对数积分系列

Question

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

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

Solution
Let

\[\Large J=\displaystyle \int_0^{3-2\sqrt{2}} \dfrac{\ln x}{(x-1)\sqrt{x^2-6x+1}}dx \]

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

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

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

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

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

\[\Large \displaystyle x-1=\dfrac{2}{\sqrt{2}\cos y-1} \]

Therefore,

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

On the other hand,

\[\Large \sqrt{2}\cos x+1=\sqrt{2}\left(\cos x+\cos\left(\dfrac{\pi}{4}\right)\right)=2\sqrt{2}\cos\left(\dfrac{x}{2}+\dfrac{\pi}{8}\right)\cos\left(\dfrac{\pi}{8}-\dfrac{x}{2}\right) \]

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

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

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

Therefore,

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

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,

\[\Large J=\displaystyle \int_\tfrac{\pi}{8}^{\tfrac{\pi}{4}} \ln(\cot(x))dx+\int_0^{\tfrac{\pi}{8}}\ln(\cot(x))dx \]

Therefore,

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

Perform the change of variable \(y=\tan x\),

\[\Large \boxed{\displaystyle\color{blue} {J=-\int_0^1 \dfrac{\ln(x)}{1+x^2}dx=G}} \]

\(G\) is the Catalan constant.

Let,

\[\Large K=\displaystyle \int_0^{3-2\sqrt{2}} \dfrac{\ln x}{\sqrt{x^2-6x+1}}dx \]

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

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

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

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

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

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 \(y=\dfrac{\pi}{8}-\dfrac{x}{2}\)

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

Therefore,

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

On the other hand,

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

Therefore,

\[\Large{\displaystyle\color{blue} {K=\int_0^{\tfrac{\pi}{4}}\dfrac{(1+(\tan x)^2)\ln(\tan x)}{1+\tan x}dx} } \]

perform the change of variable \(y=\tan x\),

\[\Large\boxed{\displaystyle\color{blue}{K=\int_0^1 \dfrac{\ln x}{1+x}dx=-\dfrac{\pi^2}{12}}} \]