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Question Number 139612 by 676597498 last updated on 29/Apr/21

$$showthat$$$${\int }_0^{\mathrm{\infty }}\frac{cos\left(\sqrt{x}\right)}{e^{2\pi \sqrt{x}}-1}dx=1-\frac{e}{{(e-1)}^2}$$

Answered by Dwaipayan Shikari last updated on 29/Apr/21

$$2{\int }_0^{\mathrm{\infty }}u\frac{cos\left(u\right)}{e^{2\pi u}-1}du$$$$=2\underset{n=1}{\sum ^{\mathrm{\infty }}}{\int }_0^{\mathrm{\infty }}{ue}^{-2\pi nu}cos\left(u\right)du=\underset{n=1}{\sum ^{\mathrm{\infty }}}{\int }_0^{\mathrm{\infty }}{ue}^{-u(2\pi n-i)}+{ue}^{-(2\pi n+i)}du$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(2\pi n-i)}^2}+\frac{1}{{(2\pi n+i)}^2}=\frac{1}{4}(\underset{n=-\mathrm{\infty }}{\sum ^{\mathrm{\infty }}}\frac{1}{{(\pi n+\frac{i}{2})}^2}-\frac{1}{{\left(\frac{i}{2}\right)}^2})$$$$=\frac{1}{4}.\frac{1}{{sin}^2\left(\frac{i}{2}\right)}+1=1-\frac{1}{{(e^{\frac{1}{2}}-e^{-\frac{1}{2}})}^2}=1-\frac{e}{{(e-1)}^2}$$

Commented by 676597498 last updated on 29/Apr/21

$$thanks$$$$plsexplainthesecondline$$

Commented by Dwaipayan Shikari last updated on 30/Apr/21

$${\int }_0^{\mathrm{\infty }}{ue}^{-u(2\pi n-i)}du+{ue}^{-u(2\pi n+i)}duu=\frac{y}{(2\pi n-i)}u=\frac{k}{(2\pi n+i)}$$$$=\frac{1}{{(2\pi n-i)}^2}{\int }_0^{\mathrm{\infty }}{ye}^{-y}dy+\frac{1}{{(2\pi n+i)}^2}{\int }_0^{\mathrm{\infty }}{ke}^{-k}dk$$$$=\frac{\mathrm{\Gamma }\left(2\right)}{{(2\pi n-i)}^2}+\frac{\mathrm{\Gamma }\left(2\right)}{{(2\pi n+i)}^2}=\frac{1}{{(2\pi n-i)}^2}+\frac{1}{{(2\pi n+i)}^2}$$

Commented by 676597498 last updated on 30/Apr/21

$$sorrysiriunderstoodthis$$$$imeantthethirdlinesir$$

Commented by Dwaipayan Shikari last updated on 30/Apr/21

$$Generally$$$$sinx=x\underset{n=1}{\prod ^{\mathrm{\infty }}}(1-\frac{x^2}{{\pi }^2{n}^2})$$$$log\left(sinx\right)=log\left(x\right)+\mathrm{\Sigma }log(1-\frac{x}{\pi n})+\mathrm{\Sigma }log(1+\frac{x}{\pi n})$$$$Differentiatebothsidesrespecttox$$$$\frac{cosx}{sinx}=\frac{1}{x}+\mathrm{\Sigma }\frac{-\frac{1}{\pi n}}{1-\frac{x}{\pi n}}+\mathrm{\Sigma }\frac{\frac{1}{\pi n}}{1+\frac{x}{\pi n}}$$$$cotx=\frac{1}{x}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{x-\pi n}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{x+\pi n}$$$$cotx=\underset{n=-\mathrm{\infty }}{\sum ^{\mathrm{\infty }}}\frac{1}{x+\pi n}\Rightarrow \frac{1}{{sin}^{2}x}=\underset{n=-\mathrm{\infty }}{\sum ^{\mathrm{\infty }}}\frac{1}{{(x+\pi n)}^2}\left(Differentiateagain\right)$$

Commented by mnjuly1970 last updated on 30/Apr/21

$$thanksalot...$$

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