{2} \log \left( \frac{a^2+b^2}{a^2} \right) + \gamma +\log(a) \right) \arctan \left( \frac{b}{a} \right)\\ &=- \left( \frac{ \log(a^2+b^2) }{2} +\gamma \right) \arctan \left( \frac{b} {a} \right) \end{align}

Prove that

\displaystyle \begin{align} \int^1_0 \frac{\mathrm{Li}_p(x)\,\, \mathrm{Li}_q(x)\, }{x}\, dx&= \sum_{n=1}^{p-1}(-1)^{n-1}\zeta(p-n+1)\zeta(q+n) -\frac{1}{2}\sum_{n=1}^{{p+q}-2}(-1)^{p-1}\zeta(n+1)\zeta({p+q}-n)\\ &+(-1)^{p-1}\left(1+\frac{{p+q}}{2} \right)\zeta({p+q}+1)\end{align}​

proof

We can see that

$$\int^1_0 \frac{\mathrm{Li}_p(x)\,\, \mathrm{Li}_q(x)\, }{x}\, dx = \sum_{k\geq 1}\sum_{n\geq 1}\frac{1}{k^{q}n^{p}(n+k)}$$

Let us first look at the following

$$\mathscr{C}(\alpha , k) =\sum_{n\geq 1}\frac{1}{n^{\alpha}(n+k)}\,\,\, ; \,\,\,\,\mathscr{C}(1, k)=\frac{H_k}{k}$$

We can find a general formula to find the sum

\begin{align}
\mathscr{C}(\alpha , k) &=\sum_{n\geq 1}\frac{1}{k\, n^{\alpha-1}}\left( \frac{1}{n}-\frac{1}{n+k}\right)\\ &= \frac{1}{k}\zeta(\alpha)-\frac{1}{k}\mathscr{C}(\alpha-1 , k)\\ &= \frac{1}{k}\zeta(\alpha)-\frac{1}{k^2}\zeta(\alpha-1)+\frac{1}{k^2}\mathscr{C}(\alpha-2 , k)\\ &= \sum_{n=1}^{\alpha-1}(-1)^{n-1}\frac{\zeta(\alpha-n+1)}{k^n}+(-1)^{\alpha-1}\frac{H_k}{k^\alpha}
\end{align}

Hence we have the general formula

$$\mathscr{C}(\alpha , k) = \sum_{n=1}^{\alpha-1}(-1)^{n-1}\frac{\zeta(\alpha-n+1)}{k^n}+(-1)^{\alpha-1}\frac{H_k}{k^\alpha}$$

Dividing by $k^{\beta}$ and summing w.r.t to $k$

$$\sum_{k\geq 1}\frac{\mathscr{C}(\alpha , k)}{k^{\beta}} = \sum_{n=1}^{\alpha-1}(-1)^{n-1}\zeta(\alpha-n+1)\zeta(\beta+n)+(-1)^{\alpha-1}\sum_{k\geq 1}\frac{H_k}{k^{\alpha+\beta}}$$

Now we use that

$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$

Hence we have

$$\sum_{n=1}^\infty \frac{H_n}{n^{\alpha+\beta}}= \left(1+\frac{{\alpha+\beta}}{2} \right)\zeta({\alpha+\beta}+1)-\frac{1}{2}\sum_{k=1}^{{\alpha+\beta}-2}\zeta(k+1)\zeta({\alpha+\beta}-k)$$

And the generalization is the following formula

\begin{align}
\sum_{k\geq 1}\frac{\mathscr{C}(\alpha , k)}{k^{\beta}} &= \sum_{n=1}^{\alpha-1}(-1)^{n-1}\zeta(\alpha-n+1)\zeta(\beta+n) -\frac{1}{2}\sum_{n=1}^{{\alpha+\beta}-2}(-1)^{\alpha-1}\zeta(n+1)\zeta({\alpha+\beta}-n)\\ &+(-1)^{\alpha-1}\left(1+\frac{{\alpha+\beta}}{2} \right)\zeta({\alpha+\beta}+1)\end{align}

We conclude by putting that

$$\displaystyle \sum_{k\geq 1}\sum_{n\geq 1}\frac{1}{k^{q}n^{p}(n+k)} = \sum_{k\geq 1}\frac{\mathscr{C}(p, k)}{k^{q}}$$

Gold Member
MHB
4.Integration using special functions (continued)

4.10. Euler sums (continued)

We can relate the generalized harmonic number to the polygamma function

$$\displaystyle H^{(p)}_k = \zeta(p) +(-1)^{p-1}\frac{\psi_{p-1}(k+1)}{ (p-1)!}$$​

proof

$$H^{(p)}_k = \sum^k_{n=1}\frac{1}{n^p} =\zeta(p)-\sum^\infty_{n=k+1}\frac{1}{n^p}$$

Now let $n=i+k+1$

$$H^{(p)}_k = \sum^k_{n=1}\frac{1}{n^p} =\zeta(p)-\sum_{i\geq 0}\frac{1}{(i+k+1)^p}$$

We know that

$$(-1)^{p}\frac{\psi_{p-1}(k+1)}{ (p-1)!} =\sum_{i\geq 0} \frac{1}{(i+k+1)^{p}}\,\,p\geq 1$$

Hence we have

$$H^{(p)}_k = \zeta(p) +(-1)^{p-1}\frac{\psi_{p-1}(k+1)}{ (p-1)!}$$

We can use that to obtain a nice integral representation.

$$\displaystyle \sum_{k\geq 1}\frac{H^{(p)}_k}{k^q} = \zeta(p)\zeta(q) +(-1)^{p}\frac{1}{ (p-1)!}\int^1_0\frac{\mathrm{Li}_q(x)\log(x)^{p-1}}{1-x}\,dx$$​

Note that

$$\psi_0(a+1)= \int^1_0\frac{1-x^a}{1-x}\,dx$$

By differentiating with respect to $a$ , $p$ times we have

$$\psi_p(a+1) = \frac{\partial}{\partial a^p}\int^1_0\frac{1-x^a}{1-x}\,dx$$

$$\psi_p(a+1) = -\int^1_0\frac{x^a\log(x)^{p}}{1-x}\,dx$$

Let $a=k$

$$\psi_{p-1}(k+1) = -\int^1_0\frac{x^k\log(x)^{p-1}}{1-x}\,dx$$

Substituting that in our formula

$$H^{(p)}_k = \zeta(p) +(-1)^{p}\frac{1}{ (p-1)!}\int^1_0\frac{x^k\log(x)^{p-1}}{1-x}\,dx$$

Now divide by $k^q$ and sum with respect to $k$

$$\sum_{k\geq 1}\frac{H^{(p)}_k}{k^q} = \zeta(p)\zeta(q) +(-1)^{p}\frac{1}{ (p-1)!}\int^1_0\frac{\mathrm{Li}_q(x)\log(x)^{p-1}}{1-x}\,dx$$

Symmetric formula

$$\displaystyle \sum_{k\geq 1} \frac{H^{(p)}_k}{k^q}+\sum_{k\geq 1} \frac{H^{(q)}_k}{k^p} =\zeta(p)\zeta(q)+\zeta(p+q)$$​

proof

Take the left side and swap the finite and infinite sums

$$\sum_{i\geq 1} \,\sum_{k\geq i}\frac{1}{i^p} \frac{1}{k^q}=\sum_{i\geq 1} \,\sum_{k\geq 1}\frac{1}{i^p} \frac{1}{k^q}-\sum_{i\geq 1}\frac{1}{i^p} \,\sum_{1\leq k \leq i-1} \frac{1}{k^q}$$

The second sum can be written as

$$\sum_{i\geq 1}\frac{1}{i^p} \,\sum_{1\leq k \leq i-1} \frac{1}{k^q} = \sum_{i\geq 1}\frac{1}{i^p} \,\left(\sum_{1\leq k \leq i}\frac{1}{k^q}-\frac{1}{i^p}\right)$$

By separating and changing the index we get

$$\sum_{k\geq 1}\frac{H^{(q)}_k}{k^p}-\zeta(p+q)$$

Hence we have

$$\sum_{k\geq 1} \frac{H^{(p)}_k}{k^q} =\zeta(p)\zeta(q)-\sum_{k\geq 1}\frac{H^{(q)}_k}{k^p}+\zeta(p+q)$$

$$\sum_{k\geq 1} \frac{H^{(p)}_k}{k^q}+\sum_{k\geq 1}\frac{H^{(q)}_k}{k^p} =\zeta(p)\zeta(q)+\zeta(p+q)$$

for the special case $p=q=n$

$$\sum_{k\geq 1} \frac{H^{(n)}_k}{k^n} =\frac{\zeta^2(n)+\zeta(2n)}{2}$$

Last edited:
Gold Member
MHB
Examples

$$\displaystyle \sum_{k\geq 1}\frac{H_k^{(3)}}{k^2} = \frac{11\zeta(5)}{2}-2\zeta(2)\zeta(3)$$​

Using the symmetry formula

$$\sum_{k\geq 1}\frac{H_k^{(3)}}{k^2} = \zeta(2)\zeta(3)+\zeta(5)-\sum_{k\geq 1}\frac{H_k^{(2)}}{k^3}$$

Using the integral formula on the second sum

$$\sum_{k\geq 1}\frac{H_k^{(2)}}{k^3} =\zeta(2)\zeta(3)+ \int^1_0 \frac{\mathrm{Li}_3(x)\log(x)}{1-x}\,dx$$

Using integration by parts on the integral

$$\int^1_0 \frac{\mathrm{Li}_3(x)\log(x)}{1-x}\,dx =- \int^1_0\frac{\mathrm{Li}_2(x)\mathrm{Li}_2(1-x)}{x}\,dx$$

Let us think of solving

$$\int^1_0\frac{\mathrm{Li}_2(x)\mathrm{Li}_2(1-x)}{x}\,dx$$

Using the duplication formula

$$\mathrm{Li}_2(1-x) = \zeta(2)-\mathrm{Li}_2(x)-\log(x)\log(1-x)$$

$$\int^1_0\frac{\mathrm{Li}_2(x)(\zeta(2)-\mathrm{Li}_2(x)-\log(x)\log(1-x))}{x}\,dx$$

