Is there Closed form :$\int^1_0 \frac{\arctan(x)}{x}\ln\left({\frac{1-x}{1+x}}\right)\mathrm{d}x$

Question

I was trying solve this integral$$I= \int_{0}^{\pi/2} \frac{x}{\sin(x)} \ln\left(\frac{1+\cos(x) - \sin(x)}{1+\cos(x) + \sin(x)} \right) dx $$ let :$t=\tan(x/2)$ $$I = 2\int^1_0 \frac{\arctan(t)}{t}\ln\left({\frac{1-t}{1+t}}\right)dt$$

I have a problem evaluating it.

I try :$y=\frac{1-t}{1+t}$

\begin{align*} I &= 4\int^1_0\frac{\left({\frac{\pi}{4}-\arctan(y)}\right)\ln(y)}{1-y^2} dy=\pi\int^1_0 \frac{\ln(y)}{1-y^2}dy -4\int^1_0 \frac{\arctan(y)\ln(y)}{1-y^2}dy \\ &= -\frac{\pi^3}{8}-4\int^1_0 \frac{\arctan(y)\ln(y)}{1-y^2}dy\end{align*}

this integral complicated $\int^1_0 \frac{\arctan(y)\ln(y)}{1-y^2}dy$

I try to :

$$I = -\int^1_{-1}\int^1_0 \frac{\arctan(t)}{(1-at)}dtda$$

I don't know how to continue evaluating the integration.

Answer 1

Substituting $\frac{1-x}{1+x} \to x$ transforms the integral into

$$I = \int_{0}^{1}\frac{\arctan\left(x\right)}{x}\ln\left(\frac{1-x}{1+x}\right)dx \stackrel{\frac{1-x}{1+x}\to x}{=} 2\int_{0}^{1}\frac{\arctan\left(\frac{1-x}{1+x}\right)}{1-x^{2}}\ln\left(x\right)dx$$ Note that $\arctan\left(\frac{1-x}{1+x}\right) = \frac{\pi}{4} - \arctan x$ by the addition formula for arctan. So, $$ I = \frac{\pi}{2}\underbrace{\int_{0}^{1}\frac{\ln\left(x\right)}{1-x^{2}}}_{I_1}dx-2\underbrace{\int_{0}^{1}\frac{\arctan\left(x\right)\ln\left(x\right)}{1-x^{2}}dx}_{I_2} \tag1$$ $I_1$ can be evaluated by expanding $\frac1{1-x^2}$ as a geometric series: $$I_1 = \sum_{n=0}^\infty\int\limits_0^1x^{2n}\ln (x)dx = -\sum_{n=0}^\infty\frac{1}{(2n+1)^2} = -\frac{3\zeta(2)}{4} = -\frac{\pi^2}{8}$$ For $I_2$, note that $$\arctan(x) = \int\limits_0^x\frac{1}{1+t^2}dt \stackrel{t \to tx}{=}\int\limits_0^1\frac{x}{1+t^2x^2}dt$$ Substituting this in to $I_2$ gives us $$I_2 = \int_{0}^{1}\int_{0}^{1}\frac{x\ln\left(x\right)}{\left(1-x^{2}\right)\left(1+t^{2}x^{2}\right)}dtdx$$ Doing a partial fraction decomposition and exchanging the order of integration gives \begin{align} I_2 &= \int_{0}^{1}\frac{1}{t^{2}+1}\int_{0}^{1}\left(\frac{t^{2}x\ln\left(x\right)}{t^{2}x^{2}+1}-\frac{\ln\left(x\right)}{2\left(x-1\right)}-\frac{\ln x}{2\left(x+1\right)}\right)dxdt \\&=\int_{0}^{1}\frac{t^{2}}{t^{2}+1}\int_{0}^{1}\frac{x\ln\left(x\right)}{t^{2}x^{2}+1}dxdt-\frac{1}{2}\int_{0}^{1}\frac{1}{t^{2}+1}\int_{0}^{1}\frac{\ln\left(x\right)}{x-1}dxdt\\&-\frac{1}{2}\int_{0}^{1}\frac{1}{t^{2}+1}\int_{0}^{1}\frac{\ln\left(x\right)}{x+1}dxdt \end{align} The last two integrals can simply be integrated with geometric series, giving $$\int_{0}^{1}\frac{\ln\left(x\right)}{x-1}dx = \zeta(2)$$ $$\int_{0}^{1}\frac{\ln\left(x\right)}{x+1}dx = -\frac{\zeta(2)}2$$ And of course, $$\int_{0}^{1}\frac{1}{t^{2}+1} = \frac \pi4$$ Substituting this in gives us $$I_2 = -\frac{\pi}{16}\zeta\left(2\right)+\int_{0}^{1}\frac{t^{2}}{t^{2}+1}\underbrace{\int_{0}^{1}\frac{x\ln\left(x\right)}{\left(t^{2}x^{2}+1\right)}dx}_{\mathcal{J}}dt \tag2$$

