How would we integrate this without using double integrals?

[Hiero]

Homework Statement

[/B]
Evaluate $\int_0^∞ \frac{\tan^{-1}(\pi x) - \tan^{-1}x}{x}dx$

2. Relevant information
This problem comes after a chapter on “multiple integrals” and so, in this context, I realized we could rewrite the single integral as a double integral:
$$ \int_0^∞ \int_1^{\pi} \frac{dydx}{1+(xy)^2} $$
Then we can reverse the boundaries (make it dxdy) and we nicely get an answer of π*ln(π)/2

The Attempt at a Solution

That is a truly clever solution... but I’m pretty sure if it were not for the context of this problem (being at the end of a chapter on “multiple integrals”) I am 99% sure I would not have thought of that beautiful approach.

If I came across that integral “in the wild” I would try to solve it directly, so I thought I would try that and see how it works out... spoiler: it didn’t work out.

My first attempt deals with the first and second term separately, but we soon realize that this leads to infinity minus infinity.

My next idea was to express the integrand as an infinite polynomial, then integrate term by term.

If we use the fact that $\tan^{-1}(kx) = \int_0^x \frac{kdt}{1+(kt)^2}$ and the fact that $\frac{1}{1-z} = \Sigma_{n=0}^∞ [z^n]$ then we get: $\tan^{-1}(kx) = \Sigma_{n=0}^∞[(-1)^n\frac{x^{2n+1}}{2n+1}k^{2n+1}]$ and we therefore get:

$$ \frac{\tan^{-1}(\pi x) - \tan^{-1}x}{x} = \Sigma_{n=0}^∞[(-1)^n\frac{x^{2n}}{2n+1}(\pi^{2n+1}-1)]$$

Now we can integrate term by term to express the original integral as an infinite series:

$$\int \frac{\tan^{-1}(\pi x) - \tan^{-1}x}{x}dx = C + \Sigma_{n=0}^∞[(-1)^n\frac{x^{2n+1}}{(2n+1)^2}(\pi^{2n+1}-1)]$$

At the lower bound, x=0, all of the terms in the sum are zero (except of course for C, the constant of integration).

Now for the upper bound, we evaluate at (say) t and take the limit as t → ∞. But the thing about this sum is that it is divergent for x > 1/π ... yet this sum represents the integral of both terms...

At this point (if I came across this problem “in the wild”) I would conclude that the integral truly is divergent... but our beautiful solution using double integrals (and the book) says otherwise.

What’s going on here? Why am I getting a divergent answer?Much thanks to those who take the time to follow all this!

 

[Charles Link]

One comment, and I think I have this part correct: When you use $\frac{1}{1-z}=1+z^2+z^3+...$, you do this step without taking into account that later on, you allow $|z|$ to be greater than $1$ in the integral that you are performing. This takes it outside the region where the series form is valid, and thereby, this step would be invalid. Notice even if $|z|>1$, the left side of this equation still gives a valid number, but the series does not converge. $\\$ In any case, an interesting integral that you presented. This is the first time that I have seen this one.

 

[Hiero]

That’s a great point @Charles Link thank you for clarifying that.

So then we want |kt| < 1 from
$$\tan^{-1}(kx) = \int_0^x \frac{kdt}{1+(kt)^2} = k\int_0^x \Sigma_{n=0}^∞[(-k^2t^2)^n]dt $$ $$\tan^{-1}(kx)= \Sigma_{n=0}^∞[(-1)^n\frac{(kx)^{2n+1}}{2n+1}]$$
Which then implies |x| < 1/k
[EDIT: we can actually use the stronger condition |x| ≤ 1/k]

So even though the domain of arctan(kx) is all ℝ, the “Maclaurin series” for arctan(kx) only applies for |x| ≤ 1/k

That’s a bit of a silly mistake! Thanks for pointing that out.

So that’s the end of that approach. If anyone can solve it a new way please do! But I don’t think the arctan(x)/x has any elementary anti-derivative so I’m all out of ideas.

 

[fresh_42]

So that’s the end of that approach. If anyone can solve it a new way please do! But I don’t think the arctan(x)/x has any elementary anti-derivative so I’m all out of ideas.

No, it hasn't. It doesn't even converge. I first tried to split the quotient, but you need both in one integral. The first thing which I thought about was the Weierstraß substitution. But except some funny expressions that went nowhere - at least for me.

 

[Ray Vickson]

Homework Statement

[/B]
Evaluate $\int_0^∞ \frac{\tan^{-1}(\pi x) - \tan^{-1}x}{x}dx$

2. Relevant information
This problem comes after a chapter on “multiple integrals” and so, in this context, I realized we could rewrite the single integral as a double integral:
$$ \int_0^∞ \int_1^{\pi} \frac{dydx}{1+(xy)^2} $$
Then we can reverse the boundaries (make it dxdy) and we nicely get an answer of π*ln(π)/2

The Attempt at a Solution

That is a truly clever solution... but I’m pretty sure if it were not for the context of this problem (being at the end of a chapter on “multiple integrals”) I am 99% sure I would not have thought of that beautiful approach.

