# Compute in closed form : $\int_0^{\frac{π}{4}} x\ln(\tan x)\left(1-\frac{1}{\cos^2 x}\right)dx$

Question :

Compute in closed form without use series

$$I =\displaystyle\int_0^{\pi / 4} x\ln\left(\tan x\right)\left(1-\frac{1}{\cos^2 x}\right)\,dx$$

I think use : $$y=\tan x$$ then $$dy=\frac{1}{\cos^2 x}$$

So : $$I =\displaystyle\int_0^{1} \arctan x\ln\left(x\right)x^{2}\,dx$$

But I find integration arctan

Please give me ideas to approach it .

• I edited it to a point where the Mathjax at least renders, but I don't know if it matches what you were trying to do. Please proof-read before posting. – user296602 Apr 30 at 21:32
• Your new integral should be in terms of y if you used substitution, as in 'arctan(y)ln(y)(y^2)*(1/1-y^2) if I'm not mistaken – ricky Apr 30 at 21:46
• Roze: You have forgotten denominator (1+x^2) and multiply by -1. – FDP Apr 30 at 23:23

\begin{align}I =\displaystyle\int_0^{\pi / 4} x\ln\left(\tan x\right)\left(1-\frac{1}{\cos^2 x}\right)\,dx\end{align} Perform the change of variable $$y=\tan x$$, \begin{align}I&=\int_0^1 \arctan x\ln x \left(\frac{1}{1+x^2}-1\right)\,dx\\ &=-\int_0^1 \frac{x^2\arctan x\ln x}{1+x^2}\,dx\\ &=\Big[\left(\arctan x-x\right)\arctan x\ln x\Big]_0^1+\int_0^1 \left(x-\arctan x\right)\left(\frac{\ln x}{1+x^2}+\frac{\arctan x}{x}\right)\,dx\\ &=\int_0^1 \frac{x\ln x}{1+x^2}\,dx+\int_0^1 \arctan x\,dx-\int_0^1 \frac{\arctan x\ln x}{1+x^2}\,dx-\int_0^1 \frac{\arctan^2 x}{x}\,dx\\ &=\left[x\arctan x-\frac{1}{2}\ln(1+x^2)\right]_0^1+\frac{1}{4}\int_0^1\frac{2x\ln(x^2)}{1+x^2}\,dx-\int_0^1 \frac{\arctan x\ln x}{1+x^2}\,dx-\\ &\Big[\ln x\arctan^2 x\Big]_0^1+2\int_0^1 \frac{\arctan x\ln x}{1+x^2}\,dx\\ &=\frac{\pi}{4}-\frac{1}{2}\ln 2+\frac{1}{4}\times -\frac{\pi^2}{12}+\int_0^1 \frac{\arctan x\ln x}{1+x^2}\,dx\\ \end{align}

One knows that, \begin{align}\int_0^1 \frac{\arctan x\ln x}{1+x^2}\,dx=\frac7{16}\zeta(3)-\frac{1}{4}\pi\text{G}\end{align}

Therefore,

\begin{align}\boxed{I=\frac{\pi}{4}-\frac{1}{2}\ln 2-\frac{1}{48}\pi^2-\frac{1}{4}\pi\text{G}+\frac7{16}\zeta(3)}\end{align}

With Maple I get $$I = -\frac{\pi\,G}{4}-\frac{{\pi}^{2}}{48}+{\frac {25\,\zeta \left( 3 \right) }{64}}+\frac{\pi}{4}-\frac{\ln\left( 2 \right)}{2} - \frac{\text{Re}\; {\rm Li}_3( i) }{2} \approx 0.03971$$ where $$G$$ is Catalan's constant $$G = \sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^2} \tag{1}$$ and $$\text{Li}_3$$ is the polylogarithm $$\text{Li}_s(z) = \sum_{k=1}^\infty \frac{z^k}{k^s} \tag{2}$$

Is it counted as "without series" when I use (1) and (2)? And $$\zeta(3)$$ is also defined as a series.

• The polylog has a closed-form as, $${\rm Li}_3(i)=-\frac{3\zeta(3)}{32}+\frac{\pi^3}{32}i$$ – Tito Piezas III Jul 2 at 2:39