Struggling to figure out why $\int\limits_{0}^{1}\frac{\ln \left(x^{2}\right)}{\left(1+x^{2}\right)^{2}}dx = -G-\frac{\pi}{4}$.

Question

Struggling to figure out why $$\int\limits_{0}^{1}\frac{\ln \left(x^{2}\right)}{\left(1+x^{2}\right)^{2}}dx = -G-\frac{\pi}{4}$$

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I've taken note of the exponent in the logarithm, tried substituting $x=\tan\theta$, and even differentiating under the integral sign with the function $J(a)=\int\limits_{0}^{1}\frac{\ln\left(ax\right)}{\left(1+x^{2}\right)^{2}}$, but all to no avail.

The trigonometric substitution leads me to $$\int\limits_{0}^{\frac{\pi}{4}}\ln\left(\tan^{2}\theta\right)\cos^{2}\theta d\theta$$ which is verifiably equivalent to the original expression, but it also looks ugly and I don't know where to go from here (Surely not integration by parts, since then it either gets very ugly or births a divergent integral).

I haven't tried complex analysis but since the bounds are 0 to 1 I doubt that would work.

I know that $-G=\int\limits_{0}^{1}\frac{\ln x}{1+x^{2}}dx$ but I can't tell how to transform my expressions into this one. There must be some sneaky but potent substitution that I'm missing.

Judging by the Catalan's constant in the solution, I assume there's also some series, presumably for the logarithm, that could be used. Could anyone share some insights, please?

This is my first time using this forum, apologies in advance if I did anything wrong.

Answer 1

It seems to me the simplest way is to use trigonometric substitution and IBP. In fact, by IBP, \begin{eqnarray} &&\int_{0}^{\frac{\pi}{4}}\ln\left(\tan^{2}\theta\right)\cos^{2}\theta d\theta\\ &=&\frac12\int_{0}^{\frac{\pi}{4}}\ln\left(\tan\theta\right)d(2\theta+\sin(2\theta))\\ &=&\frac12\ln\left(\tan\theta\right)(2\theta+\sin(2\theta))\bigg|_{0}^{\frac{\pi}{4}}-\frac12\int_{0}^{\frac{\pi}{4}}(2\theta+\sin(2\theta))\cot\theta\sec^2\theta d\theta\\ &=&-\frac12\int_{0}^{\frac{\pi}{4}}\left(\frac{2\theta}{\sin\theta\cos\theta}+2\right) d\theta\\ &=&-\frac\pi4-\frac12\int_{0}^{\frac{\pi}{2}}\frac{\theta}{\sin\theta}d\theta\\ &=&-\frac\pi4-G. \end{eqnarray}

Answer 2

\begin{align}\int_{0}^{1}\frac{\ln \left(x^{2}\right)}{\left(1+x^{2}\right)^{2}}dx =\int_{0}^{1}\frac{\ln x}{1+x^{2}} +\frac{(1-x^2)\ln x}{\left(1+x^{2}\right)^{2}} \ dx \end{align} where $\int_{0}^{1}\frac{\ln x}{1+x^{2}}dx=-G$ and $$\int_{0}^{1}\frac{(1-x^2)\ln x}{\left(1+x^{2}\right)^{2}} \ dx =\int_{0}^{1}\ln x\ d\left(\frac{x}{1+x^{2}}\right) \overset{ibp}= -\frac{\pi}{4} $$

Answer 3

$$ \begin{aligned} \int_0^1 \frac{\ln \left(x^2\right)}{\left(1+x^2\right)^2} d x&=2 \int_0^1 \frac{\ln x}{\left(1+x^2\right)^2} d x \\ & =\int_0^1 \frac{\ln x}{x} d\left(\frac{x^2}{1+x^2}\right) \\ & =\left[\frac{x \ln x}{\left(1+x^2\right)}\right]_0^1-\int_0^1\left(-\frac{\ln x}{x^2}+\frac{1}{x^2}\right) \frac{x^2}{1+x^2} d x \\ & =\int_0^1 \frac{\ln x}{1+x^2} d x-\int_0^1 \frac{d x}{1+x^2} \\ & =-G-\frac{\pi}{4} \end{aligned} $$