If I came across that integral “in the wild” I would try to solve it directly, so I thought I would try that and see how it works out... spoiler: it didn’t work out.

Much thanks to those who take the time to follow all this!

When I submit the problem
$$J = \int_0^\infty \frac{\arctan(\pi x) - \arctan(x)}{x} \, dx$$
to Maple ("in the wild"), it immediately comes back with the answer $J = (1/2) \pi \ln(\pi).$ However, without additional probing, Maple does not automatically explain the steps it took. To remedy this we can use the "infolevel" command to get more information:

> restart;

> infolevel[int]:=3;

infolevel[int] := 3

> f:=(arctan(Pi*x)-arctan(x))/x;

arctan(Pi x) - arctan(x)
f := ------------------------
x

> J:=int(f,x=0..infinity);
Definite Integration: Integrating expression on x=0..infinity
Definite Integration: Using the integrators [distribution, piecewise, series, o, polynomial, ln, lookup, cook, ratpoly, elliptic, elliptictrig, meijergspecial, improper, asymptotic, ftoc, ftocms, meijerg, contour]
LookUp Integrator: unable to find the specified integral in the table
int/elliptic: trying elliptic integration
int/indef1: first-stage indefinite integration
int/indef2: second-stage indefinite integration
int/indef2: trying integration by parts
int/indef1: first-stage indefinite integration
int/indef1: first-stage indefinite integration
int/indef1: first-stage indefinite integration
int/indef2: second-stage indefinite integration
int/indef2: trying integration by parts
int/indef1: first-stage indefinite integration
int/indef1: first-stage indefinite integration
int/indef2: second-stage indefinite integration
int/ln: case of integrand containing ln
int/hermiter/horowitzLog: integrating

ln(_X)
-------
2
_X + 1

int/hermiter/horowitzLog: Horowitz' method yields

/
| ln(_X)
| ------- d_X
| 2
/ _X + 1

int/hermiter/horowitzLog: result is

-1/2 I ln(_X) ln(1 + _X I) + 1/2 I ln(_X) ln(1 - _X I)

- 1/2 I dilog(1 + _X I) + 1/2 I dilog(1 - _X I)

int/indef1: first-stage indefinite integration
int/indef2: second-stage indefinite integration
int/indef2: applying change of variables
int/indef1: first-stage indefinite integration
int/indef1: first-stage indefinite integration
int/indef1: first-stage indefinite integration
int/indef1: first-stage indefinite integration
int/indef1: first-stage indefinite integration
int/indef1: first-stage indefinite integration
int/indef1: first-stage indefinite integration
int/indef1: first-stage indefinite integration
int/indef1: first-stage indefinite integration
int/indef1: first-stage indefinite integration
int/indef1: first-stage indefinite integration
int/indef1: first-stage indefinite integration
int/indef1: first-stage indefinite integration
int/indef1: first-stage indefinite integration
Definite Integration: Method ftoc succeeded.
Definite Integration: Finished sucessfully. J := 1/2 ln(Pi) Pi

So: Maple transformed the problem using a Hermite/Horowitz method for the integral over a finite interval $[0,N],$ then took the limit as the $N \to \infty.$

 

[Hiero]

So: Maple transformed the problem using a Hermite/Horowitz method for the integral over a finite interval $[0,N],$ then took the limit as the $N \to \infty.$

Thanks for taking the time! I haven’t heard of this Hermite/Horowitz method before. Is it a numerical method? If so, how does it recognize the precise answer? If it’s not a numerical method, can it be done by hand? Or is it some kind of intricate algorithm?

Of course I tried googling it but what I found seemed to imply that it’s an algorithm for dealing with a ratio of polynomials?

I lack numerical experience so using software didn’t even cross my mind, but that definitely is the “wild” approach!

 

[Ray Vickson]

Thanks for taking the time! I haven’t heard of this Hermite/Horowitz method before. Is it a numerical method? If so, how does it recognize the precise answer? If it’s not a numerical method, can it be done by hand? Or is it some kind of intricate algorithm?

Of course I tried googling it but what I found seemed to imply that it’s an algorithm for dealing with a ratio of polynomials?

I lack numerical experience so using software didn’t even cross my mind, but that definitely is the “wild” approach!

