Micromass' big integral challenge

[micromass]

Integrals are pretty interesting, and there are a lot of different methods to solve them. In this thread, I will give as a challenge 10 integrals. Here are the rules:

• For a solution to count, the answer must not only be correct, but a detailed solution must also be given.
• A correct answer would consist of the correct number (which may be infinite) or the statement that the given integral does not exist.
• Any use of outside sources is allowed, but do not look up the integral directly. For example, it is ok to go check calculus textbooks for methods, but it is not allowed to type in the integral in wolframalpha.
• If you previously encountered this integral and remember the solution, then you cannot participate with that specific integral.
• All mathematical methods are allowed.

Anyway, here we go:

1. \int_0^1 \frac{\text{ln}(1+x)}{1+x^2}dx = \frac{\pi}{4}\text{ln}(\sqrt{2}) SOLVED BY Ssnow
2. \int_0^\infty \frac{|\sin(x)|}{x}dx = +\infty SOLVED BY PeroK
   
3. \int_0^\infty \frac{\text{atan}(2016x) - \text{atan}(1916x)}{x}dx = \frac{\pi}{2}\text{ln} \left(\frac{2016}{1916}\right) SOLVED BY Samy_A
   
4. \int_0^{\pi/2} \frac{\sqrt{\sin(x)}}{\sqrt{\sin(x)}+\sqrt{\cos(x)}}dx = \frac{\pi}{4} SOLVED BY Ssnow
   
5. \int_0^1 \sqrt{-\text{ln}(x)}dx = \frac{\sqrt{\pi}}{2} SOLVED BY Ssnow
   
6. \int_0^1 \frac{1-4x^5}{(x^5 - x + 1)^2}dx = 1 SOLVED BY Samy_A
7. \int_0^{\pi/2} \text{acos}\left(\frac{\cos(x)}{1+2\cos(x)}\right)dx
8. \int_0^\infty \frac{\sin^{9}(x)}{x}dx = \frac{35}{256}\pi SOLVED BY vela
9. \int_0^\infty \frac{x^{1916}}{x^{2016} + 1}dx = \frac{1}{2016} \pi \frac{1}{\sin{\frac{1917 \pi}{2016}}} SOLVED BY fresh_42
10. \int_{\sqrt{2}}^\infty \frac{1}{x + x^{\sqrt{2}}}dx = -\ln \frac{\sqrt{2}}{{(1+{\sqrt 2}^{\sqrt{2}-1}})^{\sqrt{2}+1}} SOLVED BY Samy_A

 

[Ssnow]

I want propose the solution of 5. $\int_{0}^{1}\sqrt{-\ln{x}}dx=\frac{\sqrt{\pi}}{2}$

I started changing $-\ln{x}=t^{2}$ so $x=e^{-t^{2}}$ and $dx=-2te^{-t^{2}}dt$, changing also the domain:

$\int_{0}^{\infty}2t^{2}e^{-t^{2}}dt$

doing integration by parts $t\rightarrow 1$ and $\int 2te^{-t^{2}}dt \rightarrow -e^{-t^{2}}$ so

$=-\lim_{z\rightarrow +\infty}te^{-t^{2}}|_{0}^{z}-\int_{0}^{\infty}-e^{-t^{2}}dt=0+\int_{0}^{\infty}e^{-t^{2}}dt=\frac{\sqrt{\pi}}{2}$

The last is half the Gaussian integral. I declare I never see this integral before ... It is correct?

 

[micromass]

I want propose the solution of 5. $\int_{0}^{1}\sqrt{-\ln{x}}dx=\frac{\sqrt{\pi}}{2}$

I started changing $-\ln{x}=t^{2}$ so $x=e^{-t^{2}}$ and $dx=-2te^{-t^{2}}dt$, changing also the domain:

$\int_{0}^{\infty}2t^{2}e^{-t^{2}}dt$

doing integration by parts $t\rightarrow 1$ and $\int 2te^{-t^{2}}dt \rightarrow -e^{-t^{2}}$ so

$=-\lim_{z\rightarrow +\infty}te^{-t^{2}}|_{0}^{z}-\int_{0}^{\infty}-e^{-t^{2}}dt=0+\int_{0}^{\infty}e^{-t^{2}}dt=\frac{\sqrt{\pi}}{2}$

The last is half the Gaussian integral. I declare I never see this integral before ... It is correct?

Correct! Congratulations!

