# How to evaluate $\int_{0}^{1}\frac{\arctan x}{x} \log{\left(\frac{ 1+ x}{\sqrt{1+x^2}}\right)}\mathrm dx$

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

I tried to integrate by parts, but no way so far, help me, thanks.

• Utilizing $x\mapsto\frac{1-x}{1+x}$ allows to extract $\frac{\mathrm G}2\log2$. Again applying $x\mapsto\frac{1-x}{1+x}$ then yields to something of the form $$\int_0^1\frac{\arctan(x)}x\log(1+x^2)\mathrm dx$$ which can be further simplified via IBP. Essentially this reduces the problem to finding $$\int_0^1\frac x{1+x^2}\operatorname{Ti}_2(x)\mathrm dx$$ but I do not know how to continued from hereon. – mrtaurho Jun 12 '19 at 12:20
• Alternatively one can find an anti-derivative of $\frac{\log(1+x^2)}x$ using polylogarithms which boils down to finding $$\int_0^1\frac{\operatorname{Li}_2(-x^2)}{1+x^2}\mathrm dx=\int_0^\frac\pi4\operatorname{Li}_2(-\tan^2(x))\mathrm dx$$ and I'm almost entirely sure that I have seen something similiar to the latter one here on MSE before. – mrtaurho Jun 12 '19 at 12:38
• Mathematica can do mrtaurho's integral. Constants are well-known, except for Li$_3((1+i)/2)).$ – skbmoore Jun 12 '19 at 15:14
• Integral is \begin{align}2\Im\left(\text{Li}_3\left(\frac{1+i}{2}\right)\right)+\text{G}\ln 2-\frac{3}{64}\pi^3-\frac{1}{16}\pi\ln^2 2\end{align} – FDP Jun 12 '19 at 16:57
• @FofX It seems like all of the other responses on this webpage involve sophisticated mathematics. I would be very interested in knowing the context of the problem. Is this problem from a book or a class? If so, what theorems or preliminary (solved) problems do you think might be pertinent to this problem? – user2661923 Aug 16 '20 at 21:01

From here we have that $$\frac12 \int_0^1 \frac{\arctan x \ln(1+x^2)}{x} dx =\frac13 \int_0^1 \frac{\arctan x \ln(1+x)}{x}dx$$ $$\Rightarrow I=\int_{0}^{1}\frac{\arctan x}{x} \ln{\left(\frac{1+ x}{\sqrt{1+x^2}}\right)} dx=\frac23 \int_{0}^{1}\frac{\arctan x \ln(1+x)}{x} dx$$ I have encountered this integral too last year and asked it on AoPS, you can take a look at Knas solution from there, giving: I=\begin{align}2\Im\left(\text{Li}_3\left(\frac{1+i}{2}\right)\right)+\text{G}\ln 2-\frac{3}{64}\pi^3-\frac{1}{16}\pi\ln^2 2\end{align}

• Thank you for your beautyful answer! – FofX Jun 13 '19 at 6:03

From here , we have $$\ \displaystyle \ 3\int_0^1\frac{\arctan x\ln(1+x^2)}{x}\ dx-2\int_0^1\frac{\arctan x\ln(1+x)}{x}\ dx=0$$

or $$\ I=\displaystyle\int_{0}^{1}\frac{\arctan x}{x} \ln{\left(\frac{ 1+ x}{\sqrt{1+x^2}}\right)}\ dx=\int_0^1\frac{\arctan x\ln(1+x^2)}{x}\ dx$$

using $$\ \displaystyle\arctan x\ln(1+x^2)=-2\sum_{n=0}^{\infty}\frac{(-1)^n H_{2n}} {2n+1}x^{2n+1}$$ ( proved here) , we get \begin{align} I&=-2\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{2n+1}\int_0^1x^{2n}\ dx\\ &=-2\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{(2n+1)^2}\\ &=-2\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^2}+2\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}\\ &=-2\Im\sum_{n=1}^\infty\frac{i^nH_n}{n^2}+\frac{\pi^3}{16} \end{align} using the generating function with $$x=i$$ $$\sum_{n=1}^\infty\frac{x^nH_n}{n^2}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$$ we get $$\qquad\displaystyle\Im\sum_{n=1}^\infty\frac{i^nHn}{n^2}=-\frac{\pi}{16}\ln^22-\frac12G\ln2-\Im\operatorname{Li}_3(1-i)$$

Plugging this result, we get $$\quad\boxed{\displaystyle I=\frac{\pi^3}{16}+\frac{\pi}{8}\ln^22+G\ln2+2\Im\operatorname{Li}_3(1-i)}$$

different approach to evaluate $$\displaystyle\int_0^1 \frac{\arctan x\ln(1+x)}{x}\ dx$$ :

from here , we have $$\displaystyle\int_0^1\frac{\arctan x\ln(1+x^2)}{x}\ dx-2\int_0^1\frac{\arctan x\ln(1-x)}{x}\ dx=\frac{\pi^3}{16}\tag{1}$$

and from here , we have $$\displaystyle \ 3\int_0^1\frac{\arctan x\ln(1+x^2)}{x}\ dx-2\int_0^1\frac{\arctan x\ln(1+x)}{x}\ dx=0\tag{2}$$

by combining $$(1)$$ and $$(2)$$, we obtain that $$\displaystyle\int_0^1\frac{\arctan x\ln(1+x)}{x}\ dx=3\int_0^1\frac{\arctan x\ln(1-x)}{x}\ dx+\frac{3\pi^3}{32}\tag{3}$$

we have \begin{align} \int_0^1 \frac{\arctan x\ln(1-x)}{x}\ dx&=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\int_0^1 x^{2n}\ln(1-x)\ dx\\ &=-\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^2}=-\text{Im}\sum_{n=1}^\infty\frac{i^nH_n}{n^2}\\ \end{align} and using the generating function with $$\ x=i$$ $$\sum_{n=1}^\infty\frac{x^nH_n}{n^2}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$$ we get $$\ \displaystyle\int_0^1 \frac{\arctan x\ln(1-x)}{x}\ dx=\frac{\pi}{16}\ln^22+\frac12G\ln2+\text{Im}\operatorname{Li}_3(1-i)\tag{4}$$

plugging $$(4)$$ in $$(3)$$, we get $$\int_0^1 \frac{\arctan x\ln(1+x)}{x}\ dx=\frac{3\pi^3}{32}+\frac{3\pi}{16}\ln^22+\frac32G\ln2+3\text{Im}\operatorname{Li}_3(1-i)$$