prove-Use-the-residus-form-0-xsinx-x-2-1-2-dx-pi-4e-

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

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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