The first integral

$$\zeta(2)\int^1_0 \frac{\mathrm{Li}_2(x)}{x}\,dx = \zeta(2)\zeta(3)$$

The third integral

$$\int^1_0\frac{\mathrm{Li}_2(x)\log(x)\log(1-x)}{x}\,dx = \frac{1}{2}\int^1_0\frac{\mathrm{Li}^2_2(x)}{x}\,dx$$

Finally we get

$$\int^1_0 \frac{\mathrm{Li}_3(x)\log(x)}{1-x}\,dx =\frac{3}{2}\int^1_0\frac{\mathrm{Li}^2_2(x)}{x}\,dx-\zeta(2)\zeta(3)$$

So

$$\sum_{k\geq 1}\frac{H_k^{(2)}}{k^3} = \frac{3}{2}\int^1_0\frac{\mathrm{Li}^2_2(x)}{x}\,dx$$

Hence we finally get that

$$\sum_{k\geq 1}\frac{H_k^{(3)}}{k^2}=\zeta(2)\zeta(3)+\zeta(5)-\frac{3}{2}\int^1_0\frac{\mathrm{Li}^2_2(x)}{x}\,dx$$

Let us solve the integral

$$\int^1_0 \frac{\mathrm{Li}^2_2(x)}{x}\,dx$$

By series expansion

$$\sum_{k,n\geq 1}\frac{1}{(nk)^2}\int^1_0x^{n+k-1}\,dx =\sum_{k,n\geq 1}\frac{1}{(nk)^2(n+k)}$$

By some manipulations we get

$$\sum_{k\geq 1}\frac{1}{k^3}\sum_{n\geq 1}\frac{k}{n^2(n+k)}= \sum_{k\geq 1}\frac{1}{k^3}\sum_{n\geq 1}\frac{1}{n^2}-\sum_{k\geq 1}\frac{1}{k^3}\sum_{n\geq 1}\frac{1}{n(n+k)}$$

This can be simplified to conclude that

$$\int^1_0 \frac{\mathrm{Li}^2_2(x)}{x}\,dx = \zeta(2)\zeta(3)-\sum_{k\geq 1}\frac{H_k}{k^4}$$

Now using that

$$\sum_{k\geq 1}\frac{H_k}{k^4} = 3\zeta(5)-\zeta(2)\zeta(3)$$

Hence

$$\int^1_0 \frac{\mathrm{Li}^2_2(x)}{x}\,dx = 2\zeta(2)\zeta(3)-3\zeta(5)$$

Finally we get

$$\sum_{k\geq 1}\frac{H_k^{(3)}}{k^2}=\zeta(2)\zeta(3)+\zeta(5)-\frac{3}{2}\left( 2\zeta(2)\zeta(3)-3\zeta(5)\right)=\frac{11\zeta(5)}{2}-2\zeta(2)\zeta(3)$$

Last edited:
Gold Member
MHB
Advanced Integration Techniques (back with new lessons)

4.Integration using special functions (continued)

4.11. Sine Integral function

We define the following

$$\mathrm{Si}(z) = \int^z_0 \frac{\sin(x) }{x}\, dx$$

A closely related function is the following

$$\mathrm{si}(z) = -\int^\infty_z \frac{\sin(x) }{x}\, dx$$

These function are related through the equation

$$\mathrm{Si}(z) = \mathrm{si}(z)+\frac{\pi}{2}$$

A closely related function is the $\mathrm{sinc}$ function

$$\mathrm{sinc}=\begin{cases} 1 & x=0 \\ \frac{\sin(x)}{x} & x \neq 0 \end{cases}$$

Using that we conclude that

$$\frac{d}{dx} \mathrm{Si}(x) = \mathrm{sinc}(x)$$

For the integration we conclude that

$$\int \mathrm{Si}(x)\,dx = \cos(x)+ x \,\mathrm{Si(x)}$$

Examples

Prove that

$$\int^\infty_0 \sin(x) \mathrm{si}(x) \, dx = -\frac{\pi}{4}$$

Using integration by parts we get

$$-\int^\infty_0 \frac{\sin(x)\cos(x)}{x} dx = -\frac{1}{2} \int^\infty_0 \frac{\sin(2x)}{x}\,dx$$

Let $2x = t$

$$-\frac{1}{2}\int^\infty_0 \frac{\sin(t)}{t} dx = -\frac{\pi}{4}$$

Prove that

$$\int^\infty_0 x^{\alpha -1}\,\mathrm{si}(x) \, dx =- \frac{\Gamma(\alpha)}{\alpha}\sin\left(\frac{\pi \alpha}{2}\right)$$

Using the integral representation

$$-\int^\infty_0 x^{\alpha -1}\,\int^\infty_x \frac{\sin(t)}{t}\,dt \, dx$$

Let $xy = t$

$$-\int^\infty_0 x^{\alpha -1}\,\int^\infty_1 \frac{\sin(xy)}{y}\,dy \, dx$$

Switching the integrals we get

$$-\int^\infty_1 \frac{1}{y} \int^\infty_0 x^{\alpha -1}\, \sin(xy) \, dx \,dy$$

Now let $xy = t$

$$-\int^\infty_1 \frac{1}{y^{\alpha+1}} \int^\infty_0 t^{\alpha -1}\, \sin(t) \, dx \,dy$$

The Mellin transform of the sine function is defined as

$$\mathcal{M}_s(\sin (x)) = \int^\infty_0 x^{s-1}\sin(x)\,dx = \Gamma (s) \sin\left( \frac{\pi s}{2} \right)$$

Hence we conclude that

$$-\Gamma (\alpha) \sin\left( \frac{\pi \alpha}{2} \right)\int^\infty_1 \frac{1}{y^{\alpha+1}} = -\frac{\Gamma (\alpha)}{\alpha} \sin\left( \frac{\pi \alpha}{2} \right)$$

Prove that

$$\int^\infty_0 e^{-\alpha \, x}\,\mathrm{si}(x) \, dx =- \frac{\arctan (\alpha)}{\alpha}$$

Using the integral representation

$$-\int^\infty_0 e^{-\alpha \, x}\,\int^\infty_x \frac{\sin(t)}{t}\,dt \, dx$$

Let $xy = t$

$$-\int^\infty_0 e^{-\alpha \, x}\,\int^\infty_1 \frac{\sin(xy)}{y}\,dy \, dx$$

Switching the integrals

$$-\int^\infty_1 \frac{1}{y}\int^\infty_0 e^{-\alpha \, x}\, \sin(xy)\,dx \, dy$$

The inner integral is the laplace transform of the sine function

$$\mathcal{L}_s(\sin(at)) = \frac{a}{s^2+a^2}$$

Hence we conclude that

$$-\int^\infty_1 \frac{1}{y^2+\alpha^2} \,dy = -\frac{\arctan(\alpha)}{\alpha}$$

Gold Member
MHB
Re: Advanced Integration Techniques (back with new lessons)

4.Integration using special functions (continued)

4.11. Sine Integral function (continued)

Prove that

$$\int^\infty_0 \mathrm{si}(x) \log(x) \,dx =\gamma+1$$

We know that

$$\int^\infty_0 x^{\alpha -1}\,\mathrm{si}(x) \, dx =- \frac{\Gamma(\alpha)}{\alpha}\sin\left(\frac{\pi \alpha}{2}\right)$$

Differentiate with respect to $\alpha$

$$\int^\infty_0 x^{\alpha -1}\,\mathrm{si}(x) \,\log(x) dx = \frac{\Gamma(\alpha)}{\alpha^2}\sin\left(\frac{\pi \alpha}{2}\right)-\frac{\Gamma(\alpha)\psi(\alpha)}{\alpha}\sin\left(\frac{\pi \alpha}{2}\right) +\frac{\pi}{2}\cos\left(\frac{\pi \alpha}{2}\right)$$

Let $\alpha \to 1$

$$\int^\infty_0 \mathrm{si}(x) \,\log(x) dx = 1- \psi(1) = 1-(-\gamma) = 1+\gamma$$

Find the integral

$$\int^\infty_0 \mathrm{si}(x) \, \sin(px) \, dx$$

Using integration by parts we get

$$-\left[\frac{\mathrm{si}(x) \cos(px)}{p} \right]^\infty_0+\frac{1}{p}\int^\infty_0 \frac{\sin(x)}{x} \, \cos(px) \, dx$$

Taking the limits

$$\lim_{x \to 0} \frac{\mathrm{si}(x) \cos(px)}{p} = \frac{\mathrm{si}(0)}{p} = -\frac{\pi}{2p}$$

$$\lim_{x \to \infty} \frac{\mathrm{si}(x) \cos(px)}{p} = 0$$

Hence we get

$$-\frac{\pi}{2p}+\frac{1}{p}\int^\infty_0 \frac{\sin(x)}{x} \, \cos(px) \, dx$$

The integral

$$\int^\infty_0 \frac{\sin(x)}{x} \, \cos(px) \, dx = \frac{1}{2}\int^\infty_0 \frac{\sin((p+1)x) -\sin((p-1)x)}{x}dx$$

Separate the integrals

$$I = \frac{1}{2}\int^\infty_0 \frac{\sin((p+1)x)}{x}dx -\frac{1}{2}\int^\infty_0 \frac{\sin((p-1)x)}{x}dx$$

If $p-1>0$ we get

$$I = \frac{\pi}{4}-\frac{\pi}{4} = 0$$

If $p-1<0$

$$I = \frac{1}{2}\int^\infty_0 \frac{\sin((p+1)x)}{x}dx +\frac{1}{2}\int^\infty_0 \frac{\sin((p-1)x)}{x}dx = \frac{\pi}{2}$$

If $p=1$ we have

$$I = \frac{1}{2}\int^\infty_0 \frac{\sin(2x)}{x}dx +0 = \frac{\pi}{4}$$

Finally we get

$$\int^\infty_0 \mathrm{si}(x)\sin(px) dx= \begin{cases} -\frac{\pi}{2p} & p > 1 \\ -\frac{\pi}{4p} & p < 1 \\ 0 & p = 1 \end{cases}$$

Last edited:
Gold Member
MHB
Re: Advanced Integration Techniques (back with new lessons)

4.Integration using special functions (continued)

