let I = ∫_0 ^∞ e^(−tx) ∣sint∣dt with x>0 find the value of I .


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Question Number 35589 by abdo mathsup 649 cc last updated on 20/May/18

$l e t I = \int_{0}^{\infty} e^{- t x} ∣ s i n t ∣ d t w i t h x > 0$ $f i n d t h e v a l u e o f I .$
Commented by abdo mathsup 649 cc last updated on 21/May/18
$I = Σ_(n=0) ^∞ ∫_(nπ) ^((n+1)π) e^(−tx) ∣sint∣dt changement t =nπ + u give I = Σ_(n=0) ^∞ ∫_0 ^π e^(−nπx) e^(−xu) ∣sinu∣du = Σ_(n=0) ^∞ e^(−nπx) ∫_0 ^π e^(−xu) sinu du but A(x)=∫_0 ^π e^(−xu) sin(u)du =Im( ∫_0 ^π e^(−xu +iu) du) =Im( ∫_0 ^π e^((−x+i)u) du) but ∫_0 ^π e^((−x+i)u) du =[ (1/(−x+i)) e^((−x+i)u) ]_0 ^π =((−1)/(x−i)){ e^(−xπ +iπ) −1}= ((1 +e^(−πx) )/(x−i)) =((x+i)/(x^2 +1))( 1+e^(−πx) )⇒ A(x)= ((1+e^(−πx) )/(1+x^2 )) Σ_(n=0) ^∞ e^(−nπx) = Σ_(n=0) ^∞ (e^(−πx) )^n = (1/(1−e^(−πx) )) so I = (1/(1−e^(−πx) )) ((1 +e^(−πx) )/(1+x^2 )) ⇒ I = ((1+e^(−πx) )/((1+x^2 )(1−e^(−πx) ))) .$
$I = \sum_{n = 0}^{\infty} \int_{n π}^{(n + 1) π} e^{- t x} ∣ s i n t ∣ d t c h a n g e m e n t$ $t = n π + u g i v e$ $I = \sum_{n = 0}^{\infty} \int_{0}^{π} e^{- n π x} e^{- x u} ∣ s i n u ∣ d u$ $= \sum_{n = 0}^{\infty} e^{- n π x} \int_{0}^{π} e^{- x u} s i n u d u$ $b u t A (x) = \int_{0}^{π} e^{- x u} s i n (u) d u = I m (\int_{0}^{π} e^{- x u + i u} d u)$ $= I m (\int_{0}^{π} e^{(- x + i) u} d u) b u t$ $\int_{0}^{π} e^{(- x + i) u} d u = {[\frac{1}{- x + i} e^{(- x + i) u}]}_{0}^{π}$ $= \frac{- 1}{x - i} {e^{- x π + i π} - 1} = \frac{1 + e^{- π x}}{x - i}$ $= \frac{x + i}{x^{2} + 1} (1 + e^{- π x}) \Rightarrow A (x) = \frac{1 + e^{- π x}}{1 + x^{2}}$ $\sum_{n = 0}^{\infty} e^{- n π x} = \sum_{n = 0}^{\infty} {(e^{- π x})}^{n} = \frac{1}{1 - e^{- π x}} s o$ $I = \frac{1}{1 - e^{- π x}} \frac{1 + e^{- π x}}{1 + x^{2}} \Rightarrow I = \frac{1 + e^{- π x}}{(1 + x^{2}) (1 - e^{- π x})} .$
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