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## Log-trig integral again

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### Log-trig integral again

Sat Aug 20, 2016 12:15 pm

Posts: 22
Enjoy,

$\displaystyle \int_0^{\tfrac{\pi}{4}} \dfrac{x\ln\left(\tfrac{\cos x+\sin x}{cos x-sin x}\right)}{\cos x(\cos x+\sin x)}dx=\dfrac{\pi^3}{192}+\dfrac{G\ln 2}{2}$

### Re: Log-trig integral again

Sun Sep 25, 2016 9:22 pm

Posts: 22
$\displaystyle J=\int_0^{\tfrac{\pi}{4}} \dfrac{x\ln\left(\tfrac{\cos x+\sin x}{cos x-sin x}\right)}{\cos x(\cos x+\sin x)}dx$

For $x\in [0,\tfrac{\pi}{4}]$,

$\displaystyle \dfrac{x\ln\left(\dfrac{\cos x+\sin x}{\cos x-\sin x}\right)}{\cos x(\cos x+\sin x)}=\dfrac{x\ln\left(\dfrac{1+\tan x}{1-\tan x}\right)}{(\cos x)^2(1+\tan x)}=\dfrac{x(1+(\tan x)^2)\ln\left(\dfrac{1+\tan x}{1-\tan x}\right)}{1+\tan x}$

Perform the change of variable $y=\tan x$ in the following integral,

$J=\displaystyle \int_{0}^{\tfrac{\pi}{4}} \dfrac{x(1+(\tan x)^2)\ln\left(\dfrac{1+\tan x}{1-\tan x}\right)}{1+\tan x}dx=\int_0^1 \dfrac{\arctan x\ln\left(\dfrac{1+x}{1-x}\right)}{1+x}dx$

Perform the change of variable $y=\dfrac{1-x}{1+x}$,

$\displaystyle J=\int_0^1 \dfrac{\ln x\arctan\left(\dfrac{x-1}{x+1}\right)}{1+x}dx=\int_0^1 \dfrac{\arctan x\ln x}{1+x}dx-\dfrac{\pi}{4}\int_0^1 \dfrac{\ln x}{1+x}dx$

$\displaystyle \int_0^1 \dfrac{\ln x}{1+x}dx=-\dfrac{\pi^2}{12}$

Define for $x\in [0,1]$,

$\displaystyle F(x)=\int_0^x \dfrac{\log t}{1+t}dt=\int_0^1 \dfrac{x\log(xy)}{1+xy}dy$

Note that,

$F(1)=-\dfrac{\pi^2}{12}$

\begin{align} \displaystyle \int_0^1 \dfrac{\arctan x\ln x}{1+x}dx&=\Big[F(x)\arctan x\Big]_0^1-\int_0^1 \dfrac{F(x)}{1+x^2}dx\\ \displaystyle &=-\dfrac{\pi^3}{48}-\int_0^1 \dfrac{x\log(xy)}{(1+xy)(1+x^2)}dxdy\\ \displaystyle &=-\dfrac{\pi^3}{48}-\int_0^1 \dfrac{x\log(x)}{(1+xy)(1+x^2)}dxdy-\int_0^1 \dfrac{x\log(y)}{(1+xy)(1+x^2)}dxdy\\ \displaystyle &=-\dfrac{\pi^3}{48}-\int_0^1\left[\dfrac{\log x\log(1+xy)}{1+x^2}\right]_{y=0}^{y=1} dx-\\ & \int_0^1 \left[-\dfrac{\log y\log(1+xy)}{1+y^2}+\dfrac{\log y\log(1+x^2)}{2(1+y^2)}+\dfrac{y\log y\arctan x}{1+y^2}\right]_{x=0}^{x=1}dy\\ &=-\dfrac{\pi^3}{48}-\int_0^1 \dfrac{\log x\log(1+x)}{1+x^2}dx+\int_0^1\dfrac{\log y\log(1+y)}{1+y^2}dy-\dfrac{\log 2}{2}\int_0^1 \dfrac{\log y}{1+y^2}dy-\\ &\dfrac{\pi}{4}\times \int_0^1 \dfrac{y\log y}{1+y^2}dy\\ &=\dfrac{G\ln 2}{2}-\dfrac{\pi^3}{64} \end{align}

Therefore,

$J=\dfrac{G\ln 2}{2}+\dfrac{\pi^3}{192}$