4.11. Sine Integral function (continued)

Prove that for $0<a<2$

$$\int^\infty_0 \mathrm{si}^2(x) \, \cos(ax) \, dx = \frac{\pi}{2a}\log(a+1)$$

Using integration by parts we get

$$\left[\frac{\mathrm{si}^2(x) \sin(ax)}{a} \right]^\infty_0-\frac{2}{a}\int^\infty_0 \frac{\mathrm{si}(x)\sin(x)}{x} \, \sin(ax) \, dx$$

Taking the limits

$$\lim_{x \to 0} \frac{\mathrm{si}^2(x) \sin(ax)}{a} = 0$$

$$\lim_{x \to \infty} \frac{\mathrm{si}^2(x) \sin(ax)}{a} = 0$$

Let the integral

$$I(a) = \int^\infty_0 \frac{\mathrm{si}(x)\sin(x)}{x} \, \sin(ax) \, dx$$

Differentiate with respect to $a$

$$I'(a) = \int^\infty_0 \mathrm{si}(x)\sin(x) \, \cos(ax) \, dx$$

Now use the product to sum trigonometric rules

$$I'(a) =\frac{1}{2} \int^\infty_0 \mathrm{si}(x)(\sin((a+1)x)-\sin((a-1)x)) \, dx$$

From the previous exercise we have

$$\int^\infty_0 \mathrm{si}(x)\sin((a+1)x) dx = \frac{-\pi}{4(a+1)}\,\,\,;\, a>0$$

$$\int^\infty_0 \mathrm{si}(x)\sin((a+1)x) dx = 0\,\,\,;\, a<2$$

Hence we conclude that for $0<a<2$

$$I'(a) =-\frac{\pi}{4(a+1)}$$

Integrate with respect to $a$

$$I(a) =-\frac{\pi}{4}\log(a+1)+C$$

Let $a \to 0$

$$I(0) =0+C \,\,\, \to \,\,\, C = 0$$

Hence we have

$$\int^\infty_0 \frac{\mathrm{si}(x)\sin(x)}{x} \, \sin(ax) \, dx = -\frac{\pi}{4}\log(a+1)$$

Which implies that

$$\int^\infty_0 \mathrm{si}^2(x) \, \cos(ax) \, dx = \frac{-2}{a}\left( -\frac{\pi}{4}\log(a+1)\right) = \frac{\pi}{2a}\log(a+1)$$

Find the integral, for $a \neq 1$

$$\int^\infty_0 \mathrm{si}(x) \cos(ax)\,dx$$

Use integration by parts to obtain

$$\frac{1}{a}\int^\infty_0\frac{\sin(x) \sin(ax)}{x}\,dx$$

Let the integral

$$I(t) = \int^\infty_0e^{-tx}\frac{\sin(x) \sin(ax)}{x}\,dx$$

Differentiate with respect to $t$

$$I'(t) = -\int^\infty_0e^{-tx}\sin(x) \sin(ax)\,dx$$

Use product to sum rules

$$I'(t) = \frac{1}{2}\int^\infty_0e^{-tx}(\cos((a+1)x)-\cos((a-1)x))\,dx$$

Now we can use the Laplace transform

$$I'(t) = \frac{1}{2}\left(\frac{t}{t^2+(a+1)^2}-\frac{t}{t^2+(a-1)^2} \right)$$

Integrate with respect to $t$

$$I(t) = -\frac{1}{4} \log\left(\frac{t^2+(a+1)^2}{t^2+(a-1)^2} \right) +C$$

After verifying the constant goes to 0, we have

$$\int^\infty_0e^{-tx}\frac{\sin(x) \sin(ax)}{x}\,dx = -\frac{1}{4} \log\left(\frac{t^2+(a+1)^2}{t^2+(a-1)^2} \right)$$

Let $t \to 0$

$$\int^\infty_0\frac{\sin(x) \sin(ax)}{x}\,dx = -\frac{1}{4} \log\left(\frac{a+1}{a-1} \right)^2$$

We conclude that

$$\int^\infty_0 \mathrm{si}(x) \cos(ax)\,dx = -\frac{1}{4a} \log\left(\frac{a+1}{a-1} \right)^2$$

Gold Member
MHB
(AIT) Cosine Integral function

4.Integration using special functions (continued)

4.12. Cosine Integral function

Define

$$\mathrm{ci}(x) =- \int^\infty_x \frac{\cos(t)}{t}\,dt$$

A related function is the following

$$\mathrm{Cin}(x) = \int^x_0 \frac{1-\cos(t)}{t}\,dt$$

The derivative is

$$\frac{d}{dx}\mathrm{ci}(x) = \frac{\cos(x)}{x}$$

The integral

$$\int \mathrm{ci}(x) \,dx = x\mathrm{ci}(x)-\sin(x)$$

Prove the following

$$\mathrm{Cin}(x) = -\mathrm{ci}(x)+\log(x)+\gamma$$

Start by

$$\mathrm{Cin}(x) = \int^x_0 \frac{1-\cos(t)}{t}\,dt$$

Rewrite as

$$\mathrm{Cin}(x) =\int^\infty_0 \frac{1-\cos(t)}{t}\,dt- \int^\infty_x \frac{1-\cos(t)}{t}\,dt$$

Which simplifies to

$$\mathrm{Cin}(x) =\lim_{z \to \infty } \left[\int^z_0 \frac{1-\cos(t)}{t}\,dt- \log(z)\right] -\mathrm{ci}(x)+\log(x)$$

The limit goes to the Euler Maschorinit constant

$$\mathrm{Cin}(x) =\gamma -\mathrm{ci}(x)+\log(x)$$

Find the integral

$$\int^\infty_0 \mathrm{ci}(x) \, \cos(px) \, dx$$

Using integration by parts we get

$$\left[\frac{\mathrm{ci}(x) \sin(px)}{p} \right]^\infty_0-\frac{1}{p}\int^\infty_0 \frac{\cos(x)}{x} \, \sin(px) \, dx$$

Taking the limits

$$\lim_{x \to 0} \frac{\mathrm{ci}(x) \sin(px)}{p}=0$$

$$\lim_{x \to \infty}\frac{\mathrm{ci}(x) \sin(px)}{p} = 0$$

Hence we get

$$-\frac{1}{p}\int^\infty_0 \frac{\cos(x)}{x} \, \sin(px) \, dx$$

The integral

$$\int^\infty_0 \frac{\cos(x)}{x} \, \sin(px) \, dx = \frac{1}{2}\int^\infty_0 \frac{\sin((p-1)x) +\sin((p+1)x)}{x}dx$$

Separate the integrals

$$I = \frac{1}{2}\int^\infty_0 \frac{\sin((p+1)x)}{x}dx +\frac{1}{2}\int^\infty_0 \frac{\sin((p-1)x)}{x}dx$$

If $p-1>0$ we get

$$I = \frac{\pi}{4}+\frac{\pi}{4} = \frac{\pi}{2}$$

If $p-1<0$

$$I = \frac{\pi}{4}-\frac{\pi}{4} = 0$$

If $p=1$ we have

$$I = \frac{1}{2}\int^\infty_0 \frac{\sin(2x)}{x}dx +0 = \frac{\pi}{4}$$

Finally we get

$$\int^\infty_0 \mathrm{ci}(x) \, \cos(px) \, dx = \begin{cases} -\frac{\pi}{2p} & p > 1 \\ -\frac{\pi}{4p} & p = 1 \\ 0 & p < 1 \end{cases}$$

Find the integral for, $p>1$

$$\int^\infty_0 \mathrm{ci}(px) \mathrm{ci}(x)\,dx$$

Let

$$I(p) = \int^\infty_0 \mathrm{ci}(px) \mathrm{ci}(x)\,dx$$

Differentiate with respect to $p$

$$I'(p) = \frac{1}{p}\int^\infty_0 \mathrm{cos}(px) \mathrm{ci}(x)\,dx$$

If $p>1$ from the previous example we conclude that

$$I'(p) = \frac{1}{p}\left(\frac{-\pi}{2p}\right) = -\frac{\pi}{2p^2}$$

Integrate with respect to $p$

$$I(p) = \frac{\pi}{2p} + C$$

Take the limit $p \to \infty$, so $C = 0$.

Gold Member
MHB
Re: (AIT) Cosine Integral function

Prove that

$$\int^\infty_0 x^{\alpha-1}\mathrm{ci}(x)\,dx = -\frac{\Gamma(\alpha)}{\alpha}\cos\left( \frac{\alpha\pi}{2}\right)$$

Use the integral representation

$$-\int^\infty_0 x^{\alpha-1}\int^\infty_x \frac{\cos(t)}{t}\,dt\,dx$$

Let $t = yx$

$$-\int^\infty_0 x^{\alpha-1}\int^\infty_1 \frac{\cos(yx)}{y}\,dy\,dx$$

Switch the integrals

$$-\int^\infty_1\frac{1}{y}\int^\infty_0 x^{\alpha-1} \cos(yx)\,dx\,dy$$

Using the Mellin transform we get

$$-\Gamma(\alpha)\cos\left( \frac{\alpha\pi}{2}\right)\int^\infty_1\frac{1}{y^{1+\alpha}}\,dy = -\frac{\Gamma(\alpha)}{\alpha}\cos\left( \frac{\alpha\pi}{2}\right)$$

Prove that

$$\int^\infty_0 \mathrm{ci}(x) \log(x) \,dx = \frac{\pi}{2}$$

From the previous example we know

$$\int^\infty_0 x^{\alpha-1}\mathrm{ci}(x)\,dx = -\frac{\Gamma(\alpha)}{\alpha}\cos\left( \frac{\alpha\pi}{2}\right)$$

Differentiate with respect to $\alpha$

$$\int^\infty_0 x^{\alpha-1}\mathrm{ci}(x) \log(x)\,dx = \frac{\Gamma(\alpha)}{\alpha^2}\cos\left( \frac{\alpha\pi}{2}\right)-\frac{\Gamma(\alpha) \psi(\alpha)}{\alpha}\cos\left( \frac{\alpha\pi}{2}\right)+\frac{\pi}{2}\frac{\Gamma(\alpha)}{\alpha}\sin\left( \frac{\alpha\pi}{2}\right)$$

Take the limit $\alpha \to 1$

$$\int^\infty_0 \mathrm{ci}(x) \log(x)\,dx = 0-0+\frac{\pi}{2}\sin\left( \frac{\pi}{2}\right) = \frac{\pi}{2}$$

