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## Logarithmic integrals, some more again

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### Logarithmic integrals, some more again

Fri Oct 07, 2016 8:57 pm

Posts: 22
Compute:

$\displaystyle \int_0^{3-2\sqrt{2}} \dfrac{\ln x}{(x-1)\sqrt{x^2-6x+1}}dx$

$\displaystyle \int_0^{3-2\sqrt{2}} \dfrac{\ln x}{\sqrt{x^2-6x+1}}dx$

Enjoy.

### Re: Logarithmic integrals, some more again

Wed Oct 12, 2016 9:26 pm

Posts: 22
Let
$J=\displaystyle \int_0^{3-2\sqrt{2}} \dfrac{\ln x}{(x-1)\sqrt{x^2-6x+1}}dx$

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

$J=\displaystyle \int_{3+2\sqrt{2}}^{+\infty} \dfrac{\ln x}{(x-1)\sqrt{x^2-6x+1}}dx$

Perform the change of variable $\displaystyle x=\dfrac{\sqrt{2}\cos y+1}{\sqrt{2}\cos y-1}$

$dx=\dfrac{2\sqrt{2}\sin y}{\left(\sqrt{2}\cos y-1\right)^2}dy$

$\displaystyle x^2-6x+1=\dfrac{8(\sin y)^2}{\left(\sqrt{2}\cos y-1\right)^2}$

$\displaystyle x-1=\dfrac{2}{\sqrt{2}\cos y-1}$

Therefore,

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

On the other hand,

$\sqrt{2}\cos x+1=\sqrt{2}\left(\cos x+\cos\left(\dfrac{\pi}{4}\right)\right)=2\sqrt{2}\cos\left(\dfrac{x}{2}+\dfrac{\pi}{8}\right)\cos\left(\dfrac{\pi}{8}-\dfrac{x}{2}\right)$
$\sqrt{2}\cos x-1=\sqrt{2}\left(\cos x+\cos\left(\dfrac{\pi}{4}\right)\right)=2\sqrt{2}\sin\left(\dfrac{x}{2}+\dfrac{\pi}{8}\right)\sin\left(\dfrac{\pi}{8}-\dfrac{x}{2}\right)$

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

$\displaystyle \dfrac{\sqrt{2}\cos x +1}{\sqrt{2}\cos x -1}=\cot\left(\dfrac{x}{2}+\dfrac{\pi}{8}\right)\cot\left(\dfrac{\pi}{8}-\dfrac{x}{2}\right)$

Therefore,

$J=\displaystyle \dfrac{1}{2}\int_0^{\tfrac{\pi}{4}} \ln\left(\cot\left(\dfrac{x}{2}+\dfrac{\pi}{8}\right)\right)dx+\dfrac{1}{2}\int_0^{\tfrac{\pi}{4}} \ln\left(\cot\left(\dfrac{\pi}{8}-\dfrac{x}{2}\right)\right)dx$

In the first integral perform the change of variable $y=\dfrac{\pi}{8}+\dfrac{x}{2}$
In the second integral perform the change of variable $y=\dfrac{\pi}{8}-\dfrac{x}{2}$

One obtains,
$J=\displaystyle \int_\tfrac{\pi}{8}^{\tfrac{\pi}{4}} \ln(\cot(x))dx+\int_0^{\tfrac{\pi}{8}}\ln(\cot(x))dx$

Therefore,

$\displaystyle J=\int_0^{\tfrac{\pi}{4}}\ln(\cot(x))dx$

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

$\boxed{\displaystyle J=-\int_0^1 \dfrac{\ln(x)}{1+x^2}dx=G}$

$G$ is the Catalan constant.

Let,
$K=\displaystyle \int_0^{3-2\sqrt{2}} \dfrac{\ln x}{\sqrt{x^2-6x+1}}dx$

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

$K=-\displaystyle \int_{3+2\sqrt{2}}^{+\infty} \dfrac{\ln x}{x\sqrt{x^2-6x+1}}dx$

Perform the change of variable $\displaystyle x=\dfrac{\sqrt{2}\cos y+1}{\sqrt{2}\cos y-1}$,

$K=\displaystyle -\int_0^{\tfrac{\pi}{4}}\dfrac{\ln\left(\dfrac{\sqrt{2}\cos x +1}{\sqrt{2}\cos x -1}\right)}{\sqrt{2}\cos x +1}dx=-\int_0^{\tfrac{\pi}{4}}\dfrac{\ln\left(\dfrac{\sqrt{2}\cos x +1}{\sqrt{2}\cos x -1}\right)}{2\sqrt{2}\cos\left(\dfrac{x}{2}+\dfrac{\pi}{8}\right)\cos\left(\dfrac{\pi}{8}-\dfrac{x}{2}\right)}dx$

