prsategazd

2022-01-02

I want to solve this integral :
${\int }_{0}^{\frac{\pi }{2}}\mathrm{cot}x\mathrm{ln}\left(\mathrm{sec}x\right)dx$
I tried the following substitution : $\mathrm{ln}\left(\mathrm{sec}x\right)=t$ which means $dt=\mathrm{tan}xdx$
$I={\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{cot}x}{\mathrm{tan}x}tdt={\int }_{0}^{\mathrm{\infty }}\frac{t}{{\mathrm{tan}}^{2}x}dt$
I'm really disturbed by the ${\mathrm{tan}}^{2}x$, I tried also to substitute $\mathrm{sec}x=t$ but it's not helpful either. Any helpful approach to solve this problem ?

Charles Benedict

Substitute $t={\mathrm{tan}}^{2}x$
${\int }_{0}^{\frac{\pi }{2}}\mathrm{cot}x\mathrm{ln}\left(\mathrm{sec}x\right)dx={\frac{14}{\int }}_{0}^{1}\frac{\mathrm{ln}\left(1+t\right)}{t\left(1+t\right)}dt+{\frac{14}{\int }}_{1}^{\mathrm{\infty }}\frac{\mathrm{ln}\left(1+t\right)}{t\left(1+t\right)}dt$
$=\frac{1}{4}{\int }_{0}^{1}\frac{\mathrm{ln}\left(1+t\right)}{t}dt-{\frac{14}{\int }}_{0}^{1}\frac{\mathrm{ln}t}{1+t}dt$
$={\frac{12}{\int }}_{0}^{1}\frac{\mathrm{ln}\left(1+t\right)}{t}dt=\frac{12}{\cdot }\frac{{\pi }^{2}}{12}=\frac{{\pi }^{2}}{24}$

Fasaniu

Another possibility, using the Basel problem formula $\sum _{k\ge 1}{k}^{-2}=\frac{{\pi }^{2}}{6}$
We have
$I=-{\int }_{0}^{\frac{\pi }{2}}\frac{\mathrm{cos}x}{\mathrm{sin}x}\mathrm{ln}\mathrm{cos}xdx=-{\frac{12}{\int }}_{0}^{\frac{\pi }{2}}\frac{\mathrm{cos}x}{\mathrm{sin}x}\mathrm{ln}\left(1-{\mathrm{sin}}^{2}x\right)dx$
$={\frac{12}{\sum }}_{k\ge 1}\frac{1}{k}{\int }_{0}^{\frac{\pi }{2}}\mathrm{cos}\left(x\right){\mathrm{sin}}^{2k-1}\left(x\right)dx$
The integral in the summation is evaluated as
${\int }_{0}^{\frac{\pi }{2}}\mathrm{cos}\left(x\right){\mathrm{sin}}^{2k-1}\left(x\right)dx=\frac{1}{2k}{\int }_{0}^{\frac{\pi }{2}}d\left({\mathrm{sin}}^{2k}\left(x\right)\right)dx=\frac{1}{2k}$
Hence
$I={\frac{14}{\sum }}_{k\ge 1}\frac{1}{{k}^{2}}=\frac{{\pi }^{2}}{24}$

Vasquez

${\int }_{0}^{2\pi }\mathrm{cot}x\mathrm{ln}\left(\mathrm{sec}x\right)dx=\frac{1}{4}{\int }_{0}^{\frac{\pi }{2}}\frac{-2\mathrm{cos}x\mathrm{sin}x\mathrm{ln}\left({\mathrm{cos}}^{2}x\right)}{1-{\mathrm{cos}}^{2}x}dx$
$\stackrel{\stackrel{t={\mathrm{cos}}^{2}x}{dt=-2\mathrm{cos}x\mathrm{sin}xdx}}{=}\frac{1}{4}{\int }_{0}^{1}\frac{\mathrm{ln}t}{t-1}dt=\frac{1}{4}\cdot \frac{{\pi }^{2}}{6}=\frac{{\pi }^{2}}{24}$

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