Prove that

$$\int^\infty_0 \mathrm{ci}(x) e^{-\alpha x}\,dx = -\frac{1}{\alpha}\log\sqrt{1+\alpha^2}$$

Use the integral representation

$$-\int^\infty_0 e^{-\alpha x}\int^\infty_1 \frac{\cos(yx)}{y}\,dy\,dx$$

Switch the integrals

$$-\int^\infty_1 \frac{1}{y}\int^\infty_0 e^{-\alpha x} \cos(yx)\,dx\,dy$$

Use the Laplace transformation

$$-\int^\infty_1 \frac{\alpha}{y(\alpha^2+y^2)}\,dy =- \frac{1}{2a}\log(1+\alpha^2) = -\frac{1}{\alpha}\log\sqrt{1+\alpha^2}$$

Gold Member
MHB
Re: (AIT) Cosine Integral function

Find the integral

$$\int^\infty_0 \mathrm{si}(qx) \mathrm{ci}(x) \, dx$$

Using the integral representation

$$\int^\infty_0 \mathrm{si}(qx) \int^\infty_1 \frac{\cos(yx)}{y}\,dy\, dx$$

Switch the integrals

$$\int^\infty_1 \frac{1}{y} \int^\infty_0 \mathrm{si}(qx) \cos(yx)\, dx\,dy$$

We also showed that

$$\frac{1}{2}\int^\infty_1 \frac{1}{y^2}\log\left( \frac{y+q}{y-q}\right)\,dy$$

We can prove that the anti-derivative

$$\left[\frac{\log(y)}{q}-\frac{1}{2q}\log(y^2-q^2)-\frac{1}{2y}\log\left( \frac{y+q}{y-q}\right)\right]^\infty_1$$

Which simplifies

$$\left[-\frac{1}{2q}\log\left(\frac{y^2-q^2}{y^2} \right)-\frac{1}{2y}\log\left( \frac{y+q}{y-q}\right)\right]^\infty_1$$

The limits

$$\lim_{y \to \infty }\frac{1}{2q}\log\left(\frac{y^2-q^2}{y^2} \right)+\frac{1}{2y}\log\left( \frac{y+q}{y-q}\right) = 0$$

The limit $y \to 1$
$$\frac{1}{2q}\log\left(1-q^2 \right)+\frac{1}{2}\log\left( \frac{1+q}{1-q}\right)$$

Can be written as

$$\frac{1}{4q}\log\left(1-q^2 \right)^2+\frac{1}{4}\log\left( \frac{1+q}{1-q}\right)^2$$

Prove that

$$\int^\infty_0 \frac{\mathrm{ci}(\alpha x)}{x+\beta}\,dx = -\frac{1}{2}\left\{ \mathrm{si(\alpha \beta)^2+\mathrm{ci}(\alpha \beta)^2}\right\}$$

Let the following

$$I(\alpha) = \int^\infty_0 \frac{\mathrm{ci}(\alpha x)}{x+\beta}\,dx$$

Differentiate with respect to $\alpha$

$$I'(\alpha) = \frac{1}{\alpha}\int^\infty_0 \frac{\cos(\alpha x)}{x+\beta}\,dx$$

Let $x+\beta = t$

$$I'(\alpha) = \frac{1}{\alpha}\int^\infty_\beta \frac{\cos(\alpha (t-\beta)}{t}\,dt$$

Use sum to product rules

$$I'(\alpha) = \frac{1}{\alpha}\int^\infty_\beta \frac{\cos(\alpha t) \cos(\alpha \beta) + \sin(\alpha t) \sin(\alpha \beta)}{t}\,dt$$

Separate the integrals

$$I'(\alpha) = \frac{\cos(\alpha \beta)}{\alpha}\int^\infty_\beta \frac{\cos(\alpha t) }{t}\,dt+ \frac{\sin(\alpha \beta)}{\alpha}\int^\infty_\beta\frac{\sin(\alpha t) }{t}\,dt$$

This simplifies to

$$I'(\alpha) = -\frac{\cos(\alpha \beta)}{\alpha}\mathrm{ci}(\alpha \beta)-\frac{\sin(\alpha \beta)}{\alpha}\mathrm{si}(\alpha \beta)$$

Integrate with respect to $\alpha$

$$I(\alpha) = -\frac{1}{2}\left\{ \mathrm{si(\alpha \beta)^2+\mathrm{ci}(\alpha \beta)^2}\right\}+ C$$

If $\alpha \to \infty$ we have $C =0$.

Gold Member
MHB
(AIT) Logarithm integral function

4.Integration using special functions (continued)

4.13. Logarithm Integral function

Define

$$\mathrm{li}(x) = \int^x_0 \frac{dt}{\log(t)}$$

Prove that

$$\int^1_0 \mathrm{li}(x)\,dx = -\log(2)$$

Let the following

$$I(a) = \int^1_0 \int^x_0 \frac{e^{-a\log(t)}dt}{\log(t)}\,dx$$

Differentiate with respect to $a$

$$I'(a) = - \int^1_0 \int^x_0 t^{-a}\,dt\,dx$$

$$I'(a) = \frac{1}{a-1} \int^1_0 x^{1-a}\,dx$$

Which reduces to

$$I'(a) = \frac{1}{(a-1)(2-a)} = \frac{1}{2-a}-\frac{1}{1-a}$$

Integrate with respect to $a$

$$I(a) = \log\left( \frac{1-a}{2-a}\right)+C$$

Take the limit $a \to \infty$ we get $C = 0$

$$\int^1_0 \int^x_0 \frac{e^{-a\log(t)}dt}{\log(t)}\,dx =\log\left( \frac{1-a}{2-a}\right)$$

Let $a \to 0$

$$\int^1_0 \mathrm{li}(x)\,dx =\log\left( \frac{1}{2}\right) = -\log(2)$$

Find the following

$$\int^1_0 x^{p-1}\mathrm{li}(x)\,dx$$

Let the following

$$I(a) = \int^1_0 x^{p-1}\int^x_0 \frac{e^{-a\log(t)}dt}{\log(t)}\,dx$$

Differentiate with respect to $a$

$$I'(a) = - \int^1_0 x^{p-1}\int^x_0 t^{-a}\,dt\,dx$$

$$I'(a) = \frac{1}{a-1} \int^1_0 x^{p-a}\,dx$$

Which reduces to

$$I'(a) = \frac{1}{(a-1)(p-a+1)} =\frac{1}{p} \left\{\frac{1}{p-a+1}-\frac{1}{1-a} \right\}$$

Integrate with respect to $a$

$$I(a) = \frac{1}{p}\log\left( \frac{1-a}{p-a+1}\right)+C$$

Take the limit $a \to \infty$ we get $C = 0$

$$\int^1_0 x^{p-1}\int^x_0 \frac{e^{-a\log(t)}dt}{\log(t)}\,dx =\frac{1}{p}\log\left( \frac{1-a}{p-a+1}\right)$$

Let $a \to 0$

$$\int^1_0 x^{p-1}\int^x_0 \frac{e^{-a\log(t)}dt}{\log(t)}\,dx =\frac{1}{p}\log\left( \frac{1}{p+1}\right) = -\frac{1}{p}\log\left( p+1\right)$$

Find the following

$$\int^1_0\mathrm{li}\left(\frac{1}{x}\right) \sin(a\log(x))\,dx$$

Let the following

$$I(b) = \int^1_0 \sin(a\log(x))\int^{\frac{1}{x}}_0 \frac{e^{-b\log(t)}dt}{\log(t)}\,dx$$

Differentiate with respect to $b$

$$I'(b) =-\int^1_0 \sin(a\log(x))\int^{\frac{1}{x}}_0 t^{-b}\,dt\,dx$$

$$I'(b) =\frac{1}{b-1}\int^1_0 x^{b-1}\sin(a\log(x))\,dx$$

Let $\log(x) = -t$

$$I'(b) =\frac{1}{1-b}\int^\infty_0 e^{-tb}\sin(at)\,dt$$

Now use the Laplace transform

$$I'(b) =\frac{1}{1-b}\frac{a}{a^2+b^2}$$

Integrate with respect to $b$

$$I(b) = \frac{a\log(a^2+b^2) -2a\log(b-1)+2\arctan(b/a)}{2a^2+2}+C$$

Let $b \to \infty$

$$0 = \frac{\pi}{2(a^2+1)}+C$$

Hence we have

$$\int^1_0 \sin(a\log(x))\int^{\frac{1}{x}}_0 \frac{e^{-b\log(t)}dt}{\log(t)}\,dx= \frac{a\log(a^2+b^2) -2a\log(b-1)+2\arctan(b/a)}{2a^2+2}-\frac{\pi}{2(a^2+1)}$$

Let $b \to 0$

$$\int^1_0 \sin(a\log(x))\mathrm{li}(x)\,dx= \frac{a\log(a^2)}{2a^2+2}-\frac{\pi}{2(a^2+1)} = \frac{1}{a^2+1}\left(a\log(a) -\frac{\pi}{2}\right)$$

Gold Member
MHB
Re: (AIT) Logarithm integral function

Find the following integral

$$\int^1_0\frac{\mathrm{li}(x)}{x} \log^{p-1}\left(\frac{1}{x}\right)\,dx$$

Let the following

$$I(a) = \int^1_0\frac{1}{x} \left[\int^x_0 \frac{e^{-a\log(t)}}{\log(t)}dt \right]\log^{p-1}\left(\frac{1}{x}\right)\,dx$$

Differentiate with respect to $a$

$$I'(a) = -\int^1_0\frac{1}{x} \left[\int^x_0t^{-a}dt \right]\log^{p-1}\left(\frac{1}{x}\right)\,dx$$

$$I'(a) = \frac{1}{a-1}\int^1_0x^{-a}\log^{p-1}\left(\frac{1}{x}\right)\,dx$$

Let $-\log(x) = t$

$$I'(a) = \frac{-1}{1-a}\int^\infty_0 e^{-(1-a)t}t^{p-1}\,dx$$

$$I'(a) = \frac{-1}{1-a}\frac{\Gamma(p)}{(1-a)^p} = -\frac{\Gamma(p)}{(1-a)^{p+1}}$$

