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.

  • $\begingroup$ First idea (not sure it leads anywhere): Since $\arctan^\prime x = \frac{1}{1+x^2}$, you have something of the form $\int_0^1 f(x) f^\prime(x) \ln\frac{1+x^2}{1+x} dx$. Have you tried integration by parts? $\endgroup$
    – Clement C.
    Mar 11, 2016 at 3:19
  • $\begingroup$ @ClementC. IBP seems useless, but use your first idea I got$$\int_{0}^{\frac{\pi }{4}}x\ln\left ( 1+\tan^{2}x \right )\mathrm{d}x$$ $\endgroup$ Mar 11, 2016 at 3:25
  • $\begingroup$ $1+\tan^2(x)=(1/\cos(x))^2$. can u finish it off grom there? $\endgroup$
    – tired
    Mar 11, 2016 at 10:47
  • $\begingroup$ And the endresult is $\frac{3}{64}\pi^2\log(2)-\frac{\pi}{8}C$ so u know what u are looking for $\endgroup$
    – tired
    Mar 11, 2016 at 11:07
  • $\begingroup$ In a hurry so won't post full answer, but here is a way to go: make the substitution $x\mapsto \frac{1-x}{1+x}$ and then take the avarage of the two representations you got. Next, expand the log term into two; you'll get two integrals, the first with $\ln(1+x^2)$ and the second with $\ln(1+x)$. In the second integral, make the above substition again and again take avarage. In the first, make a trigonometric sub. It may be helpful to remember that $\tan^{-1} \frac{1-x}{1+x} =\frac{\pi}{4}-\tan^{-1} x$. $\endgroup$ Mar 11, 2016 at 11:40

2 Answers 2


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 $$\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( \frac{1+x^{2}}{1+x} \right )\mathrm{d}x=\frac{1}{64} \pi (3\pi \ln2-8 G).$$

  • $\begingroup$ Nice, big (+1)! $\endgroup$
    – tired
    Mar 12, 2016 at 17:21
  • $\begingroup$ @xpaul What made you decide to make the substitution $t=\arctan x$ instead of something else like integration by parts? My first idea was to integrate by parts and simplify each integral, which took a really long time. $\endgroup$
    – Frank W
    Jul 4, 2018 at 19:53

\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}


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