Focusing on the inner integral, $$\mathcal{J}=\int_{0}^{1}\frac{x\ln\left(x\right)}{\left(t^{2}x^{2}+1\right)}dx$$ Due to the bounds on $x$ and $t$, we can expand $\cal J$ as a geometric series $$\mathcal J = \sum_{n=0}^{\infty}\left(-1\right)^{n}t^{2n}\int_{0}^{1}x^{2n+1}\ln\left(x\right)dx = -\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}t^{2n}}{\left(2n+2\right)^{2}}=\frac{1}{4t^{2}}\sum_{n=0}^{\infty}\frac{\left(-t^{2}\right)^{\left(n+1\right)}}{\left(n+1\right)^{2}}$$ The sum is a representation of the dilogarithm, giving us $$\mathcal{J} = \frac{\operatorname{Li}_2\left(-t^2\right)}{4t^2}$$ Substituting into $(2)$ gives $$I_2=-\frac{\pi}{16}\zeta\left(2\right)+\frac{1}{4}\int_{0}^{1}\frac{\operatorname{Li}_{2}\left(-t^{2}\right)}{t^{2}+1}dt$$ The dilogarithm integral can be evaluated here, giving us (after a bit of simplifying) $$I_2 = \frac{\pi^{3}}{64}+\frac{\pi}{16}\ln^{2}\left(2\right)+\frac{1}{2}\ln\left(2\right)G + \Im\operatorname{Li}_3(1-i)$$ where $G$ is Catalan's constant and $\operatorname{Li}_3(x)$ is the trilogarithm. Finally, substituing into $(1)$ gives us

$$I = -\frac{3}{32}\pi^{3}-\frac{\pi}{8}\ln^{2}\left(2\right)-\ln\left(2\right)G-2\Im\operatorname{Li}_3(1-i)$$

Answer 2

The antiderivative $$I=\int \frac{\tan ^{-1}(x)}{x}\,\log \left(\frac{1-x}{1+x}\right)\,dx$$ is not too bad if you use the logarithmic representation of the arc tangent $$I=\frac i 2 \int \frac 1 x\, \log \left(\frac{i+x}{i-x}\right)\,\log \left(\frac{1-x}{1+x}\right)\,dx$$

Expand the logarithms and remember that $$\int \frac 1 x\,\log (x+a) \,\log (x+b)\,dx$$ has a closed form in terms of logarithms and polylogarithms $\text{Li}_2(.)$ and $\text{Li}_3(.)$

So, we have the antiderivative.

Considering now $$J=\int_\epsilon^{1-\epsilon} \frac{\tan ^{-1}(x)}{x}\,\log \left(\frac{1-x}{1+x}\right)\,dx$$ and using series expansion, we have $$J=-\Bigg(C \log (2)+\frac{3 \pi ^3}{32}+\frac{1}{8} \pi \log ^2(2) +i\Big(\text{Li}_3(1+i)-\text{Li}_3(1-i) \Big) \Bigg)+$$ $$\frac{\pi }{4} \log \left(\frac{2 e}{\epsilon }\right)\,\epsilon +\frac 18 \Bigg(7+(\pi-2)\log \left(\frac{2}{\epsilon }\right) \Bigg) \,\epsilon^2+O\left(\epsilon ^3\right)$$ which is a quite good approximation.

For $\epsilon=\frac 14$, this truncated series gives $-0.518364$ instead of $-0.519366$ corresponding to a relative error of $0.2$%.

Answer 3

The integrand of this integral can be easily expanded into a Taylor series:

$$\frac{\arctan(x)}{x}\ln\left({\frac{1-x}{1+x}}\right)=-2\sum_{k=0}^{\infty}\frac{H_{k}}{2k+1}x^{4k+1}$$

where

$$H_{k}=\sum_{m=0}^{2k}\frac{(-1)^{m}}{2m+1}$$

The result of the integration:

$$I=-\sum_{k=0}^{\infty}\frac{H_{k}}{(2k+1)^{2}}$$

Adding the first 10 terms of this series, the absolute error is about $0.018$

A physicist's approach:

Here i use an approximation:

$$\frac{\arctan(x)}{x}\approx \frac{\pi}{392}(129-24x-7x^{2})$$

In this case, the integral is easily computed analytically:

$$I\approx -\pi\frac{760\ln 2 -79}{1176}$$

In this case the absolute error is $3\cdot 10^{-6}$

Answer 4

The following solution works for only even powers of $\ln$:

$$\int^1_0 \frac{\arctan(t)}{t}\ln^{2a}\left({\frac{1-t}{1+t}}\right)dt=\frac{\pi}{2}\int_0^1\frac{\ln^{2a}(t)}{1-t^2}dt-2\int_0^1\frac{\arctan(t)\ln^{2a}(t)}{1-t^2}dt$$

$$=2(2a)!\sum_{k=0}^{a}2^{2k-2a}\eta(2a-2k)\beta(2k+2).$$

where the first integral is the integral form of the Dirichlet lambda function and the second integral is given in page 14 of this preprint.

Examples:

$$\int^1_0 \frac{\arctan(t)}{t}\ln^{2}\left({\frac{1-t}{1+t}}\right)dt=\frac{\pi^2}{12}\beta(2)+2\beta(4)$$

$$\int^1_0 \frac{\arctan(t)}{t}\ln^{4}\left({\frac{1-t}{1+t}}\right)dt=\frac{7\pi^4}{240}\beta(2)+\pi^2\beta(4)+24 \beta(6)$$

$$\int^1_0 \frac{\arctan(t)}{t}\ln^{6}\left({\frac{1-t}{1+t}}\right)dt=\frac{31\pi^6}{1344}\beta(2)+\frac{7\pi^2}{8}\beta(4)+30 \pi^2\beta(6)+720\beta(8)$$