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Question Number 28613 by abdo imad last updated on 27/Jan/18

let give x>0 and S(x)= ∫_0 ^∞ ((sint)/(e^(xt) −1))dt . developp S at form of series.

letgivex>0andS(x)=∫0∞sintext−1dt. developpSatformofseries.

Commented byabdo imad last updated on 01/Feb/18

S(x)= ∫_0 ^∞ ((e^(−xt) sint)/(1−e^(−xt) ))dt= ∫_0 ^∞ ( Σ_(n=0) ^∞ e^(−nxt) )e^(−xt) sintdt = Σ_(n=0) ^∞ ∫_0 ^∞ e^(−(n+1)xt) sint dt=Σ_(n=0) ^∞ A_n (x) A_n (x)= ∫_0 ^∞ e^(−(n+1)xt) sint dt=−Im( ∫_0 ^∞ e^(−((n+1)x +i)t) dt)but ∫_0 ^∞ e^(−((n+1)x+i)t) dt =−(1/((n+1)x+i))[ e^(−((n+1)x+i)t) ]_0 ^(+∞) = (1/((n+1)x+i))=(((n+1)x−i)/((n+1)^2 x^2 +1))=(((n+1)x)/(1+(n+1)^2 x^2 ))−(i/(1+(n+1)^2 x^2 )) A_n (x)= (1/(1+(n+1)^2 x^2 )) ⇒S(x)= Σ_(n=0) ^∞ (1/(1+(n+1)^2 x^2 )) with x>0

S(x)=∫0∞e−xtsint1−e−xtdt=∫0∞(∑n=0∞e−nxt)e−xtsintdt =∑n=0∞∫0∞e−(n+1)xtsintdt=∑n=0∞An(x) An(x)=∫0∞e−(n+1)xtsintdt=−Im(∫0∞e−((n+1)x+i)tdt)but ∫0∞e−((n+1)x+i)tdt=−1(n+1)x+i[e−((n+1)x+i)t]0+∞ =1(n+1)x+i=(n+1)x−i(n+1)2x2+1=(n+1)x1+(n+1)2x2−i1+(n+1)2x2 An(x)=11+(n+1)2x2⇒S(x)=∑n=0∞11+(n+1)2x2withx>0

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