How to evaluate $\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( \frac{1+x^{2}}{1+x} \right )\mathrm{d}x$

Question

How to evaluate $$\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( \frac{1+x^{2}}{1+x} \right )\mathrm{d}x$$ I completely have no idea how to find the result.Mathematic gave me the following answer part of the integral $$\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( 1+x^{2} \right )\mathrm{d}x=-\frac{1}{4}\mathbf{G}\pi +\frac{\pi ^{2}}{16}\ln 2+\frac{21}{64}\zeta \left ( 3 \right )$$ where $\mathbf{G}$ donates the Catalan's Constant.

But it can't evaluate the other part.So I'd like to know how to evaluate the original integral or the above integral.

Answer 1

Let $t=\arctan x$. Then \begin{eqnarray} &&\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( \frac{1+x^{2}}{1+x} \right )\mathrm{d}x\\ &=&-\int_{0}^{\frac{\pi}{4}}t\ln[\cos t(\cos t+\sin t))]\mathrm{d}t\\ &=&-\frac{1}{2}\int_{0}^{\frac{\pi}{4}}t\ln[\cos t^2(\cos t+\sin t)^2]\mathrm{d}t\\ &=&-\frac{1}{2}\int_{0}^{\frac{\pi}{4}}t\ln[\frac{1+\cos 2t}{2}(1+\sin 2t)]\mathrm{d}t\\ &=&-\frac{1}{2}\int_{0}^{\frac{\pi}{4}}t\ln[(1+\cos 2t)(1+\sin 2t)]\mathrm{d}t+\frac{1}{64}\pi^2\ln2\\ &=&-\frac{1}{8}\int_{0}^{\frac{\pi}{2}}t\ln[(1+\cos t)(1+\sin t)]\mathrm{d}t+\frac{1}{64}\pi^2\ln2\\ &=&-\frac{\pi}{32}\int_{0}^{\frac{\pi}{2}}\ln[(1+\cos t)(1+\sin t)]\mathrm{d}t+\frac{1}{64}\pi^2\ln2\\ &=&-\frac{\pi}{16}\int_{0}^{\frac{\pi}{2}}\ln(1+\cos t)\mathrm{d}t+\frac{1}{64}\pi^2\ln2\\ \end{eqnarray} Noting \begin{eqnarray} \int_{0}^{\frac{\pi}{2}}\ln(1+\cos t)\mathrm{d}t&=&\int_{0}^{\frac{\pi}{2}}\ln(2\cos^2\frac{t}{2})\mathrm{d}t\\ &=&\frac{\pi}{2}\ln 2+2\int_{0}^{\frac{\pi}{2}}\ln(\cos\frac{t}{2})\mathrm{d}t\\ &=&\frac{\pi}{2}\ln 2+4\int_{0}^{\frac{\pi}{4}}\ln(\cos t)\mathrm{d}t\\ &=&2G-\frac{\pi}{2}\ln 2 \end{eqnarray} and hence $$\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( \frac{1+x^{2}}{1+x} \right )\mathrm{d}x=\frac{1}{64} \pi (3\pi \ln2-8 G).$$

Answer 2

\begin{align}J&=\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( \frac{1+x^{2}}{1+x} \right )\mathrm{d}x\\ &\overset{y=\frac{1-x}{1+x}}=\frac{\pi}{4}\int_0^1\frac{1}{1+y^2}\ln\left ( \frac{1+y^{2}}{1+y} \right )\mathrm{d}y-J\\ 2J&=\frac{\pi}{4}\int_0^1\frac{\ln\left ( 1+y^{2} \right )}{1+y^2}\mathrm{d}y-\frac{\pi}{4}\int_0^1\frac{\ln\left ( 1+y \right )}{1+y^2}\mathrm{d}y\\ K&=\int_0^1\frac{\ln\left ( 1+y^{2} \right )}{1+y^2}\mathrm{d}y\\&\overset{u=\frac{1}{y}}=\int_1^{\infty}\frac{\ln\left(1+u^2\right)}{1+u^2}du-2\underbrace{\int_1^{\infty}\frac{\ln u}{1+u^2}du}_{v=\frac{1}{u}}\\ 2K&=\int_0^\infty\frac{\ln\left ( 1+y^{2} \right )}{1+y^2}\mathrm{d}y+2\int_0^{1}\frac{\ln u}{1+u^2}du\\ &\overset{z=\sqrt{\frac{1-\frac{u}{\sqrt{{{u}^{2}}+1}}}{1+\frac{u}{\sqrt{{{u}^{2}}+1}}}}}=-2\int_0^1 \frac{\ln\left(\frac{4z^2}{(1+z^2)^2}\right)}{1+z^2}dz-2\text{G}\\ &=2\text{G}-\pi\ln 2+4K\\ K&=\boxed{\frac{1}{2}\pi\ln 2-\text{G}}\\ L&=\int_0^1\frac{\ln\left ( 1+y \right )}{1+y^2}\mathrm{d}y\\ &\overset{u=\frac{1-y}{1+y}}=\int_0^1 \frac{\ln\left(\frac{2}{1+u}\right)}{1+u^2}du\\ 2L&=\boxed{\frac{1}{4}\pi\ln 2}\\ \end{align} Therefore,
\begin{align}\boxed{\displaystyle J=\dfrac{3}{64}\pi^2\ln 2 -\dfrac{1}{8}\pi\text{G} }\end{align}