How to calculate $\int_{-\infty}^{+\infty} \frac{x \sin x}{x^2+4x+20}\, dx$? I get the answer that it equals $\frac{\pi(2\cos2+\sin2)}{2e^4}$ by mathematica.
But I don't know how it can equal this form.
My idea is to construct equations so that problems can be transformed.Like these:
$$\int_{-\infty}^{+\infty} \frac{x \sin x}{x^2+4x+20}\, dx=\int_{-\infty}^{+\infty} \frac{(y+4) \sin (y+4)}{y^2+4y+20}\, dy$$
$$\int_{-\infty}^{+\infty} \frac{x \cos x}{x^2+4x+20}\, dx=-\int_{-\infty}^{+\infty} \frac{(y+4) \cos (y+4)}{y^2+4y+20}\, dy$$
So only by calculating $\int_{-\infty}^{+\infty} \frac{ \sin x}{x^2+4x+20}\, dx$ and $\int_{-\infty}^{+\infty} \frac{ \cos x}{x^2+4x+20}\, dx$ then we can get the values of $\int_{-\infty}^{+\infty} \frac{x \sin x}{x^2+4x+20}\, dx$ and $\int_{-\infty}^{+\infty} \frac{x \cos x}{x^2+4x+20}\, dx$.
What's more I found that 
$$\int_{-\infty}^{+\infty} \frac{ \sin x}{x^2+4x+20}\, dx=- \frac{\pi \sin 2}{4e^4}$$
$$\int_{-\infty}^{+\infty} \frac{ \cos x}{x^2+4x+20}\, dx= \frac{\pi \cos 2}{4e^4}$$
So I guess I can solve the problem only by calculating
$$\int_{-\infty}^{+\infty} \frac{ \sin x}{x^2+ c}\, dx$$ and
$$\int_{-\infty}^{+\infty} \frac{ \cos x}{x^2+ c}\, dx$$
Can you help me?
 A: As you've already noted, the substitution $y=x-2$ gives$$\int_{\mathbb{R}}\frac{x\sin xdx}{x^{2}+4x+20} =2C_{0}\sin 2-2S_{0}\cos 2+S_{1}\cos 2-C_{1}\sin 2$$with $$C_{n}:=\int_{\mathbb{R}}\frac{y^{n}\cos ydy}{y^{2}+16},\,S_{n}:=\int_{\mathbb{R}}\frac{y^{n}\sin ydy}{y^{2}+16},$$i.e. $$C_{n}+iS_{n}=\int_{\mathbb{R}}\frac{\exp iydy}{y^{2}+16}.$$The characteristic function of the Cauchy distribution is$$\int_{\mathbb{R}}\frac{\gamma}{\pi}\frac{\exp itydy}{\left(y-y_{0}\right)^{2}+\gamma^{2}}=\exp\left(iy_{0}t-\gamma\left|t\right|\right),$$so in particular$$\int_{\mathbb{R}}\frac{\exp itydy}{y^{2}+16}=\frac{\pi}{4}\exp-4\left|t\right|.$$Differentiating with respect to t,$$\int_{\mathbb{R}}\frac{y\exp itydy}{y^{2}+16}=\pi i\operatorname{sgn}t\exp-4\left|t\right|.$$Hence$$C_{0}+iS_{0}=\frac{\pi}{4e^{4}},\,C_{1}+iS_{1}=\frac{\pi i}{e^{4}}\implies C_{1}=S_{0}=0,\,C_{0}=\frac{\pi}{4e^{4}},\,S_{1}=\frac{\pi}{e^{4}}.$$Finally,$$\int_{\mathbb{R}}\frac{x\sin xdx}{x^{2}+4x+20}=\frac{\pi}{2e^{4}}\left(2\cos2+\sin2\right).$$
