let x>0 calculate f(x) =∫_0 ^(+∞) e^(−t) ∣sin(xt)∣ dt


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Question Number 42188 by maxmathsup by imad last updated on 19/Aug/18

$l e t x > 0 c a l c u l a t e f (x) = \int_{0}^{+ \infty} e^{- t} ∣ s i n (x t) ∣ d t$
Commented bymaxmathsup by imad last updated on 23/Aug/18
$changement xt =u give f(x) =∫_0 ^∞ e^(−(u/x)) ∣sin(u)∣(du/x) =(1/x) ∫_0 ^∞ e^(−(u/x)) ∣sin(u)∣du ⇒xf(x)=Σ_(n=0) ^∞ ∫_(nπ) ^((n+1)π) e^(−(u/x)) ∣sinu∣ du =_(u =nπ +t) Σ_(n=0) ^∞ ∫_0 ^π e^(−((nπ+t)/x)) ∣sin(nπ+t)∣dt =Σ_(n=0) ^∞ ∫_0 ^π e^(−((nπ)/x)) e^(t/x) sint dt =Σ_(n=0) ^∞ e^(−((nπ)/x)) ∫_0 ^π e^(t/x) sint dt but ∫_0 ^π e^(t/x) sint dt = Im( ∫_0 ^π e^((t/x)+it) dt) and ∫_0 ^π e^(((1/x)+i)t) dt =[(1/((1/x)+i)) e^(((1/x)+i)t) ]_0 ^π =(x/(1+xi)){ e^((π/x) +iπ) −1}=((x(1−xi))/(1+x^2 )){−e^(π/x) −1} =((x(−1+xi)(1+e^(π/x) ))/(1+x^2 )) =((1+e^(π/x) )/(1+x^2 ))(−x +x^2 i) ⇒∫_0 ^π e^(t/x) sint dt=(x^2 /(1+x^2 ))(1+e^(π/x) ) ⇒ xf(x)=(x^2 /(1+x^2 ))(1+e^(π/x) )Σ_(n=0) ^∞ (e^(−(π/x)) )^n =(x^2 /(1+x^2 ))(1+e^(π/x) ) (1/(1−e^(−(π/x)) )) ⇒ xf(x) =(x^2 /(1+x^2 )) ((1+e^(π/x) )/(1−e^(−(π/x)) )) ⇒ f(x)= (x/(1+x^2 )) ((1+e^(π/x) )/(1−e^(−(π/x)) )) .$
$c h a n g e m e n t x t = u g i v e f (x) = \int_{0}^{\infty} e^{- \frac{u}{x}} ∣ s i n (u) ∣ \frac{d u}{x}$ $= \frac{1}{x} \int_{0}^{\infty} e^{- \frac{u}{x}} ∣ s i n (u) ∣ d u \Rightarrow x f (x) = \sum_{n = 0}^{\infty} \int_{n π}^{(n + 1) π} e^{- \frac{u}{x}} ∣ s i n u ∣ d u$ $=_{u = n π + t} \sum_{n = 0}^{\infty} \int_{0}^{π} e^{- \frac{n π + t}{x}} ∣ s i n (n π + t) ∣ d t$ $= \sum_{n = 0}^{\infty} \int_{0}^{π} e^{- \frac{n π}{x}} e^{\frac{t}{x}} s i n t d t = \sum_{n = 0}^{\infty} e^{- \frac{n π}{x}} \int_{0}^{π} e^{\frac{t}{x}} s i n t d t b u t$ $\int_{0}^{π} e^{\frac{t}{x}} s i n t d t = I m (\int_{0}^{π} e^{\frac{t}{x} + i t} d t) a n d \int_{0}^{π} e^{(\frac{1}{x} + i) t} d t$ $= {[\frac{1}{\frac{1}{x} + i} e^{(\frac{1}{x} + i) t}]}_{0}^{π} = \frac{x}{1 + x i} {e^{\frac{π}{x} + i π} - 1} = \frac{x (1 - x i)}{1 + x^{2}} {- e^{\frac{π}{x}} - 1}$ $= \frac{x (- 1 + x i) (1 + e^{\frac{π}{x}})}{1 + x^{2}} = \frac{1 + e^{\frac{π}{x}}}{1 + x^{2}} (- x + x^{2} i) \Rightarrow \int_{0}^{π} e^{\frac{t}{x}} s i n t d t = \frac{x^{2}}{1 + x^{2}} (1 + e^{\frac{π}{x}}) \Rightarrow$ $x f (x) = \frac{x^{2}}{1 + x^{2}} (1 + e^{\frac{π}{x}}) \sum_{n = 0}^{\infty} {(e^{- \frac{π}{x}})}^{n} = \frac{x^{2}}{1 + x^{2}} (1 + e^{\frac{π}{x}}) \frac{1}{1 - e^{- \frac{π}{x}}} \Rightarrow$ $x f (x) = \frac{x^{2}}{1 + x^{2}} \frac{1 + e^{\frac{π}{x}}}{1 - e^{- \frac{π}{x}}} \Rightarrow f (x) = \frac{x}{1 + x^{2}} \frac{1 + e^{\frac{π}{x}}}{1 - e^{- \frac{π}{x}}} .$
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