prove: Use the residus form ∫_0 ^(+∞) ((xsinx)/((x^2 +1)^2 ))dx=(π/(4e))


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Question Number 175012 by lapache last updated on 16/Aug/22

$p r o v e : U s e t h e r e s i d u s f o r m$ $\int_{0}^{+ \infty} \frac{x s i n x}{{(x^{2} + 1)}^{2}} d x = \frac{π}{4 e}$
	Answered by Mathspace last updated on 16/Aug/22
	$letI=∫_0 ^∞ ((xsinx)/((x^2 +1)^2 )) ⇒ 2I=∫_(−∞) ^(+∞) ((xsinx)/((x^2 +1)^2 ))dx =Im(∫_(−∞) ^(+∞) ((xe^(ix) )/((x^2 +1)^2 ))dx) Ψ(z)=((ze^(iz) )/((z^2 +1)^2 )) poles of Ψ? Ψ(z)=((ze^(iz) )/((z−i)^2 (z+i)^2 )) so the poles are +^− i(doubles) ∫_(−∞) ^(+∞) Ψ(z)dz=2iπRes(Ψ,i) Res(Ψ,i)=lim_(z→i) (1/((2−1)!)){(z−i)^2 Ψ(z)}^((1)) =lim_(z→i) {((ze^(iz) )/((z+i)^2 ))}^((1)) =lim_(z→i) (((e^(iz) +ize^(iz) )(z+i)^2 −2(z+i)ze^(iz) )/((z+i)^4 )) =lim_(z→i) (((1+iz)e^(iz) (z+i)−2ze^(iz) )/((z+i)^3 )) =((−2ie^(−1) )/((2i)^3 ))=((−2ie^(−1) )/(−8i))=(1/(4e)) ∫_(−∞) ^(+∞) Ψ(z)dz=2iπ.(1/(4e))=((iπ)/(2e)) so 2I=(π/(2e)) ⇒★I=(π/(4e))★$
	$l e t I = \int_{0}^{\infty} \frac{x s i n x}{{(x^{2} + 1)}^{2}} \Rightarrow$ $2 I = \int_{- \infty}^{+ \infty} \frac{x s i n x}{{(x^{2} + 1)}^{2}} d x$ $= I m (\int_{- \infty}^{+ \infty} \frac{{x e}^{i x}}{{(x^{2} + 1)}^{2}} d x)$ $Ψ (z) = \frac{{z e}^{i z}}{{(z^{2} + 1)}^{2}} p o l e s o f Ψ ?$ $Ψ (z) = \frac{{z e}^{i z}}{{(z - i)}^{2} {(z + i)}^{2}}$ $s o t h e p o l e s a r e \bar{+} i (d o u b l e s)$ $\int_{- \infty}^{+ \infty} Ψ (z) d z = 2 i π R e s (Ψ, i)$ $R e s (Ψ, i) = {l i m}_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} Ψ (z)}^{(1)}$ $= {l i m}_{z \to i} {\frac{{z e}^{i z}}{{(z + i)}^{2}}}^{(1)}$ $= {l i m}_{z \to i} \frac{(e^{i z} + {i z e}^{i z}) {(z + i)}^{2} - 2 (z + i) {z e}^{i z}}{{(z + i)}^{4}}$ $= {l i m}_{z \to i} \frac{(1 + i z) e^{i z} (z + i) - 2 {z e}^{i z}}{{(z + i)}^{3}}$ $= \frac{- 2 {i e}^{- 1}}{{(2 i)}^{3}} = \frac{- 2 {i e}^{- 1}}{- 8 i} = \frac{1}{4 e}$ $\int_{- \infty}^{+ \infty} Ψ (z) d z = 2 i π . \frac{1}{4 e} = \frac{i π}{2 e}$ $s o 2 I = \frac{π}{2 e} \Rightarrow ★ I = \frac{π}{4 e} ★$
	Commented by Mathspace last updated on 16/Aug/22

	$t h a n k y o u$
	Commented by Tawa11 last updated on 16/Aug/22

	$Great sir$
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