Integrate with respect to $a$

$$I(a) = -\frac{\Gamma(p)}{p(1-a)^p}$$

Let $a \to 0$, Hence

$$\int^1_0\frac{\mathrm{li}(x)}{x} \log^{p-1}\left(\frac{1}{x}\right)\,dx = -\frac{\Gamma(p)}{p}$$

Prove that

$$\int^\infty_1 \mathrm{li}\left(\frac{1}{x}\right) \log^{p-1}(x)\,dx = -\frac{\pi}{\sin(\pi p)} \Gamma(p)$$

Let the following

$$I(a) = \int^\infty_1 \mathrm{li}\left(x^{-a}\right) \log^{p-1}(x)\,dx = -\frac{\pi}{\sin(\pi p)} \Gamma(p)$$

Differentiate with respect to $a$

$$\frac{d}{da}\mathrm{li}\left(x^{-a}\right)=\frac{d}{da} \int^{x^{-a}}_0\frac{dt}{\log(t)}= \frac{x^{-a}}{a}$$

Hence we have

$$I'(a) = \frac{1}{a}\int^\infty_1 x^{-a}\log^{p-1}(x)\,dx$$

Let $\log(x) = t$

$$I'(a) = \frac{1}{a}\int^\infty_0 e^{-(a-1)t}t^{p-1}\,dt$$

Using the Laplace transform

$$I'(a) =\Gamma(p) \frac{1}{a(a-1)^p}$$

Take the integral

$$\int^\infty_1 I'(a)da = \Gamma(p)\int^\infty_1 \frac{1}{a(a-1)^p} da$$

The left hand-side

$$I(\infty)-I(1) = \Gamma(p)\int^\infty_1 \frac{1}{a(a-1)^p} da$$

Now since $I(\infty) = 0$

$$I(1) = -\Gamma(p)\int^\infty_1 \frac{1}{a(a-1)^p} da$$

Which implies that

$$\int^\infty_1 \mathrm{li}\left(\frac{1}{x}\right) \log^{p-1}(x)\,dx = -\Gamma(p)\int^\infty_1 \frac{1}{a(a-1)^p} da$$

Now let $t = a-1$

$$\int^\infty_0 \frac{t^{-p}}{t+1} dt$$

Using the beta integral $x+y = 1$ and $x-1 = -p$ which implies that $x = 1-p,y=p$

Hence we have

$$\int^\infty_0 \frac{t^{-p}}{t+1} dt = \beta(p,1-p) = \Gamma(p)\Gamma(1-p) = \frac{\pi}{\sin(\pi p)}$$

Finally we get

$$\int^\infty_1 \mathrm{li}\left(\frac{1}{x}\right) \log^{p-1}(x)\,dx = -\frac{\pi}{\sin(\pi p)} \Gamma(p)$$

Gold Member
MHB
(AIT) Clausen functions

4.Integration using special functions (continued)

4.14. Clausen functions

$$\mathrm{cl}_m(\theta) = \begin{cases} \sum_{k=1}^\infty \frac{\sin(k\theta)}{k^m} & m \text{ is even} \\ \sum_{k=1}^\infty \frac{\cos(k\theta)}{k^m} & m \text{ is odd} \end{cases}$$

Duplication formula

$$\mathrm{cl}_m(2 \theta) = 2^{m-1}(\mathrm{cl}_m(\theta)-(-1)^m \mathrm{cl}_m(\pi-\theta))$$

Proof

If $m$ is even then

$$\mathrm{cli}_m(\pi-\theta) = \sum_{k=1}^\infty \frac{\sin(k\pi -k\theta)}{k^m} = -\sum_{k=1}^\infty (-1)^k \frac{\sin(k\theta)}{k^m}$$

$$\sum_{k=1}^\infty (-1)^k \frac{\sin(k\theta)}{k^m} + \sum_{k=1}^\infty \frac{\sin(k\theta)}{k^m} = \frac{1}{2^{m-1}}\sum_{k=1}^\infty \frac{\sin(2k\theta)}{k^m}$$

This implies that

$$\mathrm{cl}_m(2 \theta) = 2^{m-1}(\mathrm{cl}_m(\theta)- \mathrm{cl}_m(\pi-\theta))$$

If $m$ is odd then

$$\mathrm{cli}_m(\pi-\theta) = \sum_{k=1}^\infty \frac{\cos(k\pi -k\theta)}{k^m} = \sum_{k=1}^\infty (-1)^k \frac{\cos(k\theta)}{k^m}$$

$$\sum_{k=1}^\infty \frac{\cos(k\theta)}{k^m} + \sum_{k=1}^\infty (-1)^k \frac{\cos(k\theta)}{k^m} = \frac{1}{2^{m-1}}\sum_{k=1}^\infty \frac{\sin(2k\theta)}{k^m}$$

Which implies that

$$\mathrm{cl}_m(2 \theta) = 2^{m-1}(\mathrm{cl}_m(\theta)+ \mathrm{cl}_m(\pi-\theta))$$

Collecting the results we have

$$\mathrm{cl}_m(2 \theta) = 2^{m-1}(\mathrm{cl}_m(\theta)-(-1)^m \mathrm{cl}_m(\pi-\theta))$$

Find the integral

$$\int^\pi_0 \mathrm{cl}_{m}(\theta) d\theta \,\,\,\,\,\,\,\,\,\, m \text{ is even}$$

Using the series representation

$$\int^\pi_0 \sum_{k=1}^\infty \frac{\sin(k\theta)}{k^{m}} d\theta$$

Swap the integral and the series

$$\sum_{k=1}^\infty \frac{1}{k^{m}}\int^\pi_0 \sin(k\theta) d\theta$$

The integral

$$\int^\pi_0 \cos(k\theta) d\theta =-\left[ \frac{1}{k}\sin(k\theta) \right]^\pi_0=\frac{-(-1)^k+1}{k}$$

We get the summation

$$\sum_{k=1}^\infty \frac{-(-1)^k+1}{k^{m+1}} =\zeta(m+1)+\eta(m+1)$$

Now use that

$$\eta(s) = (1-2^{1-s})\zeta(s)$$

$$\sum_{k=1}^\infty \frac{-(-1)^k+1}{k^{m+1}} =\zeta(m+1)+(1-2^{-m})\zeta(m+1) = \zeta(m+1)(2-2^{-m})$$

Gold Member
MHB
Re: (AIT) Clausen functions

Find the integral for $m$ is even

$$\int^\infty_0 \mathrm{cl}_m(\theta) e^{-n\theta}\, d\theta$$

Using the series representation

$$\int^\infty_0 \sum_{k=1}^\infty \frac{\sin(k\theta)}{k^m} e^{-n\theta}\, d\theta$$

Swap the integral and the series

$$\sum_{k=1}^\infty \frac{1}{k^m}\int^\infty_0 \sin(k\theta) e^{-n\theta}\, d\theta$$

Using the Laplace transform we have

$$\sum_{k=1}^\infty \frac{1}{k^{m-1}(k^2+n^2)}$$

Add and subtract $k^2$ and divide by $n^2$

$$\frac{1}{n^2}\sum_{k=1}^\infty \frac{k^2+n^2-k^2}{k^{m-1}(k^2+n^2)}$$

Distribute the numerator

$$\frac{1}{n^2}\zeta(m-1)-\frac{1}{n^2}\sum_{k=1}^\infty\frac{1}{k^{m-3}(k^2+n^2)}$$

Continue this approach to conclude that

$$\sum_{i= 1}^j(-1)^{i-1}\frac{1}{n^{2i}}\zeta(m-(2i-1))+\frac{(-1)^j}{n^{2j}}\sum_{k=1}^\infty\frac{1}{k^{m-(2j+1)}(k^2+n^2)}$$

Let $m-2j-1 = 1$ which implies that $j = m/2-1$

$$\sum_{i= 1}^{m/2-1}(-1)^{i-1}\frac{1}{n^{2i}}\zeta(m-(2i-1))+\frac{(-1)^{m/2-1}}{n^{m-2}}\sum_{k=1}^\infty\frac{1}{k(k^2+n^2)}$$

Now let us look at the sum

$$\sum_{k=1}^\infty\frac{1}{k(k^2+n^2)} = \sum_{k=1}^\infty\frac{1}{2ink}\left\{ \frac{1}{k-in}-\frac{1}{k+in}\right\}$$

Which can be written as

$$\sum_{k=1}^\infty\frac{1}{k(k^2+n^2)} = \frac{1}{2n^2}\sum_{k=1}^\infty\frac{1}{k}\left\{ \frac{in}{k+in}+\frac{-in}{k-in}\right\}$$

According to the digamma function

$$\sum_{k=1}^\infty\frac{1}{k(k^2+n^2)} = \frac{1}{2n^2}\left\{ \gamma+\psi(1+in)+\psi(1-in)+\gamma\right\}$$

which simplifies to

$$\sum_{k=1}^\infty\frac{1}{k(k^2+n^2)} = \frac{\psi(1-in)+\psi(1+in)+2\gamma}{2n^2}$$

Now we we can verify $\psi(1-in) = \overline{\psi(1+in)}$ which suggests that

$$\psi(1+in) +\psi(1-in) = 2\Re\left\{\psi(1+in) \right\}$$

Hence we have the sum

$$\sum_{k=1}^\infty\frac{1}{k(k^2+n^2)} = \frac{2\Re\left\{\psi(1+in) \right\}+2\gamma}{2n^2}=\frac{\Re\left\{\psi(1+in) \right\}+\gamma)}{n^2}$$

This concludes to

$$\sum_{j= 1}^{m/2-1}(-1)^{j-1}\frac{\zeta(m-(2i-1))}{n^{2j}}+(-1)^{m/2-1}\frac{\Re\left\{\psi(1+in) \right\}+\gamma}{n^m}$$

Find the following integral

$$\int^\infty_0 \mathrm{cl}_m(\theta) \theta^{n-1}\, d\theta$$

If $m$ is even

$$\int^\infty_0 \sum_{k=1}^\infty \frac{\sin(k\theta)}{k^m}\theta^{n-1}\, d\theta$$

Using the series representation

$$\sum_{k=1}^\infty \frac{1}{k^m}\int^\infty_0 \sin(k\theta) \theta^{n-1}d\theta$$