 

[PeroK]

The integral in #2 must diverge. For non-negative integer $n$:

$\int_{n\pi}^{(n+1)\pi} \frac{|sin x|}{x} dx > \int_{n\pi}^{(n+1)\pi} \frac{|sin x|}{(n+1)\pi} dx = \frac{2}{(n+1)\pi}$

Hence:

$\int_{0}^{\infty} \frac{|sin x|}{x} dx > \sum_{n=0}^{\infty}\frac{2}{(n+1)\pi} = \frac{2}{\pi}\sum_{n=1}^{\infty} \frac{1}{n}$

 

[micromass]

The integral in #2 must diverge. For non-negative integer $n$:

$\int_{n\pi}^{(n+1)\pi} \frac{|sin x|}{x} dx > \int_{n\pi}^{(n+1)\pi} \frac{|sin x|}{(n+1)\pi} dx = \frac{2}{(n+1)\pi}$

Hence:

$\int_{0}^{\infty} \frac{|sin x|}{x} dx > \sum_{n=0}^{\infty}\frac{2}{(n+1)\pi} = \frac{2}{\pi}\sum_{n=1}^{\infty} \frac{1}{n}$

Correct!

 

[Samy_A]

$\displaystyle \int_0^\infty \frac{\text{atan}(2016x) - \text{atan}(1916x)}{x}dx$

I use the general result $\displaystyle \int_0^\infty \frac{f(ax)-f(bx)}{x} dx= (f(\infty) -f(0)) \log \frac{a}{b}$ (proven using Leibnitz rule, see below).

So $\displaystyle \int_0^\infty \frac{\text{atan}(2016x) - \text{atan}(1916x)}{x}dx =(\text{atan}( \infty ) - \text{atan}( 0)) \log \frac{2016}{1916} = \frac{\pi}{2} \log \frac{2016}{1916}$.

Now for $\displaystyle \int_0^\infty \frac{f(ax)-f(bx)}{x} dx= (f(\infty) -f(0)) \log \frac{a}{b}$ (*)
Set $\displaystyle I(y)= \int_0^\infty \frac{f(yx)-f(bx)}{x} dx$

Notice that $I(a)$ is the integral we are trying to determine, and $I(b)=0$.

Then $\displaystyle I'(y) = \int_0^\infty f'(yx) dx$. Let's do the substitution $z=yx$, so that $\displaystyle I'(y)= \frac{1}{y}\int_0^\infty f'(z) dz =\frac{1}{y} (f(\infty) -f(0))$. Let's name $A=f(\infty) -f(0)$.
We then have $I'(y)=A \frac{1}{y}$, a differential equation with solution $I(y)=A\log y + C$.
Now, $0=I(b)= A\log b + C$, so that $C=-A\log b$.
Finally, we have $\displaystyle I(a)= A\log a + C = A\log a -A\log b= A \log \frac{a}{b}$, proving (*).

(Remark: in this proof, I made some implicit assumptions about $f$ and $a,b$, assumptions clearly satisfied by the explicit integral we had to solve here.)

 

[micromass]

$\displaystyle \int_0^\infty \frac{\text{atan}(2016x) - \text{atan}(1916x)}{x}dx$

I use the general result $\displaystyle \int_0^\infty \frac{f(ax)-f(bx)}{x} dx= (f(\infty) -f(0)) \log \frac{a}{b}$ (proven using Leibnitz rule, see below).

So $\displaystyle \int_0^\infty \frac{\text{atan}(2016x) - \text{atan}(1916x)}{x}dx =(\text{atan}( \infty ) - \text{atan}( 0)) \log \frac{2016}{1916} = \frac{\pi}{2} \log \frac{2016}{1916}$.

Now for $\displaystyle \int_0^\infty \frac{f(ax)-f(bx)}{x} dx= (f(\infty) -f(0)) \log \frac{a}{b}$ (*)
Set $\displaystyle I(y)= \int_0^\infty \frac{f(yx)-f(bx)}{x} dx$

Notice that $I(a)$ is the integral we are trying to determine, and $I(b)=0$.

Then $\displaystyle I'(y) = \int_0^\infty f'(yx) dx$. Let's do the substitution $z=yx$, so that $\displaystyle I'(y)= \frac{1}{y}\int_0^\infty f'(z) dz =\frac{1}{y} (f(\infty) -f(0))$. Let's name $A=f(\infty) -f(0)$.
We then have $I'(y)=A \frac{1}{y}$, a differential equation with solution $I(y)=A\log y + C$.
Now, $0=I(b)= A\log b + C$, so that $C=-A\log b$.
Finally, we have $\displaystyle I(a)= A\log a + C = A\log y -A\log b= A \frac{a}{b}$, proving (*).

Very nice!

 

[Mondayman]

Looks fun. I only just started integration, but I'm going to bookmark this for a future challenge!