$K=\displaystyle -\int_0^{\tfrac{\pi}{4}} \dfrac{\ln\left(\cot\left(\dfrac{x}{2}+\dfrac{\pi}{8}\right)\right)}{2\sqrt{2}\cos\left(\dfrac{x}{2}+\dfrac{\pi}{8}\right)\cos\left(\dfrac{\pi}{8}-\dfrac{x}{2}\right)}dx-\int_0^{\tfrac{\pi}{4}} \dfrac{\ln\left(\cot\left(\dfrac{\pi}{8}-\dfrac{x}{2}\right)\right)}{2\sqrt{2}\cos\left(\dfrac{x}{2}+\dfrac{\pi}{8}\right)\cos\left(\dfrac{\pi}{8}-\dfrac{x}{2}\right)}dx$

In the first integral perform the change of variable $y=\dfrac{\pi}{8}+\dfrac{x}{2}$
In the second integral perform the change of variable $y=\dfrac{\pi}{8}-\dfrac{x}{2}$

$\displaystyle K=-\int_{\tfrac{\pi}{8}}^{\tfrac{\pi}{4}} \dfrac{\ln(\cot x)}{\sqrt{2}\cos x\cos\left(\dfrac{\pi}{4}-x\right)}dx-\int_0^{\tfrac{\pi}{8}} \dfrac{\ln(\cot x)}{\sqrt{2}\cos x\cos\left(\dfrac{\pi}{4}-x\right)}dx$

Therefore,

$\displaystyle K=-\int_0^{\tfrac{\pi}{4}} \dfrac{\ln(\cot x)}{\sqrt{2}\cos x\cos\left(\dfrac{\pi}{4}-x\right)}dx$

On the other hand,

\begin{align} \sqrt{2}\cos x\cos\left(\dfrac{\pi}{4}-x\right)&=\sqrt{2}\cos x\left(\dfrac{\sqrt{2}}{2}\cos x+\dfrac{\sqrt{2}}{2}\sin x\right)\\ &=(\cos x)^2+\cos x\sin x\\ &=\dfrac{1}{1+(\tan x)^2}+\dfrac{\tan x}{1+(\tan x)^2}\\ &=\dfrac{1+\tan x}{1+(\tan x)^2} \end{align}

Therefore,

$\displaystyle K=\int_0^{\tfrac{\pi}{4}}\dfrac{(1+(\tan x)^2)\ln(\tan x)}{1+\tan x}dx$

perform the change of variable $y=\tan x$,

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

### Re: Logarithmic integrals, some more again

Fri Oct 14, 2016 10:33 am

Posts: 22
Enjoy this one,

$\displaystyle \int_0^{\tfrac{\sqrt{6}-\sqrt{2}-1}{\sqrt{6}-\sqrt{2}+1}} \dfrac{\ln x}{(x-1)\sqrt{x^2-2(15+8\sqrt{3})x+1}}dx=\dfrac{2}{3}(2-\sqrt{3})G$

### Re: Logarithmic integrals, some more again

Sun Oct 30, 2016 8:30 am

Posts: 22
Define on $]1;+\infty[$ the function $F$,

$\displaystyle F(a)=\int_0^{\tfrac{a-1}{a+1}} \dfrac{\ln x}{(x-1)\sqrt{(a^2-1)x^2-2(a^2+1)x+(a^2-1)}}dx$

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

$\displaystyle F(a)=\int_{\tfrac{a+1}{a-1}}^{+\infty} \dfrac{\ln x}{(x-1)\sqrt{(a^2-1)x^2-2(a^2+1)x+(a^2-1)}}dx$

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

$dx=\dfrac{2a\sin y}{(a\cos y-1)^2}$

$(a^2-1)x^2-2(a^2+1)x+(a^2-1)=\dfrac{4a^2\sin^2 y}{(a\cos y-1)^2}$

$x-1=\dfrac{2}{a\cos y-1}$

Let $\theta(a)\in[0;\dfrac{\pi}{2}]$ such that $\cos\left(\theta(a)\right)=\dfrac{1}{a}$

Therefore,

$\displaystyle F(a)=\dfrac{1}{2}\int_{0}^{\theta(a)} \ln\left(\dfrac{a\cos y+1}{a\cos y-1}\right)dy$

