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Question Number 110214 by mathdave last updated on 27/Aug/20

s h o w t h a t

\int_{- \infty}^{+ \infty} \frac{(1 + \frac{x}{π}) \sin (π x)}{x^{2} + 4 x + 5} d x = \frac{1}{e^{π}}

Answered by mathmax by abdo last updated on 27/Aug/20

I = \int_{- \infty}^{+ \infty} \frac{(1 + \frac{x}{π}) \sin (π x)}{x^{2} + 4 x + 5} = Im (\int_{- \infty}^{+ \infty} \frac{(1 + \frac{x}{π}) e^{i π x}}{x^{2} + 4 x + 5}) let condider the complex function

φ (z) = \frac{(1 + \frac{z}{π}) e^{i π z}}{z^{2} + 4 z + 5} poles of φ ?

z^{2} + 4 z + 5 = 0 \to Δ^{^{'}} = 4 - 5 = - 1 \Rightarrow z_{1} = - 2 + i and z_{2} = - 2 - i

residus theorem \Rightarrow \int_{- \infty}^{+ \infty} φ (z) dz = 2 i π Res (φ, - 2 + i)

we have φ (z) = \frac{(1 + \frac{z}{π}) e^{i π z}}{(z - z_{1}) (z - z_{2})} \Rightarrow Res (φ, - 2 + i) =

= \lim_{z \to z_{1}} (z - z_{1}) φ (z) = \frac{(1 + \frac{z_{1}}{π}) e^{i π z_{1}}}{2 i} = \frac{(1 + \frac{- 2 + i}{π}) e^{i π (- 2 + i)}}{2 i} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π \times \frac{(π - 2 + i) e^{- π}}{2 i π} = (π - 2 + i) e^{- π}

= (π - 2) e^{- π} + i e^{- π} \Rightarrow Im (\int \dots .) = e^{- π} \Rightarrow ★ I = \frac{1}{e^{π}} ★