 

[Ssnow]

I propose the solution for the 4. $\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin{x}}}{\sqrt{\sin{x}}+\sqrt{\cos{x}}}dx=\frac{\pi}{4}$

The integral is $\int_{0}^{\pi/2}\frac{1}{1+\sqrt{\cot{x}}}dx$, I start considering the substitution $\cot{x}=t^{2}$, so the indefinite integral is:

$\int \frac{-2t}{(1+t)(1+t^{4})}dt$

where $\cot{x}=t^{2}$ and $dx=-\frac{1}{1+t^{4}}2tdt$. I started with partial fraction:

$\frac{-2t}{(1+t)(1+t^{4})}= \frac{A}{1+t}+\frac{Bt^{3}+Ct^{2}+Dt+E}{1+t^{4}}$

obtaining $=\int \frac{1}{1+t}+\int \frac{-t^{3}+t^{2}-t-1}{1+t^{4}}$ that it is:

$\ln{(1+t)}-\frac{1}{4}\ln{(1+t^{4})} + \int \frac{t^{2}-t-1}{1+t^{4}}dt$

for the last integral I repeat the partial fraction:

$\frac{t^{2}-t-1}{1+t^{4}}=\frac{At+B}{t^{2}+t\sqrt{2}+1}+\frac{Ct+D}{t^{2}-t\sqrt{2}+1}$

and I find that $A=-\frac{1}{\sqrt{2}},B=-(1+\sqrt{2})/2\sqrt{2}, C=1/\sqrt{2},D=(1-\sqrt{2})/2\sqrt{2}$, after I must adjust the two in order to have the derivative in the numerator, so obtaining the logaritm from both:

$-\frac{1}{2\sqrt{2}}\ln{(t^{2}+t\sqrt{2}+1)}$ and from the second $+\frac{1}{2\sqrt{2}}\ln{(t^{2}-t\sqrt{2}+1)}$,

what remain are two integrals $\int \frac{-1/(2\sqrt{2})}{t^{2}+t\sqrt{2}+1}dt$ and $\int \frac{1/(2\sqrt{2})}{t^{2}-t\sqrt{2}+1}dt$, they gives two factors with sum $-\arctan{(1+\sqrt{2}t)}$

so substituting $t=\sqrt{\cot{x}}$ and valutating between $0$ and $\frac{\pi}{2}$ (taking the limit where there are logaritm with $\infty$ because the $\cot{0}=\infty$), I obtained the result $\pi/2-\arctan{1}$ that is $\frac{\pi}{4}$. It is correct?

 

[micromass]

I propose the solution for the 4. $\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin{x}}}{\sqrt{\sin{x}}+\sqrt{\cos{x}}}dx=\frac{\pi}{4}$

The integral is $\int_{0}^{\pi/2}\frac{1}{1+\sqrt{\cot{x}}}dx$, I start considering the substitution $\cot{x}=t^{2}$, so the indefinite integral is:

$\int \frac{-2t}{(1+t)(1+t^{4})}dt$

where $\cot{x}=t^{2}$ and $dx=-\frac{1}{1+t^{4}}2tdt$. I started with partial fraction:

$\frac{-2t}{(1+t)(1+t^{4})}= \frac{A}{1+t}+\frac{Bt^{3}+Ct^{2}+Dt+E}{1+t^{4}}$

obtaining $=\int \frac{1}{1+t}+\int \frac{-t^{3}+t^{2}-t-1}{1+t^{4}}$ that it is:

$\ln{(1+t)}-\frac{1}{4}\ln{(1+t^{4})} + \int \frac{t^{2}-t-1}{1+t^{4}}dt$

for the last integral I repeat the partial fraction:

$\frac{t^{2}-t-1}{1+t^{4}}=\frac{At+B}{t^{2}+t\sqrt{2}+1}+\frac{Ct+D}{t^{2}-t\sqrt{2}+1}$

and I find that $A=-\frac{1}{\sqrt{2}},B=-(1+\sqrt{2})/2\sqrt{2}, C=1/\sqrt{2},D=(1-\sqrt{2})/2\sqrt{2}$, after I must adjust the two in order to have the derivative in the numerator, so obtaining the logaritm from both:

$-\frac{1}{2\sqrt{2}}\ln{(t^{2}+t\sqrt{2}+1)}$ and from the second $+\frac{1}{2\sqrt{2}}\ln{(t^{2}-t\sqrt{2}+1)}$,

what remain are two integrals $\int \frac{-1/(2\sqrt{2})}{t^{2}+t\sqrt{2}+1}dt$ and $\int \frac{1/(2\sqrt{2})}{t^{2}-t\sqrt{2}+1}dt$, they gives two factors with sum $-\arctan{(1+\sqrt{2}t)}$

so substituting $t=\sqrt{\cot{x}}$ and valutating between $0$ and $\frac{\pi}{2}$ (taking the limit where there are logaritm with $\infty$ because the $\cot{0}=\infty$), I obtained the result $\pi/2-\arctan{1}$ that is $\frac{\pi}{4}$. It is correct?