Let $\phi = k\theta$

$$\sum_{k=1}^\infty \frac{1}{k^{m+n}}\int^\infty_0 \sin(\phi) \phi^{n-1}d\phi$$

Now using the mellin transform

$$\zeta(m+n) \sin\left( \frac{\pi}{2}n\right)\Gamma(n)$$

If $m$ is odd
$$\zeta(m+n) \cos\left( \frac{\pi}{2}n\right)\Gamma(n)$$

Gold Member
MHB
Re: (AIT) Clausen functions

Clausen Integral function

$$\mathrm{cl}_2(\theta) = -\int^\theta_0 \log\left[2\sin\left(\frac{\phi}{2}\right)\right]d\phi$$

Start by the following

$$\mathrm{Li}_2(e^{i\theta}) = \sum_{k=1}^\infty \frac{e^{ik\theta}}{k^2}= \sum_{k=1}^\infty \frac{\cos(k\theta)}{k^2}+i\sum_{k=1}^\infty \frac{\sin(k\theta)}{k^2}$$

By the integral definition of the dilogarithm

$$\mathrm{Li}_2(e^{i\theta}) -\zeta(2)= -\int^{e^{i\theta}}_1 \frac{\log(1-x)}{x}\,dx$$

Let $x = e^{i\phi}$

$$\mathrm{Li}_2(e^{i\theta}) -\zeta(2)= -i\int^{\theta}_0 \log(1-e^{i\phi})d \phi$$

Let us look at the following

$$1-e^{i\phi} = 1-\cos(\phi)-i\sin(\phi) = 2\sin^2(\phi/2)-2i\sin(\phi/2)\cos(\phi/2)$$

Which simplifies to

$$1-e^{i\phi} =2\sin(\phi/2)\left[\sin(\phi/2)-i\cos(\phi/2)\right] = 2\sin(\phi/2) e^{-(i/2) (\pi-\phi)}$$

Hence our integral

$$\mathrm{Li}_2(e^{i\theta}) -\zeta(2)= -i\int^{\theta}_0 \log \left[2\sin(\phi/2) e^{-(i/2) (\pi-\phi)}\right]d \phi$$

Use the complex integral properties

$$\mathrm{Li}_2(e^{i\theta}) -\zeta(2)= -i\int^{\theta}_0 \log \left[2\sin(\phi/2)\right]d\phi +\frac{1}{4}(\pi-\theta)^2-\frac{1}{4}\pi^2$$

By equating the imaginary parts we have our result.

We can see the special value
$$\mathrm{cl}_2\left(\frac{\pi}{2}\right) = \sum_{k=1}^\infty \frac{\sin(k\pi/2)}{k^2}= \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^2} = G$$

Where $G$ is the Catalan's constant.

Prove the following

$$\mathrm{cl}_2(2\theta) =2\mathrm{cl}_2(\theta)-\mathrm{cl}_2(\pi -\theta)$$

Use the integral representation

$$\mathrm{cl}_2(2\theta) = -\int^{2\theta}_0 \log\left[2\sin\left(\frac{t}{2}\right)\right]dt$$

Let $t= 2\phi$

$$-2\int^{\theta}_0 \log\left[2\sin\left(\phi\right)\right]d\phi$$

Use double angle identity

$$-2\int^{\theta}_0 \log\left[4\sin\left(\frac{\phi}{2}\right) \cos\left(\frac{\phi}{2}\right)\right]d\phi$$

Separate the logarithms

$$-2\int^{\theta}_0 \log\left[2\sin\left(\frac{\phi}{2}\right) \right]d\phi-2\int^{\theta}_0 \log\left[2\cos\left(\frac{\phi}{2}\right) \right]d\phi$$

We can verify that

$$\mathrm{cl}_2(\pi -\theta ) = \int^{\theta}_0 \log\left[2\cos\left(\frac{\phi}{2}\right) \right]d\phi$$

Hence

$$\mathrm{cl}_2(2\theta) =2\mathrm{cl}_2(\theta)-\mathrm{cl}_2(\pi -\theta)$$

Using that

$$\mathrm{cl}_2(3\pi) =2\mathrm{cl}_2\left(\frac{3\pi}{2}\right)-2\mathrm{cl}_2\left(-\frac{\pi}{2}\right)$$

Since $\mathrm{cl}_2(3\pi) = 0$
$$\mathrm{cl}_2\left(\frac{3\pi}{2}\right)=\mathrm{cl}_2\left(-\frac{\pi}{2}\right) = - \mathrm{cl}_2\left(\frac{\pi}{2}\right)= -G$$

Prove that

$$\int^{2\pi}_0 \mathrm{cl}_2(x)^2dx = \frac{\pi^5}{90}$$

Using the series representation

$$\sum_{k=1}^\infty \sum_{n=1}^\infty \frac{1}{(nk)^2}\int^{2\pi}_0 \sin(kx) \sin(nx) dx$$

Consider the integral

$$\int_0^{2\pi} \sin(kx)\sin(nx)\,dx = \frac{1}{2}\int_0^{2\pi} \cos((k-n)x)-\cos((k+n)x)\,dx$$

Consider two cases

If $n = k$ then

$$\frac{1}{2}\int_0^{2\pi} 1-\cos(2nx)\,dx = \pi$$

If $n \neq k$

$$\frac{1}{2}\int_0^{2\pi} \cos((k-n)x)-\cos((k+n)x)\,dx =\frac{1}{2}\left[\frac{\sin((k-n)x)}{k-n}-\frac{\sin((k+n)x)}{k+n} \right]^{2\pi}_0 =0$$

Hence we have

$$\int_0^{2\pi} \sin(kx)\sin(nx)\,dx=\begin{cases} 0 & n \neq k\\ \pi & n=k\end{cases}$$

We can write the series as

$$\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{(nk)^2} = \sum_{n\neq k}^\infty \frac{1}{(nk)^2}+ \sum_{n=1}^\infty \frac{1}{n^4}$$

Now since the integral $n\neq k$ goes to zero the result reduces to

$$\pi \sum_{n=1}^\infty \frac{1}{n^4} =\pi \zeta(4) = \frac{\pi^5}{90}$$

Gold Member
MHB
(AIT) Barnes G function

4.Integration using special functions (continued)

4.15. Barnes G function

Define the following

$$G(z+1)=(2\pi)^{z/2}\exp\left(-\frac{z+z^2(1+\gamma)}{2}\right)\prod_{n=1}^{\infty}\left\{\left(1+\frac{z}{n} \right)^n\exp\left(\frac{z^2}{2n}-z\right) \right\}$$

Difference formula

$$G(z+1) = \Gamma(z) G(z)$$

From the series representation we have

$$\frac{G(z+1)}{G(z)} = \sqrt{2 \pi} \exp \left(-z - \gamma z + \frac{\gamma}{2} \right) \prod_{k=1}^{\infty} \left(\frac{k+z}{k+z-1} \right)^{k} \exp \left( \frac{2z-1-2k}{2k} \right).$$

This can be written as

$$\frac{G(z+1)}{G(z)} = z\sqrt{2 \pi} \exp \left(-z + \frac{\gamma}{2} \right) \prod_{k=1}^{\infty} \left(\frac{k+z}{k+z-1} \right)^{k} \exp \left( -\frac{1+2k}{2k} \right)\left(1+\frac{z}{k}\right)\\ \frac{e^{- \gamma z }}{z} \prod_{k=1}^{\infty} \left(1+\frac{z}{k} \right)^{-1} e^{\left( \frac{z}{k} \right)}$$

Which simplifies to

$$\frac{G(z+1)}{G(z)} = z\Gamma(z)\sqrt{2 \pi} \exp \left(-z + \frac{\gamma}{2} \right) \prod_{k=1}^{\infty} \left(\frac{k+z}{k+z-1} \right)^{k} \exp \left( -\frac{1+2k}{2k} \right)\left(1+\frac{z}{k}\right)$$

It suffices to prove that

$$z\sqrt{2 \pi} \exp \left(-z + \frac{\gamma}{2} \right) \prod_{k=1}^{\infty} \left(\frac{k+z}{k+z-1} \right)^{k} \exp \left( -\frac{1+2k}{2k} \right)\left(1+\frac{z}{k}\right) = 1$$

or

$$\prod_{k=1}^{\infty} \left(\frac{k+z}{k+z-1} \right)^{k} \exp \left( -\frac{1+2k}{2k} \right)\left(1+\frac{z}{k}\right) = \frac{\exp \left(z -\frac{\gamma}{2} \right)}{z\sqrt{2 \pi}}$$

Start by

$$\lim_{N \to \infty }\prod_{k=1}^{N} \left(\frac{k+z}{k+z-1} \right)^{k} \exp \left( -\frac{1+2k}{2k} \right)\left(1+\frac{z}{k}\right)$$

Notice

\begin{align}
\prod_{k=1}^{N} \left(\frac{k+z}{k+z-1} \right)^{k} \left(1+\frac{z}{k}\right)&= \frac{\prod_{k=1}^{N} (k+z)^k \prod_{k=1}^{N}\left(1+\frac{z}{k}\right)}{\prod_{k=1}^{N}(k+z-1)^k} \\[1em]
&= \frac{\prod_{k=1}^{N} (k+z)^k \prod_{k=1}^{N} (k+z)}{zN!\prod_{k=1}^{N-1}(k+z)^{k+1}} \\[1em]
&= \frac{(N+z)^{N+1}\prod_{k=1}^{N-1} (k+z)^k\prod_{k=1}^{N-1} (k+z)}{zN!\prod_{k=1}^{N-1}(k+z)^{k+1}}\\[1em]
&= \frac{(N+z)^{N+1}\prod_{k=1}^{N-1} (k+z)^{k+1}}{zN!\prod_{k=1}^{N-1}(k+z)^{k+1}}\\[1em]
&= \frac{(N+z)^{N+1}}{zN!}
\end{align}

The second product

$$\prod_{k=1}^{N} \exp \left( -\frac{1+2k}{2k} \right) =\exp \left( -\sum_{k=1}^N\frac{1+2k}{2k} \right) = e^{-\frac{1}{2}H_N-N}$$

Hence we have the following
$$e^{-\frac{1}{2}H_N-N}\times \frac{(N+z)^{N+1}}{z N! }$$

According to Stirling formula we have

$$e^{-\frac{1}{2}H_N-N}\frac{(N+z)^{N+1}}{z N! } \sim e^{-\frac{1}{2}H_N-N}\frac{(N+z)^{N+1}}{z(N/e)^N}\times \frac{1}{\sqrt{2\pi N}}$$

