help with integral using residue theorem

Question

I'm solving the following integral: $$I=\int_{-\infty}^{+\infty}\frac{x\cdot\cos x}{x^2-2x+10}\mathrm{d}x$$ I used the residue theorem as follows: $$I=\mathrm{Re}\left(\int_{-\infty}^{+\infty}\frac{z\cdot e^{iz}}{z^2-2z+10}\mathrm{d}z\right)$$ The roots of the polynomials are:

$x_1=1+3i$

$x_2=1-3i$

Of which the first one lies in the integration half-circle. I computed the residue in $x_1$ as $$\frac{(1+3i)e^{1+3i}}{6i}=\frac{(1+3i)e\cdot (\cos 3+i\cdot\sin 3)}{6i}$$ Value of the integral should thus be $$\mathrm{Re}\left(2\pi i\frac{(1+3i)e\cdot (\cos 3+i\cdot\sin 3)}{6i}\right)$$ $$I=-\frac{\pi}{3}e\cdot(\cos 3-3\sin 3)$$ But according to both book and Wolfram the result should be $$\frac{\pi}{3e^3}(\cos 1-3\sin 1)$$ My result looks almost correct, but seems like I somehow interchanged 1 a 3. Can you please help me find the mistake? Thanks

Answer 1

I used the residue theorem as follows: $$I=\mathrm{Re}\left(\int_{-\infty}^{+\infty}\frac{z\cdot e^{iz}}{z^2-2z+10}\mathrm{d}z\right)$$

Presumably you're integrating over a half-circle $\gamma.$ It should be $$I=\mathrm{Re}\left(\int_\gamma\frac{z\cdot e^{iz}}{z^2-2z+10}\mathrm{d}z\right).$$

I computed the residue in $x_1$ as $$\frac{(1+3i)e^{1+3i}}{6i}=\frac{(1+3i)e\cdot (\cos 3+i\cdot\sin 3)}{6i}.$$

Here's your mistake, it should be $\dfrac{(1+3i)e^{\color{red}i(1+3i)}}{6i}$ which equals $$\dfrac{(\cos(1)-3\sin(1))+i(3\cos(1)+\sin(1))}{6e^{3}i}.$$

Answer 2

$e^{iz}|_{z=1+3i}=e^{-3+i}=e^{-3}(\cos1+i\sin1)$.