That is the correct solution! Very well done!

An easier (but much trickier!) solution. Let $I = \int_0^{\pi/2}\frac{\sqrt{\sin{x}}}{\sqrt{\sin{x}}+\sqrt{\cos{x}}}dx$. Perform substitution $y = \pi/2 - x$, then
I = \int_0^{\pi/2} \frac{\sqrt{\cos{y}}}{\sqrt{\sin{y}}+\sqrt{\cos{y}}}dy
Hence
2I = \int_0^{\pi/2} \frac{\sqrt{\sin{x}}+\sqrt{\cos{x}}}{\sqrt{\sin{x}}+\sqrt{\cos{x}}}dx = \int_0^{\pi/2} dx = \frac{\pi}{2}
Thus, $I = \frac{\pi}{4}$.

 

[Ssnow]

I propose the solution of 1. $\int_{0}^{1}\frac{\ln{(1+x)}}{1+x^{2}}=\frac{\pi}{4}\ln{\sqrt{2}}$

I started substituting $x=\tan{t}$ so :

$\int_{0}^{\frac{\pi}{4}}\frac{\ln{(1+\tan{t})}}{1+\tan^{2}{t}}\frac{dt}{\cos^{2}{t}}$

that is by the fundamental relation:

$=\int_{0}^{\frac{\pi}{4}}\ln{(1+\tan{t})}dt$

Now $1+\tan{t}=\frac{\cos{t}+\sin{t}}{\cos{t}}=\frac{sin{\left(\frac{\pi}{2}-t\right)}+\sin{t}}{\cos{t}}=\frac{2\sin{\pi/4}\cos{\left(t-\frac{\pi}{4}\right)}}{\cos{t}}=\frac{\sqrt{2}\cos{\left(t-\frac{\pi}{4}\right)}}{\cos{t}}$

so

$=\int_{0}^{\frac{\pi}{4}}\ln{\sqrt{2}}dt+\int_{0}^{\frac{\pi}{4}}\ln{\cos{(t-\pi/4)}}dt-\int_{0}^{\pi/4}\ln{\cos{t}}dt$

The second setting $\frac{\pi}{4}-t=u$ is

$=\int_{0}^{\frac{\pi}{4}}\ln{\cos{u}}du$

so it cancel with the last and the integral is $=\int_{0}^{\frac{\pi}{4}}\ln{\sqrt{2}}dt=\frac{\pi}{4}\ln{\sqrt{2}}$

 

[micromass]

I propose the solution of 1. $\int_{0}^{1}\frac{\ln{(1+x)}}{1+x^{2}}=\frac{\pi}{4}\ln{\sqrt{2}}$

I started substituting $x=\tan{t}$ so :

$\int_{0}^{\frac{\pi}{4}}\frac{\ln{(1+\tan{t})}}{1+\tan^{2}{t}}\frac{dt}{\cos^{2}{t}}$

that is by the fundamental relation:

$=\int_{0}^{\frac{\pi}{4}}\ln{(1+\tan{t})}dt$

Now $1+\tan{t}=\frac{\cos{t}+\sin{t}}{\cos{t}}=\frac{sin{\left(\frac{\pi}{2}-t\right)}+\sin{t}}{\cos{t}}=\frac{2\sin{\pi/4}\cos{\left(t-\frac{\pi}{4}\right)}}{\cos{t}}=\frac{\sqrt{2}\cos{\left(t-\frac{\pi}{4}\right)}}{\cos{t}}$

so

$=\int_{0}^{\frac{\pi}{4}}\ln{\sqrt{2}}dt+\int_{0}^{\frac{\pi}{4}}\ln{\cos{(t-\pi/4)}}dt-\int_{0}^{\pi/4}\ln{\cos{t}}dt$

The second setting $\frac{\pi}{4}-t=u$ is

$=\int_{0}^{\frac{\pi}{4}}\ln{\cos{u}}du$

so it cancel with the last and the integral is $=\int_{0}^{\frac{\pi}{4}}\ln{\sqrt{2}}dt=\frac{\pi}{4}\ln{\sqrt{2}}$

Correct again! Amazing that you came up with the substitution $x = \tan{t}$. Very clever!

 

[ShayanJ]

Amazing that you came up with the substitution $x= \tan{t}$. Very clever!

So its not just me!

Great thread micromass. Actually its starting to become an art exhibition!

 

[micromass]

Half of my challenge problems are already solved. I did not expect this to happen so soon! The people who found these solutions are truly integral masters!

