Prove $\int_0^{\infty} \frac{\arctan{(x)}}{x} \ln{\left(\frac{1+x^2}{{(1-x)}^2}\right)} \; \mathrm{d}x = \frac{3\pi^3}{16}$

Question

Prove that $$\int_0^{\infty} \frac{\arctan{(x)}}{x} \ln{\left(\frac{1+x^2}{{(1-x)}^2}\right)} \; \mathrm{d}x = \frac{3\pi^3}{16}$$ This is not a duplicate of this post, the bounds are different and the integral evaluates to a slightly different value. I tried looking at the solution from the linked post but I'm not familiar with harmonic numbers or complex analysis and the real solution is long. I tried IBP but got no where. Any advice for this monster integral (real analysis only please)?

Answer 1

Changing the bounds makes the integral way simpler, because after letting $x\to \frac{1}{x}$ we can get rid of that $\arctan x$. $$I=\int_0^{\infty} \frac{\arctan x}{x} \ln\left(\frac{1+x^2}{{(1-x)}^2}\right)dx\overset{x\to \frac{1}{x}}=\int_0^\infty \frac{\arctan \left(\frac{1}{x}\right)}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx$$ $$\Rightarrow 2I=\frac{\pi}{2} \int_0^\infty \frac{1}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx\overset{x = \tan \frac{t}{2}}=-\frac{\pi}{2}\int_0^\pi\frac{\ln(1-\sin t)}{\sin t}dt$$ Also from here we know that: $$I(a)=\int_{0}^{\pi} \frac{\ln(1+\sin a\sin x)}{\sin x}dx=a(\pi -a)$$ $$\Rightarrow I=-\frac12 \frac{\pi}{2}I\left(\frac{3\pi}{2}\right)=-\frac12 \frac{\pi}{2}\left(-\frac{3\pi^2}{4}\right)=\frac{3\pi^3}{16}$$

---

Another way to deal with the last integral (credits to this answer), is to consider: $$\mathcal J(a)=\int_0^\frac{\pi}{2}\arctan\left(\frac{\sin x -\tan\frac{a}{2}}{\cos x}\right)dx$$ And differentiate w.r.t. a, obtaining: $$\mathcal J'(a)=-\frac12\int_0^\frac{\pi}{2}\frac{\cos x}{1-\sin a\sin x}dx=\frac12 \frac{\ln(1-\sin a)}{\sin a}$$ $$\mathcal J(\pi)-\mathcal J(0)=-\frac{\pi^2}{4}-\frac{\pi^2}{8}=\frac12\int_0^\pi\frac{\ln(1-\sin a)}{\sin a}da$$ $$\Rightarrow \int_0^\pi \frac{\ln(1-\sin a)}{\sin a}da=-\frac{3\pi^2}{4}$$

Answer 2

Enforce the substitution $x\mapsto 1/x$ and use $\arctan(1/x)=\pi/2-\arctan(x)$ to find that

$$\begin{align} \color{blue}{\int_1^\infty \frac{\arctan(x)}{x}\log\left(\frac{1+x^2}{(1-x)^2}\right)\,dx}&=\int_0^1 \left(\frac{\pi/2-\arctan(x)}{x}\right)\log\left(\frac{1+x^2}{(1-x)^2}\right)\,dx\\\\ &=\frac\pi2 \int_0^1 \frac{\log(1+x^2)}{x}\,dx-\pi\int_0^1\frac{\log(1-x)}{x}\,dx\\\\ &-\color{blue}{\int_0^1\frac{\arctan(x)}{x}\log\left(\frac{1+x^2}{(1-x)^2}\right)\,dx}\\\\ \color{blue}{\int_0^\infty \frac{\arctan(x)}{x}\log\left(\frac{1+x^2}{(1-x)^2}\right)\,dx}&=\frac\pi2 \int_0^1 \frac{\log(1+x^2)}{x}\,dx-\pi\int_0^1\frac{\log(1-x)}{x}\,dx \end{align}$$

---

Now, expanding $\log(1+x)$ in its Taylors series and integrating term by term reveals that

$$\begin{align} \int_0^1 \frac{\log(1+x^2)}{x}\,dx&=\frac12\int_0^1 \frac{\log(1+x)}{x}\,dx\\\\ &=\frac12 \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2}\\\\ &=\frac{\pi^2}{24} \end{align}$$

and similarly that

$$\int_0^1\frac{\log(1-x)}{x}\,dx=-\frac{\pi^2}{6}$$

---

Putting it together, we find the coveted result

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac{\arctan(x)}{x}\log\left(\frac{1+x^2}{(1-x)^2}\right)\,dx=\frac{3\pi^3}{16}}$$