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

How to calculate this relation?

$$I=\int_{0}^{1}\frac{(\arctan x)^2}{1+x^{2}}\ln\left ( 1+x^{2} \right )\mathrm{d}x=\frac{\pi^3}{96}\ln{2}-\frac{3\pi\zeta{(3)}}{128}-\frac{\pi^2G}{16}+\frac{\beta{(4)}}{2}$$ Where G is the Catalan's constant, and $$\beta(x)=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{(2k-1)^{x}}$$ is the Dirichlet's beta function. Integrate by parts $$u=(\arctan{\,x})^2\ln{(1+x^2)}$$ $$v=\arctan{\,x}$$ We have $$3I=\frac{\pi^3}{64}\ln{2-2\int_0^{1}x(\arctan{\,x})^3}\frac{dx}{1+x^2}$$

But how to calculate the latter integral?Could anybody please help by offering useful hints or solutions?Ithing very difficult to prove.

Take $\arctan\left(x\right)=u,\,\frac{dx}{1+x^{2}}=du$. Then $$I=\int_{0}^{1}\frac{\left(\arctan\left(x\right)\right)^{2}}{1+x^{2}}\log\left(1+x^{2}\right)dx=\int_{0}^{\pi/4}u^{2}\log\left(1+\tan^{2}\left(u\right)\right)du$$ $$=-2\int_{0}^{\pi/4}u^{2}\log\left(\cos\left(u\right)\right)du$$ and now using the Fourier series $$\log\left(\cos\left(u\right)\right)=-\log\left(2\right)-\sum_{k\geq1}\frac{\left(-1\right)^{k}\cos\left(2ku\right)}{k},\,0\leq x<\frac{\pi}{2}$$ we have $$I=\frac{\log\left(2\right)\pi^{3}}{96}+2\sum_{k\geq1}\frac{\left(-1\right)^{k}}{k}\int_{0}^{\pi/4}u^{2}\cos\left(2ku\right)du$$ and the last integral is trivial to estimate $$\int_{0}^{\pi/4}u^{2}\cos\left(2ku\right)du=\frac{\pi^{2}\sin\left(\frac{\pi k}{2}\right)}{32k}-\frac{\sin\left(\frac{\pi k}{2}\right)}{4k^{3}}+\frac{\pi\cos\left(\frac{\pi k}{2}\right)}{8k^{2}}$$ so we have $$I=\frac{\log\left(2\right)\pi^{3}}{96}+\pi^{2}\sum_{k\geq1}\frac{\left(-1\right)^{k}\sin\left(\frac{\pi k}{2}\right)}{16k^{2}}-\sum_{k\geq1}\frac{\left(-1\right)^{k}\sin\left(\frac{\pi k}{2}\right)}{2k^{4}}+\pi\sum_{k\geq1}\frac{\left(-1\right)^{k}\cos\left(\frac{\pi k}{2}\right)}{4k^{3}}$$ and now observing that $$\cos\left(\frac{\pi k}{2}\right)=\begin{cases} -1, & k\equiv2\,\mod\,4\\ 1, & k\equiv0\,\mod\,4\\ 0, & \textrm{otherwise} \end{cases}$$ and $$\sin\left(\frac{\pi k}{2}\right)=\begin{cases} -1, & k\equiv3\,\mod\,4\\ 1, & k\equiv1\,\mod\,4\\ 0, & \textrm{otherwise} \end{cases}$$ we have $$I=\frac{\log\left(2\right)\pi^{3}}{96}-\frac{\pi^{2}}{16}K+\frac{\beta\left(4\right)}{2}-\frac{3\pi\zeta\left(3\right)}{128}\approx 0.064824$$ where the last sum is obtained using the relation between Dirichlet eta function and Riemann zeta function.
• @user178256 Yes there was an error in the first term. Thank you. The sign in the enunciated is $-$ but from my calculation is $+$, and I don't see the error. Mar 18, 2016 at 16:50
• @Marco Cantarini $$-\frac{\pi^3}{96}\ln{2}+\frac{beta{(4)}}{2}$$is correct Mar 18, 2016 at 16:59