Anyway, I wish to make a little adertisement now for the book containing these problems. I got these problems (except 6 which is from Apostol) from the beautiful book "Inside interesting integrals" from Paul J. Nahin. So if you want to learn tricky solutions or are up for a challenge. This is the book for you!

https://www.amazon.com/dp/1493912763/?tag=pfamazon01-20

 

[Samy_A]

$\displaystyle \int_{\sqrt{2}}^\infty \frac{1}{x + x^{\sqrt{2}}}dx$

$f(x)=\frac{1}{x + x^{\sqrt{2}}}$
Doing a sort of partial fraction decomposition, $\displaystyle f(x)=\frac{1}{x}- \frac{x^{\sqrt{2}-2}}{1+x^{\sqrt{2}-1}}$

$\displaystyle \int f(x) dx =\int \frac{1}{x} dx -\int \frac{x^{\sqrt{2}-2}}{1+x^{\sqrt{2}-1}}dx$
⇒ $\displaystyle \int f(x) dx = \ln {x} - \frac{1}{\sqrt{2}-1} \int \frac{ d(x^{\sqrt{2}-1})}{1+x^{\sqrt{2}-1}} + C$
⇒ $\displaystyle \int f(x) dx = \ln {x} - (\sqrt{2}+1)\ln(1+x^{\sqrt{2}-1}) + C$
⇒ $\displaystyle \int f(x) dx =\ln \frac{x}{{(1+x^{\sqrt{2}-1}})^{\sqrt{2}+1}}+ C$

⇒ $\displaystyle \int_{\sqrt{2}}^\infty f(x) dx = -\ln \frac{\sqrt{2}}{{(1+{\sqrt 2}^{\sqrt{2}-1}})^{\sqrt{2}+1}} \approx 1.50633$

I used that $\displaystyle \lim_{x \rightarrow +\infty} \ln \frac{x}{{(1+x^{\sqrt{2}-1}})^{\sqrt{2}+1}}=0$ so that the term at $+ \infty$ vanishes.
Proof for that limit:
set $y=x^{\sqrt{2}-1}$
⇒ $y^{\sqrt{2}+1}=x^{(\sqrt{2}-1)(\sqrt{2}+1)}=x$
⇒ $\displaystyle \lim_{x \rightarrow +\infty} \frac{x}{{(1+x^{\sqrt{2}-1}})^{\sqrt{2}+1}}= \lim_{y \rightarrow +\infty} \frac{y^{\sqrt{2}+1}}{(1+y)^{\sqrt{2}+1}} = \lim_{y \rightarrow +\infty} (\frac{y}{y+1})^{\sqrt{2}+1}=1$
and hence the limit of the log of that nice expression will be 0.

 

[micromass]

$\displaystyle \int_{\sqrt{2}}^\infty \frac{1}{x + x^{\sqrt{2}}}dx$

$f(x)=\frac{1}{x + x^{\sqrt{2}}}$
Doing a sort of partial fraction decomposition, $\displaystyle f(x)=\frac{1}{x}- \frac{x^{\sqrt{2}-2}}{1+x^{\sqrt{2}-1}}$

$\displaystyle \int f(x) dx =\int \frac{1}{x} dx -\int \frac{x^{\sqrt{2}-2}}{1+x^{\sqrt{2}-1}}dx$
⇒ $\displaystyle \int f(x) dx = \ln {x} - \frac{1}{\sqrt{2}-1} \int \frac{ d(x^{\sqrt{2}-1})}{1+x^{\sqrt{2}-1}} + C$
⇒ $\displaystyle \int f(x) dx = \ln {x} - (\sqrt{2}+1)\ln(1+x^{\sqrt{2}-1}) + C$
⇒ $\displaystyle \int f(x) dx =\ln \frac{x}{{(1+x^{\sqrt{2}-1}})^{\sqrt{2}+1}}+ C$

⇒ $\displaystyle \int_{\sqrt{2}}^\infty f(x) dx = -\ln \frac{\sqrt{2}}{{(1+{\sqrt 2}^{\sqrt{2}-1}})^{\sqrt{2}+1}} \approx 1.50633$

I used that $\displaystyle \lim_{x \rightarrow +\infty} \ln \frac{x}{{(1+x^{\sqrt{2}-1}})^{\sqrt{2}+1}}=0$ so that the term at $+ \infty$ vanishes.
Proof for that limit:
set $y=x^{\sqrt{2}-1}$
⇒ $y^{\sqrt{2}+1}=x^{(\sqrt{2}-1)(\sqrt{2}+1)}=x$
⇒ $\displaystyle \lim_{x \rightarrow +\infty} \frac{x}{{(1+x^{\sqrt{2}-1}})^{\sqrt{2}+1}}= \lim_{y \rightarrow +\infty} \frac{y^{\sqrt{2}+1}}{(1+y)^{\sqrt{2}+1}} = \lim_{y \rightarrow +\infty} (\frac{y}{y+1})^{\sqrt{2}+1}=1$
and hence the limit of the log of that nice expression will be 0.