No, it is not numerical. The Horowitz method was discovered in the 1960s, so will not be a method found in a typical calculus book (most of which present methods discovered about 200 years ago); it is implemented in computer algebra systems and falls under the topic of "symbolic integration". The indefinite integral is provably "non-elementary", meaning that it cannot be expressed as a finite number of terms involving only the standard functions. Basically, we need to invent some new functions to do some integrations, such as "erf(x)" to do the integral $\int \exp(-x^2/2) \, dx.$ Many non-elementary functions have known values at some isolated points (such as $\text{erf}(0) =0$ and $\text{erf}(\infty) = 1$) but other values need to be computed using some type of numerical algorithm. (In that respect, they are really not much different from the standard functions such as $\sin(x)$ or $e^x:$ we need to use numerical algorithms to compute these, also!)

Anyway, the indefinite integral in this problem has the form
$$\int \frac{\arctan(\pi x) - \arctan)x)}{x} \, dx = F(x) $$ where
F(x) = -ln(x)*arctan(x)-1/2*I*ln(x)*ln(1+I*x)+1/2*I*ln(x)*ln(1-I*x)-1/2*I*dilog(1+I*x)+1/2*I*dilog(1-I*x)+ln(Pi*x)*arctan(Pi*x)+1/2*I*ln(Pi*x)*ln(1+I*Pi*x)-1/2*I*ln(Pi*x)*ln(1-I*Pi*x)+1/2*I*dilog(1+I*Pi*x)-1/2*I*dilog(1-I*Pi*x)

Here, "I" is the Maple symbol for the imaginary unit $\sqrt{-1}$ and $\text{dilog}$ is the non-elementary function defined as
$$\text{dilog}(w) = \int_1^w \frac{\ln t}{1-t} \, dt.$$ The point is that lots of properties of the dilog function have been discovered over the years; in particular, asymptotic expansions are known for large arguments, thus allowing us to to compute things like $\lim_{x \to \infty} F(x) = (1/2) \pi \ln (\pi).$

 

[Charles Link]

Thank you @Ray Vickson . I am following this thread as well, and your input is extremely useful.

 

[Hiero]

Very nice exposition, thank you!

Here, "I" is the Maple symbol for the imaginary unit $\sqrt{-1}$ and $\text{dilog}$ is the non-elementary function defined as
$$\text{dilog}(w) = \int_1^w \frac{\ln t}{1-t} \, dt.$$ The point is that lots of properties of the dilog function have been discovered over the years; in particular, asymptotic expansions are known for large arguments, thus allowing us to to compute things like $\lim_{x \to \infty} F(x) = (1/2) \pi \ln (\pi).$

Hey, I’ve tried to integrate that before!

On the one hand, (this is from the same problem set as the OP)
$$\int_0^1\int_0^1\frac{dydx}{1-xy} = \int_0^1\int_0^1 \Sigma_{n=0}^∞(xy)^ndydx = \Sigma_{n=0}^∞(n+1)^{-2} = \Sigma_{n=1}^∞n^{-2}$$
(This is the “Basel problem,” and this time the generator z = xy is okay!)
That’s all the book said, but when you try to integrate you get:
$$\int_0^1\int_0^1\frac{dydx}{1-xy} = \int_0^1[\frac{-1}{x}\ln (1-xy)\big|_{y=0}^1dx=\int_0^1\frac{-\ln (1-x)}{x}dx =\int_1^0\frac{\ln t}{1-t}dt=\text{dilog}(0)$$

So from your definition I conclude that dilog(0) = π²/6 (the famous Basel solution).

But other sources say dilog(0) = 0?
(Maybe you meant to make the lower bound zero?)
EDIT: Actually they say Li₂(0) = 0 ... apparently dilog(x) = Li₂(1-x)

 

[Charles Link]

I think this google explains what may be two different definitions: https://www.mathworks.com/help/symbolic/dilog.html

 

[Fred Wright]

I use the trick of differentiation under the integral:$$I_a=\frac {d}{da}\int_0^\infty \frac{tan^{-1}(ax)dx}{x}=\int_0^\infty \frac{dx}{a^2(\frac {1}{a^2}+x^2)}=\frac{1}{a^2}\int_0^\infty \frac{dx}{(x+\frac{i}{a})(x-\frac{i}{a})}$$I extend $I_a$ to the complex plane, observe that the integral is even, enclose the pole at $\frac{i}{a}$by an infinite semi-circle contour on the upper half plane, and use the residue theorem to find: $$I_a =\frac{\pi}{2a}$$I perform the indefinite integration with respect to a:$$\int_0^\infty \frac{tan^{-1}(ax)dx}{x}=\frac {\pi}{2}log(a)+ C$$
To find the integration constant C, I set $a=1$:$$\int_0^{\infty} \frac{tan^{-1}(x)dx}{x}=\frac {\pi}{2}log(1)+ C$$
and setting $a=\pi$ I find:$$\int_0^\infty \frac{tan^{-1}(\pi x)dx}{x}-\int_0^{\infty} \frac{tan^{-1}(x)dx}{x}=\frac{\pi}{2}log(\pi)$$