Find $\int_{-\infty}^{\infty} \frac{\sin x}{x^2-2x+10}dx$ with residue

Question

So to find $\int_{-\infty}^{\infty} \frac{\sin x}{x^2-2x+10}dx$ all I tried was the residue theorem which states $$\int_{-\infty}^\infty f(x)dx=2\pi i \text{ Res}(f,z_k)$$ and then found the poles which are $1\pm 3i$ but since only $1+3i$ is above the real line I only have to find that residue. Whose residue I have found to be $$\lim_{z\to 1+3i}\frac{\sin z}{(z-(1-3i)}=\frac{\sin(1+3i)}{6i}$$ but I have no clue on how to simplify that or if that is correct or not

Answer 1

Note that $\sin(z)$ is not bounded on the contour $z=Re^{i\phi}$, $\phi\in [0,\pi]$ as $R\to \infty$. However, $e^{iz}$ is bounded on that contour. Moreover, $\text{Im}(e^{iz})=\sin(z)$.

Thus, we begin by evaluating the integral $I_R=\oint_{C_R} \frac{e^{iz}}{z^2-2z+10}\,dz$, where we shall take $R>\sqrt{10}$, and where $C_R$ is the contour comprised of $(1)$ the line segment from $-R$ to $R$, and $(2)$ the semi-circular arc $|z|=R$, $0\le \arg(z)\le \pi$.

$$\begin{align} I_R&=\oint_{C_R} \frac{e^{iz}}{z^2-2z+10}\,dz\\\\ &=\int_{-R}^R \frac{e^{ix}}{x^2-2x+10}\,dx+\int_0^\pi \frac{e^{iRe^{i\phi}}}{(Re^{i\phi})^2-2(Re^{i\phi})+10}\,iRe^{i\phi}\,d\phi\tag1 \end{align}$$

Now, as $R\to\infty$, the second integral on the right-hand side of $(1)$ approaches $0$.

And by Cauchy's integral theorem, $I_R=2\pi i \text{Res}\left(\frac{e^{iz}}{z^2-2z+10}, z=1+3i\right)=\pi\frac{e^{-3+i}}{3}$.

Putting everything together, we find that

$$\int_{-\infty}^\infty \frac{\sin(x)}{x^2-2x+10}\,dx=\frac{\pi\sin(1)}{3e^3} \\\\\int_{-\infty}^\infty \frac{\cos(x)}{x^2-2x+10}\,dx=\frac{\pi\cos(1)}{3e^3}$$