Very nice solution!

 

[vela]

\begin{align*}
\int_0^\infty \frac{\sin^9 x}{x}\,dx &= \frac 12 \int_{-\infty}^\infty \frac{\sin^9 x}{x}\,dx \\
&= \frac 12 \frac{1}{(2i)^8} \int_{-\infty}^\infty \frac{1}{2i}\frac{(e^{ix}-e^{-ix})^9}{x}\,dx \\
&= \frac {1}{512} \int_{-\infty}^\infty \frac{1}{2i}\left(\frac{e^{i9x}-9 e^{i7x} + 36 e^{i5x} - 84 e^{i3x} + 126 e^{ix} - 126 e^{-ix} + 84e^{-i3x} -36 e^{-i5x} + 9 e^{-i7x} - e^{-i9x}}{x}\right)\,dx \\
&= \frac {1}{512} \int_{-\infty}^\infty \left(\frac{\sin 9x}{x} - 9 \frac{\sin 7x}{x} + 36 \frac{\sin 5x}{x} - 84 \frac{\sin 3x}{x} + 126\frac{\sin x}{x}\right)\,dx \\
&= \frac{1}{512} (1-9+36-84+126)\int_{-\infty}^\infty \frac{\sin u}{u}\,du \\
&= \frac{35}{256}\pi
\end{align*}

 

[micromass]

\begin{align*}
\int_0^\infty \frac{\sin^9 x}{x}\,dx &= \frac 12 \int_{-\infty}^\infty \frac{\sin^9 x}{x}\,dx \\
&= \frac 12 \frac{1}{(2i)^8} \int_{-\infty}^\infty \frac{1}{2i}\frac{(e^{ix}-e^{-ix})^9}{x}\,dx \\
&= \frac {1}{512} \int_{-\infty}^\infty \frac{1}{2i}\left(\frac{e^{i9x}-9 e^{i7x} + 36 e^{i5x} - 84 e^{i3x} + 126 e^{ix} - 126 e^{-ix} + 84e^{-i3x} -36 e^{-i5x} + 9 e^{-i7x} - e^{-i9x}}{x}\right)\,dx \\
&= \frac {1}{512} \int_{-\infty}^\infty \left(\frac{\sin 9x}{x} - 9 \frac{\sin 7x}{x} + 36 \frac{\sin 5x}{x} - 84 \frac{\sin 3x}{x} + 126\frac{\sin x}{x}\right)\,dx \\
&= \frac{1}{512} (1-9+36-84+126)\int_{-\infty}^\infty \frac{\sin u}{u}\,du \\
&= \frac{35}{256}\pi
\end{align*}

Very cool solution! Shows how complex numbers make life a lot easier!

 

[ShayanJ]

Very cool solution! Shows how complex numbers make life a lot easier!

Does that mean you had a more complicated method in mind?
Although vela's method is indeed cool, I'd like to see other methods too.

 

[micromass]

Does that mean you had a more complicated method in mind?
Although vela's method is indeed cool, I'd like to see other methods too.

No, this is the method I had in mind. If you wish to find a solution without complex numbers, then you'll need to find another way to prove
\sin^9x = \sin(9x) - 9\sin(7x) + 36\sin(5x) - 84\sin(3x) + 126\sin(x)
It is not very difficult to imagine how to prove or find this formula without complex numbers, you just use the addition formula on $\sin(nx)$ a few times. But using the complex exponentials seems to be a lot more elegant!

 

[Samy_A]

\begin{align*}
\int_0^\infty \frac{\sin^9 x}{x}\,dx &= \frac 12 \int_{-\infty}^\infty \frac{\sin^9 x}{x}\,dx \\
&= \frac 12 \frac{1}{(2i)^8} \int_{-\infty}^\infty \frac{1}{2i}\frac{(e^{ix}-e^{-ix})^9}{x}\,dx \\
&= \frac {1}{512} \int_{-\infty}^\infty \frac{1}{2i}\left(\frac{e^{i9x}-9 e^{i7x} + 36 e^{i5x} - 84 e^{i3x} + 126 e^{ix} - 126 e^{-ix} + 84e^{-i3x} -36 e^{-i5x} + 9 e^{-i7x} - e^{-i9x}}{x}\right)\,dx \\
&= \frac {1}{512} \int_{-\infty}^\infty \left(\frac{\sin 9x}{x} - 9 \frac{\sin 7x}{x} + 36 \frac{\sin 5x}{x} - 84 \frac{\sin 3x}{x} + 126\frac{\sin x}{x}\right)\,dx \\
&= \frac{1}{512} (1-9+36-84+126)\int_{-\infty}^\infty \frac{\sin u}{u}\,du \\
&= \frac{35}{256}\pi
\end{align*}

Very cool indeed.

