how to solve it by residue theorem

Question

i have to solve the folowing integral by using contour integration & residue theorem $$\int_{-\infty}^{\infty}\frac{\sin x}{x^2+2x+2}\ dx$$ i factorised $x^2+2x+2=(x+1+i)(x+1-i)$ where $x=-1+i$ & $x=-1-i$ are single poles $$\int_{-\infty}^{\infty}\frac{\sin x}{(x+1+i)(x+1-i)}\ dx$$ i got stuck here, don't know which contour is to to be used? please help me solve it

Answer 1

$$I=\int \limits^{\infty }_{-\infty }\frac{\sin \left( x\right) }{x^{2}+2x+2} dx=\Im \left[ \int \limits^{\infty }_{-\infty }\frac{e^{iz}}{z^{2}+2z+2} dz\right] =\Im \left[ 2\pi i \sum Res\left( f\left( z\right) , z_{0}\right) \right] $$ you have one pole inside your contour $ z=i-1$ so your Residue will be $$2\pi i\lim \limits_{z\to i-1}\left( z+1-i\right) \frac{e^{iz}}{\left( z+1-i\right) \left( z+i+1\right) } =\Im \left( 2\pi i \left( \frac{e^{i\left( i-1\right) }}{2i} \right) \right) =\Im\left( \frac{\pi \left( \cos \left( 1\right) -i \sin \left( 1\right) \right) }{e}\right) $$ So the integral equal to $$ \int \limits^{\infty }_{-\infty }\frac{\sin \left( x\right) }{x^{2}+2x+2} dx=\boxed{\frac{-\pi \sin \left( 1\right) }{e}} $$