Prove that Ω= ∫_0 ^( ∞) (( sin(x))/(e^( x) −1)) dx =^? (1/2) ( πcoth(π) −1 ) −−− solution −−− Ω= ∫_0 ^( ∞) (( e^( −x) .sin(x))/(1− e^( −x) )) dx=∫_0 ^( ∞) (sin(x) Σ_(n=1) ^∞ e^( −nx) )dx = Σ_(n=1) ^∞ ∫_0 ^( ∞) e^( −nx) .sin(x)dx = Σ_(n=1) ^∞ (( 1)/(1 + n^( 2) )) =_(function) ^(Upsilon) (1/2) ( πcoth(π) − 1) ■ m.n Note : Υ (s )= Σ_(n=1) ^∞ (1/( s^( 2) + n^( 2) )) = (1/(2s))( πcoth(πs) −(1/(2s ))) where : s ∈ C − { ki∈ Z : k≠ 0 }


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Question Number 169115 by mnjuly1970 last updated on 24/Apr/22
$Prove that Ω= ∫_0 ^( ∞) (( sin(x))/(e^( x) −1)) dx =^? (1/2) ( πcoth(π) −1 ) −−− solution −−− Ω= ∫_0 ^( ∞) (( e^( −x) .sin(x))/(1− e^( −x) )) dx=∫_0 ^( ∞) (sin(x) Σ_(n=1) ^∞ e^( −nx) )dx = Σ_(n=1) ^∞ ∫_0 ^( ∞) e^( −nx) .sin(x)dx = Σ_(n=1) ^∞ (( 1)/(1 + n^( 2) )) =_(function) ^(Upsilon) (1/2) ( πcoth(π) − 1) ■ m.n Note : Υ (s )= Σ_(n=1) ^∞ (1/( s^( 2) + n^( 2) )) = (1/(2s))( πcoth(πs) −(1/(2s ))) where : s ∈ C − { ki∈ Z : k≠ 0 }$
$Prove that$ $Ω = \int_{0}^{\infty} \frac{s i n (x)}{e^{x} - 1} d x \overset{?}{=} \frac{1}{2} (π c o t h (π) - 1)$ $- - - s o l u t i o n - - -$ $Ω = \int_{0}^{\infty} \frac{e^{- x} . s i n (x)}{1 - e^{- x}} d x = \int_{0}^{\infty} (s i n (x) \underset{n = 1}{\sum^{\infty}} e^{- n x}) d x$ $= \underset{n = 1}{\sum^{\infty}} \int_{0}^{\infty} e^{- n x} . s i n (x) d x$ $= \underset{n = 1}{\sum^{\infty}} \frac{1}{1 + n^{2}} \underset{f u n c t i o n}{\overset{U p s i l o n}{=}} \frac{1}{2} (π c o t h (π) - 1) ◼ m . n$ $N o t e : Υ (s) = \underset{n = 1}{\sum^{\infty}} \frac{1}{s^{2} + n^{2}} = \frac{1}{2 s} (π c o t h (π s) - \frac{1}{2 s})$ $w h e r e : s \in C - {k i \in Z : k \neq 0}$
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