I gave this one a look, and my "never failing intuition" told me that it will probably be solvable with contour integration. It may well be, but this is much nicer.

 

[vela]

I gave this one a look, and my "never failing intuition" told me that it will probably be solvable with contour integration. It may well be, but this is much nicer.

Contour integration was my first thought too, which is why my first step was to change the limits of the integral. I guess I did assume that the answer to the last integral is known, but that could be evaluated using contour integration.

 

[Ssnow]

I gave this one a look, and my "never failing intuition" told me that it will probably be solvable with contour integration. It may well be, but this is much nicer.

I thought some special function as $Si(z)=\int_{0}^{z}\frac{\sin{t}}{t}dt$ but yes always pass by the complex street ...

 

[Ssnow]

Thinking your trick about previous integral I want propose the solution of 7.

Let $I=\int_{0}^{\frac{\pi}{2}}\arccos{\left(\frac{\cos{x}}{1+2\cos{x}}\right)}dx$

We observe that the function $\frac{\cos{x}}{1+2\cos{x}}$ is even so $2I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\arccos{\left(\frac{\cos{x}}{1+2\cos{x}}\right)}dx$

Now changing $x\rightarrow -y$ we have

$2I=\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}}-\arccos{\frac{\cos{y}}{1+2\cos{y}}}dy$

and by the identity $\arccos{(-x)}=\pi-\arccos{(x)}$ we have that:

$2I=\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}}\arccos{\frac{\cos{y}}{1+2\cos{y}}}-\pi dy=-2I+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\pi dy$

so $4I=\frac{\pi^{2}}{2}+\frac{\pi^{2}}{2}$ then $I=\frac{\pi^{2}}{4}$.

 

[ShayanJ]

Thinking your trick about previous integral I want propose the solution of 7.

Let $I=\int_{0}^{\frac{\pi}{2}}\arccos{\left(\frac{\cos{x}}{1+2\cos{x}}\right)}dx$

We observe that the function $\frac{\cos{x}}{1+2\cos{x}}$ is even so $2I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\arccos{\left(\frac{\cos{x}}{1+2\cos{x}}\right)}dx$

Now changing $x\rightarrow -y$ we have

$2I=\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}}-\arccos{\frac{\cos{y}}{1+2\cos{y}}}dy$

and by the identity $\arccos{(-x)}=\pi-\arccos{(x)}$ we have that:

$2I=\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}}\arccos{\frac{\cos{y}}{1+2\cos{y}}}-\pi dy=-2I+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\pi dy$

so $4I=\frac{\pi^{2}}{2}+\frac{\pi^{2}}{2}$ then $I=\frac{\pi^{2}}{4}$.

When you use $\arccos(-x)=\pi-\arccos(x)$, shouldn't you get $\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} (\arccos{(\frac{-\cos{y}}{1+2\cos{y}}})-\pi ) dy$? What did you do with that minus sign?

 

[micromass]

Thinking your trick about previous integral I want propose the solution of 7.

Let $I=\int_{0}^{\frac{\pi}{2}}\arccos{\left(\frac{\cos{x}}{1+2\cos{x}}\right)}dx$

We observe that the function $\frac{\cos{x}}{1+2\cos{x}}$ is even so $2I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\arccos{\left(\frac{\cos{x}}{1+2\cos{x}}\right)}dx$

Now changing $x\rightarrow -y$ we have

$2I=\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}}-\arccos{\frac{\cos{y}}{1+2\cos{y}}}dy$

and by the identity $\arccos{(-x)}=\pi-\arccos{(x)}$ we have that:

$2I=\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}}\arccos{\frac{\cos{y}}{1+2\cos{y}}}-\pi dy=-2I+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\pi dy$

so $4I=\frac{\pi^{2}}{2}+\frac{\pi^{2}}{2}$ then $I=\frac{\pi^{2}}{4}$.

Sorry, this is not the correct answer. Number 7 is definitely the grand prize of this thread, it is very difficult!

 

[Samy_A]

$\int_0^1 \frac{1-4x^5}{(x^5 - x + 1)^2}dx$

The derivative of quotient of polynomials, $\frac{P}{Q}$, is $\frac{QP'-PQ'}{Q²}$. The degree of the numerator being (in general) deg(P)+deg(Q)-1.
In our case, deg(Q)=5, so deg(P) should be 1, as deg(P)+deg(Q)-1=5.
That leads to an ansatz:
$\frac{1-4x^5}{(x^5 - x + 1)^2} =\frac{d}{dx} \frac{Ax+B}{x^5 - x + 1}$

Now, $\frac{d}{dx} \frac{Ax+B}{x^5 - x + 1}= \frac{(x^5 - x + 1)A-(Ax+b)(5x^4-1)}{(x^5 - x + 1)^2}=\frac{Ax^5-Ax+A-5Ax^5+Ax-5Bx^4+B}{(x^5 - x + 1)^2}$
As we have no term in $x^4$ in the numerator, $B=0$.
Hence $\frac{d}{dx} \frac{Ax+B}{x^5 - x + 1}=\frac{Ax^5-Ax+A-5Ax^5+Ax}{(x^5 - x + 1)^2}=\frac{Ax^5+A-5Ax^5}{(x^5 - x + 1)^2}$
And now a miracle occurs, A must be one, and sure enough, the numerator becomes $1-4x^5$.

