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

nice calculus

prove that : :

i :: = \int_{0}^{\infty} \frac{c o s (π x)}{e^{2 π \sqrt{x}} - 1} d x = \frac{2 - \sqrt{2}}{8}

i i :: c o m p u t e : \underset{n = - \infty}{\sum^{\infty}} \frac{1}{n^{4} + 9 n^{2} + 10} = ?

\dots m . n \dots

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

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

{R a m a n u j a n}^{'} s l e t t e r

Commented by mnjuly1970 last updated on 02/Mar/21

t h a n k y o u . .

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

\underset{n = - \infty}{\sum^{\infty}} \frac{1}{n^{4} + 9 n^{2} + 10} = \frac{1}{10} + 2 \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{4} + 9 n^{2} + 10}

= \frac{1}{10} + \frac{2}{\sqrt{41}} \underset{n = 1}{\sum^{\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{π}{\sqrt{9 - \sqrt{41}}} c o t h (\sqrt{\frac{9 - \sqrt{41}}{2}} π) - \frac{π}{\sqrt{9 + \sqrt{41}}} c o t h (\sqrt{\frac{9 + \sqrt{41}}{2}} π)) + \frac{2}{\sqrt{41}} (\frac{1}{9 + \sqrt{41}} - \frac{1}{9 - \sqrt{41}})

= \sqrt{\frac{2}{41}} (\frac{π}{\sqrt{9 - \sqrt{41}}} c o t h (\sqrt{\frac{9 - \sqrt{41}}{2}} π) - \frac{π}{\sqrt{9 + \sqrt{41}}} c o t h (\sqrt{\frac{9 + \sqrt{41}}{2}} π))

U s i n g \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} + x^{2}} = \frac{π}{2 x} c o t h (π x) - \frac{1}{2 x^{2}}