How to Integrate
$$ I = \int_{0}^{\infty} \frac{\tan^{-1}x^2}{x^2(x^4-1)}-\frac{\pi}{4(x^4-1)} \>dx\approx -0.295512 $$
Mathematica returns a result that does not match numerically with the integral approximation :
$$ \frac{i C}{4} -\frac{\pi}{4}\sqrt{4-3i} +\frac{3 \pi^2}{32} +\pi\left(\frac{1}{4}-\frac{i}{8}\right)\coth^{-1}(\sqrt{2}) \approx -0.048576 + 0.43823\,i $$
Where C denotes Catalan's Constant
Motivation
I was able to find a closed form for
$$ \int_{0}^{1} \frac{\tan^{-1}(x^2)}{x^2(x^4-1)}-\frac{\pi}{4(x^4-1)}dx $$
using double infinite sums. Upon plotting the function within the integral i saw that it could be integrated from $0$ to ${\infty}$ .
Attempts
Number 1
I tried to Split the integral as
$$ \int_{0}^{1} \frac{\tan^{-1}(x^2)}{x^2(x^4-1)}-\frac{\pi}{4(x^4-1)}dx + \int_{1}^{\infty} \frac{\tan^{-1}(x^2)}{x^2(x^4-1)}-\frac{\pi}{4(x^4-1)}dx $$
and use the Taylor Series for $\tan^{-1}(x^2) $ when $|x| >1$
Number 2
I tried using partial fractions as
$$ \frac{1}{x^4-1} = \frac{1}{4(x-1)} – \frac{1}{4(x+1)} -\frac{1}{2(x^2+1)} $$
$$ \frac{1}{x^2(x^4-1)} = \frac{1}{2(x^2+1)}-\frac{1}{x^2}-\frac{1}{4(x+1)} + \frac{1}{4(x-1)} $$
from which I obtained
$$\int_{0}^{\infty} \frac{\tan^{-1}(x^2)}{2(x^2+1)}dx = \frac{\pi^2}{16} $$
$$ \int_{0}^{\infty} \frac{\pi}{8(x^2+1)} dx= \frac{\pi^2}{8} $$
but was unable to proceed further.
Number 3
A number of basic integration techniques such as U-Sub and Integration by parts.
Number 4
Using the same technique i used to evaluate the same integral but from $0$ to $1$
I will continue to try , but for now I find myself to be stuck.
Q – Is there a closed form for I? If the solution is easy and i am missing something , could you provide hints instead?
Thank you for your help and time.
Best Answer
Note $$\frac1{x^4-1} =\frac1{2(x^2-1)} - \frac1{2(x^2+1)}\\ \frac1{x^2(x^4-1)} =\frac1{2(x^2-1)} + \frac1{2(x^2+1)}-\frac1{x^2} $$ and rewrite the integral as
\begin{align} I &= \int_{0}^{\infty} \left( \frac{\tan^{-1}x^2}{x^2(x^4-1)}-\frac{\pi}{4(x^4-1)} \right)dx\\ &=\frac{\pi^2}{16}+ \frac12\int_0^{\infty} \frac{\tan^{-1}x^2}{1+x^2}dx - \int_0^{\infty} \frac{\tan^{-1}x^2}{x^2}dx + \frac12\int_0^{\infty} \frac{\tan^{-1}x^2-\frac\pi4}{x^2-1}dx\tag1 \end{align} where $\int_0^{\infty} \frac{\tan^{-1}x^2}{1+x^2}dx=\frac{\pi^2}8$, $ \int_0^{\infty} \frac{\tan^{-1}x^2}{x^2}dx=\frac{\pi}{\sqrt2}$ and
\begin{align} &\frac12\int_0^{\infty} \frac{\tan^{-1}x^2-\frac\pi4}{x^2-1}dx =\int_0^{1} \frac{\tan^{-1}x^2-\frac\pi4}{x^2-1}dx\\ \overset{IBP}=&-\int_0^1 \frac x{1+x^4} \ln \frac{1-x}{1+x}dx \overset{x\to \frac{1-x}{1+x}}=-\int_0^\infty\frac {\ln x}{1+6x^2+ x^4}dx\\ =&-\frac1{4\sqrt2}\left( \int_0^\infty\frac{\ln x}{x^2+ (\sqrt2-1)^2}dx -\int_0^\infty\frac{\ln x}{x^2+ (\sqrt2+1)^2}dx \right)\\ =&-\frac1{8\sqrt2}\left(\frac{\ln(\sqrt2-1)}{\sqrt2-1} - \frac{\ln(\sqrt2+1)}{\sqrt2+1} \right) \end{align} Substitute above results into (1) to obtain
$$I= \frac{\pi^2}8-\frac\pi{\sqrt2}\left(1+\frac{\ln(\sqrt2-1)}{8(\sqrt2-1)} - \frac{\ln(\sqrt2+1)}{8(\sqrt2+1)} \right) $$