So, $\displaystyle \int_0^1 \frac{1-4x^5}{(x^5 - x + 1)^2}dx = \left. \frac{x}{x^5 - x + 1} \right|_0^1 =1$

(Latex fixed ...)

 

[micromass]

$\int_0^1 \frac{1-4x^5}{(x^5 - x + 1)^2}dx$

The derivative of quotient of polynomials, $\frac{P}{Q}$, is $\frac{QP'-PQ'}{Q²}$. The degree of the numerator being (in general) deg(P)+deg(Q)-1.
In our case, deg(Q)=5, so deg(P) should be 1, as deg(P)+deg(Q)-1=5.
That leads to an ansatz:
$\frac{1-4x^5}{(x^5 - x + 1)^2} =\frac{d}{dx} \frac{Ax+B}{x^5 - x + 1}$

Now, $\frac{d}{dx} \frac{Ax+B}{x^5 - x + 1}= \frac{(x^5 - x + 1)A-(Ax+b)(5x^4-1)}{(x^5 - x + 1)^2}=\frac{Ax^5-Ax+A-5Ax^5+Ax-5Bx^4+B}{(x^5 - x + 1)^2}$
As we have no term in $x^4$ in the numerator, $B=0$.
Hence $\frac{d}{dx} \frac{Ax+B}{x^5 - x + 1}=\frac{Ax^5-Ax+A-5Ax^5+Ax}{(x^5 - x + 1)^2}=\frac{Ax^5+A-5Ax^5}{(x^5 - x + 1)^2}$
And now a miracle occurs, A must be one, and sure enough, the numerator becomes $1-4x^5$.

So, $\displaystyle \int_0^1 \frac{1-4x^5}{(x^5 - x + 1)^2}dx = \left. \frac{x}{x^5 - x + 1} \right|_0^1 =1$

(Latex fixed ...)

Perfect!

 

[Ssnow]

When you use arccos(−x)=π−arccos(x) \arccos(-x)=\pi-\arccos(x) , shouldn't you get ∫−π2π2(arccos(−cosy1+2cosy)−π)dy \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} (\arccos{(\frac{-\cos{y}}{1+2\cos{y}}})-\pi ) dy ? What did you do with that minus sign?

yes, thank you now I understand the error. I think the strategy is another ...

 

[fresh_42]

#9: \int_0^\infty \frac{x^{1916}}{x^{2016} + 1}dx

(I'm not sure if it is a valid solution since I looked up the Γ-function and its formulas ... after despaired on 2016th roots of unity and many ugly looking sheets of paper ...)

For convenience I take $n=2016$ and $m=1916$.
The disturbing denominator can be transformed into an integral:

$$\int_0^\infty e^{(-x^n -1) y} dy = [- \frac{1}{x^n + 1} * e^{(-x^n -1) y } ]_{ y=0 }^{ y=\infty } = \frac{1}{x^n + 1}$$
Thus
$$\int_0^\infty \frac{x^{1916}}{x^{2016} + 1}dx = \int_0^\infty x^{m} \int_0^\infty e^{-x^n y} e^{-y} dy dx$$ and with Fubini's theorem
$$\int_0^\infty \frac{x^{1916}}{x^{2016} + 1}dx = \int_0^\infty\int_0^\infty x^m e^{-x^n y} e^{-y} dx dy$$.

Substitution $z = x^n y$ with $dx = \frac{1}{ny} x^{1-n} dz$ yields
$$\int_0^\infty \frac{x^{1916}}{x^{2016} + 1}dx = \frac{1}{n} \int_0^\infty e^{-y} (\frac{1}{y})^\frac{m+1}{n} dy \int_0^\infty e^{-z} z^\frac{m-n+1}{n} dz = \frac{1}{n} Γ(\frac{n-m-1}{n}) Γ(\frac{m+1}{n}) $$
$$\int_0^\infty \frac{x^{1916}}{x^{2016} + 1}dx = \frac{1}{n} Γ(1-\frac{m+1}{n}) Γ(\frac{m+1}{n}) = \frac{1}{2016} π \frac{1}{\sin{\frac{1917 π}{2016}}} ≈ 0.0101411901 $$