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Question Number 190622 by mnjuly1970 last updated on 07/Apr/23

$$$$$$prove:$$$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}(\frac{x}{{\underset{―}{⋮})}^2\mathrm{d}x=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k{}^2}⋖}$$

Answered by leodera last updated on 17/May/23

$$$$$$\cosh {}^2\left(x\right)={\left(\frac{e^x+e^{-x}}{2}\right)}^2=\frac{e^{2x}}{4}{(1+e^{-2x})}^2$$$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{x^2}{\cosh {}^2\left(x\right)}dx=8{\int }_0^{\mathrm{\infty }}\frac{x^2{e}^{-2x}}{{(1+e^{-2x})}^2}dx$$$$=8{\int }_0^{\mathrm{\infty }}{x}^2{e}^{-2x}\underset{k=0}{\sum ^{\mathrm{\infty }}}{e}^{-2kx}{(-1)}^k(1+k)dx$$$$=8\underset{k=0}{\sum ^{\mathrm{\infty }}}{(-1)}^k(1+k){\int }_0^{\mathrm{\infty }}{x}^2{e}^{-(2+2k)x}dx$$$$=16\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^k(1+k)}{2^3{(1+k)}^3}=2\underset{k=0}{\sum ^{\mathrm{\infty }}}{(-1)}^k\frac{1}{{(1+k)}^2}$$$$2\underset{k=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{k-1}\frac{1}{k^2}=2\eta \left(2\right)=2(1-2^{1-2})\zeta \left(2\right)$$$$=\zeta \left(2\right)=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k^2}=\frac{{\pi }^2}{6}$$$$$$

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