By some simplifications we have

$$\frac{e^{-\frac{1}{2}(H_N-\log N)}}{z}\left(1+\frac{z}{N}\right) \times \left(1+ \frac{z}{N}\right)^N\times \frac{1}{\sqrt{2\pi}} \sim \frac{\exp\left(-\frac{\gamma}{2} +z\right)}{z\sqrt{2\pi}}$$

Where we used that

$$\lim_{n \to \infty }H_n -\log(n)= \gamma$$

Reflection formula

$$\log \left\{\frac{G(1-z)}{G(1+z)} \right\} = z \log\left(\frac{\sin(\pi z)}{\pi}\right) + \frac{\mathrm{cl}_2(2\pi z)}{2\pi}$$

Start by the series expansion

$$\frac{G(1-z)}{G(1+z)} = \frac{(2\pi)^{-z/2}\exp\left(\frac{z-z^2(1+\gamma)}{2}\right)\prod_{n=1}^{\infty}\left\{\left(1-\frac{z}{n} \right)^n\exp\left(\frac{z^2}{2n}+z\right) \right\}}{(2\pi)^{z/2}\exp\left(-\frac{z+z^2(1+\gamma)}{2}\right)\prod_{n=1}^{\infty}\left\{\left(1+\frac{z}{n} \right)^n\exp\left(\frac{z^2}{2n}-z\right) \right\}}$$

This simplifies to

$$\frac{G(1-z)}{G(1+z)} = (2\pi)^{-z}e^z\prod_{n=1}^{\infty}\frac{(n-z)^n}{(n+z)^n} e^{2z}$$

Take the log of both sides

$$\log \left\{\frac{G(1-z)}{G(1+z)} \right\} = -z \log(2\pi)+z + \log\left \{ \prod_{n=1}^{\infty}\frac{(n-z)^n}{(n+z)^n} e^{2z} \right\}$$
Let the following

$$f(z) = \log\left\{\prod_{n=1}^{\infty}\frac{(n-z)^n}{(n+z)^n} e^{2z} \right\} = \sum_{n=1}^\infty n\log(n-z)-n\log(n+z)+2z$$

Differentiate with respect to $z$

$$f'(z) =\sum_{n=1}^{\infty}\frac{-n}{n-z}-\frac{n}{n+z}+2 = \sum_{n=1}^{\infty}\frac{-n(n+z)-n(n-z)+2(n^2-z^2)}{n^2-z^2}$$

Hence we have

$$\sum_{n=1}^{\infty}\frac{-2z^2}{n^2-z^2}$$

Now we can use the following

$$z\pi \cot \pi z = 1+\sum_{n=1}^{\infty}\frac{-2z^2}{n^2-z^2}$$

Hence we conclude that

$$f'(z) = z\pi \cot \pi z-1$$

Integrate with respect to $z$

$$f(z) = \int^z_0 x\pi \cot(\pi x) \, dx -z$$

Hence we have

$$\log \left\{\frac{G(1-z)}{G(1+z)} \right\} = -z \log(2\pi)+\int^z_0 z\pi \cot(\pi x) \, dx$$

Now use integration by parts for the integral
\begin{align}
\int^z_0 x\pi \cot(\pi x) \, dx
&=z\log(\sin\pi z)-\int^z_0 \log(\sin\pi x) dx\\
&=z\log(2\sin\pi z)-\int^z_0 \log(2\sin\pi x) dx\\
&=z\log(2\sin\pi z)-\frac{1}{2\pi }\int^{2\pi z}_0 \log\left(2\sin\frac{x}{2}\right) dx\\
&=z\log(2\sin\pi z)+\frac{\mathrm{cl}_2(2\pi z)}{2\pi}\\
\end{align}

That implies

$$\log \left\{\frac{G(1-z)}{G(1+z)} \right\} = z\log(2\sin\pi z)-z \log(2\pi)+ \frac{\mathrm{cl}_2(2\pi z)}{2\pi}$$

which implies our result.

Gold Member
MHB
Re: (AIT) Barnes G function

Values at positive integers

$$G(n) = \prod^{n-1}_{k=1} \Gamma(k)$$

It can be proved by induction. For $G(1) = 1$, suppose

$$G(n) = \prod^{n-1}_{k=1} \Gamma(k)$$

We want to show

$$G(n+1) = \prod^{n}_{k=1} \Gamma(k)$$

By the difference formula

$$G(n+1) = \Gamma(n)G(n) = \Gamma(n) \prod^{n-1}_{k=1} \Gamma(k) =\prod^{n}_{k=1} \Gamma(k)$$

Loggamma integral

$$\int^z_0 \log \Gamma(x) dx =\frac{z}{2}\log(2\pi)+\frac{z(z-1)}{2}+z\log \Gamma(z)-\log G(z+1)$$

Take the log to the series representation

$$\log G(z+1)=\frac{z}{2}\log(2\pi)-\frac{z+z^2(1+\gamma)}{2}+\sum_{n=1}^{\infty}n \log\left(1+\frac{z}{n} \right)+\frac{z^2}{2n}-z$$

Let the following

$$f(z) = \sum_{n=1}^{\infty}n \log\left(1+\frac{z}{n} \right)+\frac{z^2}{2n}-z$$

Differentiate with respect to $z$

$$f(z) = \sum_{n=1}^{\infty}\frac{n}{z+n}+\frac{z}{n}-1 =\sum_{n=1}^{\infty}\frac{z^2}{n(n+z)}$$

Now use the following

$$\psi(z) = -\gamma -\frac{1}{z}+\sum_{n=1}^\infty \frac{z}{n(n+z)}$$

which implies that

$$\sum_{n=1}^\infty \frac{z^2}{n(n+z)} = z\psi(z)+\gamma z+1$$

Hence we have

$$f(z) = z\psi(z)+\gamma z+1$$

Integrate with respect to $z$

$$f(z) = \int^z_0 x\psi(x) dx +\frac{\gamma z^2}{2}+z$$

which implies that

$$f(z) = z\log \Gamma(z) -\int^z_0 \log \Gamma(x) dx +\frac{\gamma z^2}{2}+z$$

Hence we have

$$\log G(z+1)=\frac{z}{2}\log(2\pi)-\frac{z+z^2(1+\gamma)}{2}+z\log \Gamma(z) -\int^z_0 \log \Gamma(x) dx +\frac{\gamma z^2}{2}+z$$

By some rearrangements we have

$$\int^z_0 \log \Gamma(x) dx =\frac{z}{2}\log(2\pi)+\frac{z(z-1)}{2}+z\log \Gamma(z)-\log G(z+1)$$

Gold Member
MHB
Re: (AIT) Barnes G function

Relation to Hyperfactorial function

Prove for $n$ is a positive integer

$$G(n+1) = \frac{(N!)^n}{H(n)}$$

Where $H(n)$ is the hyperfactorial function

$$H(n) = \prod^n_{k=1}k^k$$

proof

We can prove it by induction for $n=0$ we have, $G(1)=1$,

suppose that

$$G(n) = \frac{\Gamma(n)^{n-1}}{H(n-1)}$$

we want to find

$$G(n+1) = \Gamma(n)G(n) = \Gamma(n) \frac{\Gamma(n)^{n-1}}{H(n-1)}$$

Notice that

$$H(n-1) = \prod^{n-1}k^k = \frac{\prod^{n}k^k}{n^n} = \frac{H(n)}{n^n }$$

We deduce that

$$G(n+1) = \Gamma(n)G(n) = \frac{\Gamma(n)^{n}\times n^n}{H(n)} = \frac{(N!)^n}{H(n)}$$

A related constant

We define the Glaisher-Kinkelin constant as

$$A = \lim_{n \to \infty}\frac{H(n)}{n^{n^2/2+n/2+1/12}e^{-n^2/4}}$$

Prove that

$$\lim_{n \to \infty}\frac{G(n+1)}{(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}} = \frac{e^{1/12}}{A}$$

proof

From the previous result we have

$$\lim_{n \to \infty}\frac{(N!)^n}{H(n)(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}}$$

Now use the Stirling approximation

$$(N!)^n \sim (2\pi)^{n/2}n^{n^2+n/2}e^{-n^2+1/12}$$

Hence we deduce that

$$\lim_{n \to \infty}\frac{(2\pi)^{n/2}n^{n^2+n/2}e^{-n^2+1/12}}{H(n)}\times\frac{1}{(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}}$$

By simplifications we have

$$e^{1/12}\lim_{n \to \infty}\frac{n^{n^2/2+n/2+1/12}e^{-n^2/4}}{H(n)} = \frac{e^{1/12}}{A}$$

Exercise

$$\zeta'(2) = \frac{\pi^2}{6}\left(\log(2\pi)+\gamma-12\log A \right)$$

$$\log \left[\frac{G(n+1)}{(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}} \right] \sim \frac{1}{12}-\log A$$

Let the following

$$f(n) = \log \left[\frac{G(n+1)}{(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}} \right]$$

Use the series representation of the Barnes functions

$$f(n) = \log \left[ \frac{(2\pi)^{n/2}\exp\left(-\frac{n+n^2(1+\gamma)}{2}\right)\prod_{k=1}^{\infty}\left\{\left(1+\frac{n}{k} \right)^k\exp\left(\frac{n^2}{2k}-n\right) \right\}}{(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}} \right]$$

Which reduces to

$$f(n) = -\frac{n+n^2(1+\gamma)}{2}+\sum^\infty_{k=1}\left\{k \log \left(1+\frac{n}{k} \right)+\frac{n^2}{2k}-n \right\}\\ -\left( \frac{n^2}{2}-\frac{1}{12}\right)\log(n)+\frac{3n^2}{4}$$

Differentiate with respect to $n$

$$f'(n) = -\frac{1}{2}-n-\gamma n+n\psi(n)+\gamma n+1-n\log(n)-\frac{n}{2}+\frac{1}{12n}+\frac{3n}{2}$$

Note that we already showed that

$$\frac{d}{dn}\sum^\infty_{k=1}\left\{k \log \left(1+\frac{n}{k} \right)+\frac{n^2}{2k}-n\right\}= n\psi(n)+\gamma n+1$$

