prove-that-0-x-2-cosh-x-dx-pi-3-8-

Question Number 174685 by mnjuly1970 last updated on 08/Aug/22

prove that : Ω = ∫_0 ^( ∞) (( x^( 2) )/(cosh(x ))) dx = (π^( 3) /( 8))

$$ \\ $$$$\:\:\:\:\:{prove}\:{that}\:: \\ $$$$\: \\ $$$$\:\:\:\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:\:{x}^{\:\mathrm{2}} }{{cosh}\left({x}\:\right)}\:{dx}\:=\:\frac{\pi^{\:\mathrm{3}} }{\:\mathrm{8}} \\ $$$$ \\ $$

Answered by princeDera last updated on 08/Aug/22

(1/(cosh(x))) = ((2e^(−x) )/(1 + e^(−2x) )) Ω = ∫_0 ^∞ ((2x^2 e^(−x) )/(1+cosh (x)))dx = 2∫_0 ^∞ x^2 e^(−x) Σ_(k≥0) (−1)^k e^(−2kx) dx = 2Σ_(k≥0) (−1)^k ∫_0 ^∞ x^2 e^(−(2k+1)x) dx = 4Σ_(k≥0) (((−1)^k )/((2k+1)^3 )) 4B(3) = 4((π^3 /(32))) = (π^3 /8) where B(z) is the dirichlet beta function

$$ \\ $$$$ \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{co}{sh}\left({x}\right)}\:=\:\frac{\mathrm{2}{e}^{−{x}} }{\mathrm{1}\:+\:{e}^{−\mathrm{2}{x}} } \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{x}^{\mathrm{2}} {e}^{−{x}} }{\mathrm{1}+\mathrm{cosh}\:\left({x}\right)}{dx}\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{2}} {e}^{−{x}} \underset{{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{k}} {e}^{−\mathrm{2}{kx}} {dx} \\ $$$$=\:\mathrm{2}\underset{{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{k}} \int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{2}} {e}^{−\left(\mathrm{2}{k}+\mathrm{1}\right){x}} {dx}\:=\:\mathrm{4}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$ \\ $$$$\mathrm{4}\mathfrak{B}\left(\mathrm{3}\right)\:=\:\mathrm{4}\left(\frac{\pi^{\mathrm{3}} }{\mathrm{32}}\right)\:=\:\frac{\pi^{\mathrm{3}} }{\mathrm{8}} \\ $$$${where}\:\mathfrak{B}\left({z}\right)\:{is}\:{the}\:{dirichlet}\:{beta}\:{function} \\ $$$$ \\ $$$$ \\ $$

Commented by MME last updated on 08/Aug/22