$a\cos y+1=a\left(\cos y+\dfrac{1}{a}\right)=a(\cos y+\cos \theta(a))=2a\left(\cos\left(\dfrac{y}{2}+\dfrac{\theta(a)}{2}\right)\cos\left(\dfrac{y}{2}-\dfrac{\theta(a)}{2}\right)\right)$

$a\cos y-1=a\left(\cos y-\dfrac{1}{a}\right)=a(\cos y+\cos \theta(a))=2a\left(\sin\left(\dfrac{y}{2}+\dfrac{\theta(a)}{2}\right)\sin\left(\dfrac{\theta(a)}{2}-\dfrac{y}{2}\right)\right)$

Therefore, $y\in[0;\theta(a)[$,

$\dfrac{a\cos y+1}{a\cos y-1}=\dfrac{\cos\left(\dfrac{y}{2}+\dfrac{\theta(a)}{2}\right)\cos\left(\dfrac{y}{2}-\dfrac{\theta(a)}{2}\right)}{\sin\left(\dfrac{y}{2}+\dfrac{\theta(a)}{2}\right)\sin\left(\dfrac{\theta(a)}{2}-\dfrac{y}{2}\right)}$

Therefore,
$\displaystyle F(a)=\dfrac{1}{2}\int_0^{\theta(a)} \ln\left(\cot\left(\dfrac{y}{2}+\dfrac{\theta(a)}{2}\right)\right)dy+\dfrac{1}{2}\int_0^{\theta(a)} \ln\left(\cot\left(\dfrac{\theta(a)}{2}-\dfrac{y}{2}\right)\right)dy$

In the first intégral, perform the change of variable $x=\dfrac{\theta(a)}{2}+\dfrac{y}{2}$
In the second intégral, perform the change of variable $x=\dfrac{\theta(a)}{2}-\dfrac{y}{2}$

$\displaystyle F(a)= \int_{\tfrac{\theta(a)}{2}}^{\theta(a)}\ln(\cot x)dx+\int_0^{\tfrac{\theta(a)}{2}}\ln(\cot x)dx$

Therefore,

$\boxed{\displaystyle F(a)= \int_{0}^{\theta(a)}\ln(\cot x)dx}$

On the other hand,

$\boxed{\displaystyle \int_{0}^{\tfrac{\pi}{12}}\ln(\cot x)dx=\dfrac{2}{3}G}$

$\cos \tfrac{\pi}{12}=\dfrac{1}{4}\left(\sqrt{6}+\sqrt{2}\right)$

$\cos^2 \tfrac{\pi}{12}=\dfrac{2+\sqrt{3}}{4}$

$\dfrac{a-1}{a+1}=\dfrac{1-\tfrac{1}{a}}{1+\tfrac{1}{a}}=\dfrac{1-\dfrac{1}{4}\left(\sqrt{6}+\sqrt{2}\right)}{1+\dfrac{1}{4}\left(\sqrt{6}+\sqrt{2}\right)}=\dfrac{\tfrac{4}{\sqrt{6}+\sqrt{2}}-1}{\tfrac{4}{\sqrt{6}+\sqrt{2}}+1}=\dfrac{\sqrt{6}-\sqrt{2}-1}{\sqrt{6}-\sqrt{2}+1}$

$a^2-1=\dfrac{4}{2+\sqrt{3}}-1=\dfrac{2-\sqrt{3}}{2+\sqrt{3}}=\left(2-\sqrt{3}\right)^2$

$\dfrac{a^2+1}{a^2-1}=\dfrac{1+\tfrac{1}{a^2}}{1-\tfrac{1}{a^2}}=\dfrac{6+\sqrt{3}}{2-\sqrt{3}}=\left(6+\sqrt{3}\right)\left(2+\sqrt{3}\right)=15+8\sqrt{3}$

Therefore,

$(a^2-1)x^2-2(a^2+1)x+(a^2-1)=(a^2-1)\left(x^2-2\left( \dfrac{a^2+1}{a^2-1}\right)x+1\right)=\left(2-\sqrt{3}\right)^2\left(x^2-2(15+8\sqrt{3})x+1\right)$

and,

$\boxed{\displaystyle\int_0^{\tfrac{\sqrt{6}-\sqrt{2}-1}{\sqrt{6}-\sqrt{2}+1}} \dfrac{\ln x}{(x-1)\sqrt{x^2-2(15+8\sqrt{3})x+1}}dx=\dfrac{2}{3}(2-\sqrt{3})G}$