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Question Number 131637 by rs4089 last updated on 07/Feb/21

Answered by Dwaipayan Shikari last updated on 07/Feb/21

∫_(−∞) ^∞ ((sinx)/(x(x^2 +1)))dx=∫_(−∞) ^∞ ((sinx)/x)−((xsinx)/((x^2 +1)))dx=π−(π/e)  I(α)=∫_(−∞) ^∞ ((cosαx)/((x^2 +1)))=πe^(−α)   I′(α)=∫_(−∞) ^∞ ((xsin(αx))/(x^2 +1))dx=πe^(−α)   I′(1)=(π/e)

sinxx(x2+1)dx=sinxxxsinx(x2+1)dx=ππeI(α)=cosαx(x2+1)=πeαI(α)=xsin(αx)x2+1dx=πeαI(1)=πe

Commented by Lordose last updated on 07/Feb/21

why I′(α)=πe^(−α)  and not −απe^(−α)

whyI(α)=πeαandnotαπeα

Commented by Dwaipayan Shikari last updated on 07/Feb/21

I(α)=πe^(−α)   I′(α)=−πe^(−α)   Also I′(α)=−∫_(−∞) ^∞ ((xsin(αx))/(x^2 +1))  So πe^(−α) =∫_0 ^∞ ((x sin(αx))/(x^2 +1))dx

I(α)=πeαI(α)=πeαAlsoI(α)=xsin(αx)x2+1Soπeα=0xsin(αx)x2+1dx

Answered by mathmax by abdo last updated on 07/Feb/21

Φ=∫_(−∞) ^(+∞)  ((sinx)/(x^3 +x))dx ⇒Φ =∫_(−∞) ^(+∞)  ((sinx)/(x(x^2 +1)))dx  =∫_(−∞) ^(+∞)  sinx((1/x)−(x/(x^2 +1)))dx =∫_(−∞) ^(+∞)  ((sinx)/x)dx−∫_(−∞) ^(+∞)  ((xsinx)/(x^2  +1))dx  =π−∫_(−∞) ^(+∞)  ((xsinx)/(x^2 +1))dx =π−I  I=Im(∫_(−∞) ^(+∞)  ((xe^(ix) )/(x^2  +1))dx) but ∫_(−∞) ^(+∞)  ((xe^(ix) )/(x^2 +1))dx  =2iπRes(f,i) =2iπ.((ie^(−1) )/(2i)) =πie^(−1)  ⇒I =(π/e) ⇒ Φ =π−(π/e)

Φ=+sinxx3+xdxΦ=+sinxx(x2+1)dx=+sinx(1xxx2+1)dx=+sinxxdx+xsinxx2+1dx=π+xsinxx2+1dx=πII=Im(+xeixx2+1dx)but+xeixx2+1dx=2iπRes(f,i)=2iπ.ie12i=πie1I=πeΦ=ππe

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