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Question Number 134272 by mnjuly1970 last updated on 01/Mar/21

$$\mathrm{nice}\mathrm{calculus}$$$$\mathrm{prove}\mathrm{that}::$$$$i::={\int }_0^{\mathrm{\infty }}\frac{cos\left(\pi x\right)}{e^{2\pi \sqrt{x}}-1}dx=\frac{2-\sqrt{2}}{8}$$$$ii::compute:\underset{n=-\mathrm{\infty }}{\sum ^{\mathrm{\infty }}}\frac{1}{n^4+9n^2+10}=?$$$$...m.n...$$

Answered by Dwaipayan Shikari last updated on 01/Mar/21

Commented by Dwaipayan Shikari last updated on 01/Mar/21

$${Ramanujan}^{\prime }sletter$$

Commented by mnjuly1970 last updated on 02/Mar/21

$$thankyou..$$

Answered by Dwaipayan Shikari last updated on 01/Mar/21

$$\underset{n=-\mathrm{\infty }}{\sum ^{\mathrm{\infty }}}\frac{1}{n^4+9n^2+10}=\frac{1}{10}+2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^4+9n^2+10}$$$$=\frac{1}{10}+\frac{2}{\sqrt{41}}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{(n^2+\frac{9-\sqrt{41}}{2})}-\frac{1}{(n^2+\frac{9+\sqrt{41}}{2})}$$$$=\frac{1}{10}+\sqrt{\frac{2}{41}}(\frac{\pi }{\sqrt{9-\sqrt{41}}}coth\left(\sqrt{\frac{9-\sqrt{41}}{2}}\pi \right)-\frac{\pi }{\sqrt{9+\sqrt{41}}}coth\left(\sqrt{\frac{9+\sqrt{41}}{2}}\pi \right))+\frac{2}{\sqrt{41}}(\frac{1}{9+\sqrt{41}}-\frac{1}{9-\sqrt{41}})$$$$=\sqrt{\frac{2}{41}}(\frac{\pi }{\sqrt{9-\sqrt{41}}}coth\left(\sqrt{\frac{9-\sqrt{41}}{2}}\pi \right)-\frac{\pi }{\sqrt{9+\sqrt{41}}}coth\left(\sqrt{\frac{9+\sqrt{41}}{2}}\pi \right))$$$$Using\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2+x^2}=\frac{\pi }{2x}coth\left(\pi x\right)-\frac{1}{2x^2}$$

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