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Question Number 140715 by mnjuly1970 last updated on 11/May/21

$$$$$$......advancedcalculus......$$$$provethat:$$$$\mathit{\varphi }:={\int }_0^{\mathrm{\infty }}\frac{x^2}{{cosh}^2\left(x^2\right)}dx\stackrel{?}{=}\frac{\sqrt{2}-2}{4}\sqrt{\pi }\zeta \left(\frac{1}{2}\right)$$$$..............$$

Answered by Dwaipayan Shikari last updated on 11/May/21

$${\int }_0^{\mathrm{\infty }}\frac{x^2}{{cosh}^2\left(x^2\right)}dx=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{\sqrt{u}}{{cosh}^2\left(u\right)}du$$$$={\int }_0^{\mathrm{\infty }}\frac{\sqrt{u}}{1+cosh\left(2u\right)}=2{\int }_0^{\mathrm{\infty }}\frac{e^{-2u}\sqrt{u}}{{(1+e^{-2u})}^2}du$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{n+1}{e}^{-2nu}=\frac{1}{1+e^{-2u}}\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{n+1}{ne}^{-2nu}=\frac{-e^{2u}}{{(1+e^{-2u})}^2}$$$$=-2{\int }_0^{\mathrm{\infty }}\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{n+1}\sqrt{u}{ne}^{-2nu}du=2\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{n}n{\int }_0^{\mathrm{\infty }}\sqrt{u}{e}^{-2nu}du$$$$2nu=t\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^n{\int }_0^{\mathrm{\infty }}\frac{\sqrt{t}}{\sqrt{2n}}{e}^{-t}dt=-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{\sqrt{\pi }}{\sqrt{2}\sqrt{n}}{(-1)}^{n+1}$$$$=-\sqrt{\frac{\pi }{2}}\zeta \left(\frac{1}{2}\right)(1-\frac{1}{2^{\frac{1}{2}-1}})=\sqrt{\frac{\pi }{2}}(\sqrt{2}-1)\zeta \left(\frac{1}{2}\right)$$$$$$

Commented by mnjuly1970 last updated on 12/May/21

$$grateful....$$

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