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

s h o w t h a t

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

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

2 \int_{0}^{\infty} u \frac{c o s (u)}{e^{2 π u} - 1} d u

= 2 \underset{n = 1}{\sum^{\infty}} \int_{0}^{\infty} {u e}^{- 2 π n u} c o s (u) d u = \underset{n = 1}{\sum^{\infty}} \int_{0}^{\infty} {u e}^{- u (2 π n - i)} + {u e}^{- (2 π n + i)} d u

= \underset{n = 1}{\sum^{\infty}} \frac{1}{{(2 π n - i)}^{2}} + \frac{1}{{(2 π n + i)}^{2}} = \frac{1}{4} (\underset{n = - \infty}{\sum^{\infty}} \frac{1}{{(π n + \frac{i}{2})}^{2}} - \frac{1}{{(\frac{i}{2})}^{2}})

= \frac{1}{4} . \frac{1}{{s i n}^{2} (\frac{i}{2})} + 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

t h a n k s

p l s e x p l a i n t h e s e c o n d l i n e

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

\int_{0}^{\infty} {u e}^{- u (2 π n - i)} d u + {u e}^{- u (2 π n + i)} d u u = \frac{y}{(2 π n - i)} u = \frac{k}{(2 π n + i)}

= \frac{1}{{(2 π n - i)}^{2}} \int_{0}^{\infty} {y e}^{- y} d y + \frac{1}{{(2 π n + i)}^{2}} \int_{0}^{\infty} {k e}^{- k} d k

= \frac{Γ (2)}{{(2 π n - i)}^{2}} + \frac{Γ (2)}{{(2 π n + i)}^{2}} = \frac{1}{{(2 π n - i)}^{2}} + \frac{1}{{(2 π n + i)}^{2}}

Commented by 676597498 last updated on 30/Apr/21

s o r r y s i r i u n d e r s t o o d t h i s

i m e a n t t h e t h i r d l i n e s i r

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

G e n e r a l l y

s i n x = x \underset{n = 1}{\prod^{\infty}} (1 - \frac{x^{2}}{π^{2} n^{2}})

l o g (s i n x) = l o g (x) + Σ l o g (1 - \frac{x}{π n}) + Σ l o g (1 + \frac{x}{π n})

D i f f e r e n t i a t e b o t h s i d e s r e s p e c t t o x

\frac{c o s x}{s i n x} = \frac{1}{x} + Σ \frac{- \frac{1}{π n}}{1 - \frac{x}{π n}} + Σ \frac{\frac{1}{π n}}{1 + \frac{x}{π n}}

c o t x = \frac{1}{x} + \underset{n = 1}{\sum^{\infty}} \frac{1}{x - π n} + \underset{n = 1}{\sum^{\infty}} \frac{1}{x + π n}

c o t x = \underset{n = - \infty}{\sum^{\infty}} \frac{1}{x + π n} \Rightarrow \frac{1}{{s i n}^{2} x} = \underset{n = - \infty}{\sum^{\infty}} \frac{1}{{(x + π n)}^{2}} (D i f f e r e n t i a t e a g a i n)

Commented by mnjuly1970 last updated on 30/Apr/21

t h a n k s a l o t \dots