By simplifications we have

$$f'(n) = n\psi(n)-n\log(n)+\frac{1}{12n}+\frac{1}{2}$$

Now use that

$$\psi(n) = \log(n)-\frac{1}{2n}-2\int^\infty_0 \frac{z dz}{(n^2+z^2)(e^{2\pi z}-1)}dz$$

Hence we deduce that

$$f'(n) =-2\int^\infty_0 \frac{nz dz}{(n^2+z^2)(e^{2\pi z}-1)}dz+\frac{1}{12n}$$

Integrate with respect to $n$

$$f(n) =-\int^\infty_0 \frac{z\log(n^2+z^2)}{(e^{2\pi z}-1)}dz+\frac{1}{12}\log(n)+C$$

Take the limit $n \to 0$

$$C = \lim_{n \to 0}f(n)-\frac{1}{12}\log(n)+\int^\infty_0 \frac{z\log(z^2)}{(e^{2\pi z}-1)}dz$$

Hence we have the limit

$$\lim_{n \to 0}\frac{G(n+1)}{(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}}-\frac{1}{12}\log(n) = \lim_{n \to 0}\log \frac{G(n+1)}{(2\pi)^{n/2}n^{n^2/2} e^{-3n^2/4}} = 0$$

Hence we see that

$$C = 2\int^\infty_0 \frac{z\log(z)}{(e^{2\pi z}-1)}dz$$

Finally we have

$$f(n) =-\int^\infty_0 \frac{z\log(n^2+z^2)}{(e^{2\pi z}-1)}dz+\frac{1}{12}\log(n)+ 2\int^\infty_0 \frac{z\log(z)}{(e^{2\pi z}-1)}dz$$

$$f(n) =-\int^\infty_0 \frac{z\log\left(1+\frac{z^2}{n^2} \right)}{(e^{2\pi z}-1)}dz-\log(n^2)\int^\infty_0 \frac{z}{(e^{2\pi z}-1)}dz+\frac{1}{12}\log(n)\\+ 2\int^\infty_0 \frac{z\log(z)}{(e^{2\pi z}-1)}dz$$

Also we have

$$\int^\infty_0 \frac{z}{(e^{2\pi z}-1)}dz = \frac{1}{24}$$

That simplifies to

$$f(n)=-\int^\infty_0 \frac{z\log\left(1+\frac{z^2}{n^2} \right)}{(e^{2\pi z}-1)}dz+2\int^\infty_0 \frac{z\log(z)}{(e^{2\pi z}-1)}dz$$

Take the limit $n \to \infty$

$$2\int^\infty_0 \frac{z\log(z)}{(e^{2\pi z}-1)}dz = \frac{1}{12}-\log A$$

Now use that

\begin{align} 2\int^\infty_0 \frac{z\log(z)}{(e^{2\pi z}-1)}dz &= 2\int^\infty_0 \frac{z\log(z)}{e^{2\pi z}}\times \frac{1}{1-e^{-2\pi z}}dz\\ &= 2\sum_{n=0}^\infty \int^\infty_0 e^{-2\pi z(n+1)}z\log(z)\,dz\\ &= \sum_{n=1}^\infty \frac{\psi(2) − \log(2\pi)+\log(n)}{2\pi^2 n^2}\\ &=\frac{(\psi(2)-\log(2\pi))\zeta(2)+\zeta'(2)}{2\pi^2}\\ \end{align}

Hence we conclude that

$$\zeta'(2) =(\log(2\pi)-\psi(2))\zeta(2) + 2\pi^2\left(\frac{1}{12}-\log A \right) = \zeta(2) (\log(2\pi)+\gamma-12 \log A)$$

Gold Member
MHB
Re: (AIT) Barnes G function

Relation to hurwitz zeta function

Prove that

$$\log G(z+1) - z \log \Gamma(z) = \zeta'(-1) - \zeta'(-1,z)$$

Start by the following

$$\zeta(s,z) = \frac{z^{-s}}{2}+\frac{z^{1-s}}{s-1}+2\int^\infty_0 \frac{\sin(s\arctan(x/z))}{(z^2+x^2)^{s/2}(e^{2\pi x}-1)}\,dx$$

Hence we have

$$\zeta'(-1,z) = -\frac{z\log(z)}{2}+\frac{z^2\log(z)}{2}-\frac{z^2}{4}+\int^\infty_0 \frac{x\log(x^2+z^2)+2z\arctan(x/z)}{(e^{2\pi x}-1)}\,dx$$

Now use that

$$\psi(z) = \log(z)-\frac{1}{2z}-2\int^\infty_0 \frac{x }{(z^2+x^2)(e^{2\pi x}-1)}dx$$

Which implies that

$$\int^\infty_0 \frac{2zx }{(z^2+x^2)(e^{2\pi x}-1)}dx=z\log(z)-\frac{1}{2}-z\psi(z)$$

By taking the integral

$$\int^\infty_0 \frac{x\log(x^2+z^2) -x\log(x^2)}{(e^{2\pi x}-1)}dx=\int^z_0x\log(x)\,dx-\int^z_0x\psi(x)\,dx-\frac{z}{2}$$

Which simplifies to

$$\int^\infty_0 \frac{x\log(x^2+z^2) }{(e^{2\pi x}-1)}dx=\zeta'(-1)-\frac{z^2}{4}+\frac{1}{2} z^2 \log(z)-z\log\Gamma(z)+\int^z_0\log\Gamma(x)\,dx-\frac{z}{2}$$

Also we have

$$2\int^\infty_0 \frac{x}{(x^2+z^2)(e^{2\pi x}-1)}dx=\log(z)-\frac{1}{2z}-\psi(z)$$

By integration we have

$$2\int^\infty_0 \frac{\arctan(x/z)}{(e^{2\pi x}-1)}dx=z+\frac{\log(z)}{2}-z \log(z)+\log\Gamma(z)+C$$

Let $z \to 1$ to evaluate the constant

$$2\int^\infty_0 \frac{\arctan(x/z)}{(e^{2\pi x}-1)}dx=z+\frac{\log(z)}{2}-z \log(z)+\log\Gamma(z)-\frac{1}{2}\log(2\pi)$$

Multiply by $z$

$$2\int^\infty_0 \frac{z\arctan(x/z)}{(e^{2\pi x}-1)}dx=z^2+\frac{z\log(z)}{2}-z^2 \log(z)+z\log\Gamma(z)-\frac{z}{2}\log(2\pi)$$

Substitute both integrals our formula

$$\zeta'(-1,z) =-\frac{z\log(z)}{2}+\frac{z^2\log(z)}{2}-\frac{z^2}{4}+z^2+\frac{z\log(z)}{2}-z^2 \log(z)+z\log\Gamma(z)\\-\frac{z}{2}\log(2\pi)+\zeta'(-1)-\frac{z^2}{4}+\frac{1}{2} z^2 \log(z)-z\log\Gamma(z)+\int^z_0\log\Gamma(x)\,dx-\frac{z}{2}$$

Which reduces to

$$\int_{0}^{z} \log \Gamma(x) \, \mathrm dx = \frac{z}{2} \log(2 \pi) + \frac{z(1-z)}{2} - \zeta^{'}(-1) + \zeta^{'}(-1,z)$$

We also showed that

$$\int_{0}^{z} \log \Gamma(x) \, \mathrm dx = \frac{z}{2} \log (2 \pi) + \frac{z(1-z)}{2} + z \log \Gamma(z) - \log G(z+1)$$

By equating the equations we get our result.

Prove that

$$\zeta'(-1) = \frac{1}{12}-\log A$$

Start by

$$\zeta(s) = \lim_{m \to \infty} \left( \sum_{k=1}^{m} k^{-s} - \frac{m^{1-s}}{1-s} - \frac{m^{-s}}{2} + \frac{sm^{-s-1}}{12} \right) \ , \ \text{Re}(s) >-3.$$

Differentiate with respect to $s$

$$\zeta'(s) = \lim_{m \to \infty} \left(- \sum_{k=1}^{m} k^{-s}\log(k) - \frac{m^{1-s}}{(1-s)^2} +\frac{m^{1-s}}{1-s}\log(m)+\frac{m^{-s}}{2}\log(m)\\ + \frac{m^{-s-1}}{12}-\frac{m^{-s-1}}{12}\log(m)\right)$$

Now let $s \to -1$

$$\zeta'(-1) = \lim_{m \to \infty} \left(- \sum_{k=1}^{m} k\log(k) - \frac{m^{2}}{4} +\frac{m^{2}}{2}\log(m)+\frac{m}{2}\log(m)\\ + \frac{1}{12}-\frac{1}{12}\log(m)\right)$$

Take the exponential of both sides

$$e^{\zeta'(-1)}= e^{1/12}\lim_{m \to \infty} \frac{m^{m^2/2+m/2-1/12} e^{-m^2/4}}{ e^{\sum_{k=1}^{m} k\log(k)}} = e^{1/12}\lim_{m \to \infty} \frac{m^{m^2/2+m/2-1/12} e^{-m^2/4}}{ H(m+1)} = \frac{ e^{1/12}}{A}$$

We conclude that

$$\zeta'(-1) = \frac{1}{12}-\log A$$

Prove that

$$G\left(\frac{1}{2}\right) = 2^{1/24} \pi^{-1/4}e^{1/8}A^{-3/2}$$

We know that

$$\log G(z)+ \log \Gamma(z)- z \log \Gamma(z) = \zeta'(-1) - \zeta'(-1,z)$$

Note that

$$\zeta \left(s, \frac{1}{2} \right) = (2^s-1)\zeta(s)$$

Which implies that

$$\zeta' \left(-1, \frac{1}{2} \right) = \frac{\log(2)}{2}\zeta(-1)-\frac{1}{2}\zeta'(-1)$$

Hence we have

$$\log G\left(\frac{1}{2} \right)+ \frac{1}{2}\log \Gamma\left(\frac{1}{2} \right)= \frac{3}{2}\zeta'(-1)-\frac{\log(2)}{2}\zeta(-1)$$

Using that we have

$$G\left(\frac{1}{2}\right) = 2^{1/24}\pi^{-1/4}e^{\frac{3}{2}\zeta'(-1)}$$

Note that

$$\zeta(-1) = -\frac{1}{12}$$

This can be proved by the functional equation of the zeta function.