TITLE: How do I prove $\int_{-\infty}^{\infty}{\cos(x+a)\over (x+b)^2+1}dx={\pi\over e}{\cos(a-b)}$?
QUESTION [6 upvotes]: How do I prove these?
$$\int_{-\infty}^{\infty}{\sin(x+a)\over (x+b)^2+1}dx={\pi\over e}\color{blue}{\sin(a-b)}\tag1$$
$$\int_{-\infty}^{\infty}{\cos(x+a)\over (x+b)^2+1}dx={\pi\over e}\color{blue}{\cos(a-b)}\tag2$$
I am trying to apply the residue theorem to $(2)$
$$f(x)={\cos(x+a)\over (x+b)^2+1}$$
$(x+b)^2+1$=$(x+b-i)(x+b+i)$
$$2\pi{i}Res(f(x),-b-i)=2\pi{i}\lim_{x\rightarrow -b-i}{\cos(a-b-i)\over -2i}=-\pi\cos(a-b-i)$$
$$2\pi{i}Res(f(x),-b+i)=2\pi{i}\lim_{x\rightarrow -b+i}{\cos(a-b+i)\over 2i}=\pi\cos(a-b+i)$$
How do I suppose to evaluate $\cos(a-b-i)$ and $\cos(a-b+i)$?

REPLY [4 votes]: $$\int_{-\infty}^{\infty}{\cos(x+a)\over (x+b)^2+1}dx=\cos (a-b)\int_{-\infty}^{\infty}{\cos x\over x^2+1}dx-\sin (a-b)\int_{-\infty}^{\infty}{\sin x\over x^2+1}dx$$
Let $\lambda\in\mathbb{R}$, set
$$I(\lambda)=\int_{-\infty}^{\infty}{\cos(\lambda x)\over x^2+1}dx$$
we use integrate by parts, writing
$$u=\frac{1}{{{x}^{2}}+1}\quad,\quad dv=\cos (\lambda x)$$
we have
$$I(\lambda )=\frac{\sin (\lambda x)}{\lambda ({{x}^{2}}+1)}\left| \begin{matrix}
   \infty   \\
   -\infty   \\
\end{matrix} \right.+\frac{2}{\lambda }\int_{-\infty }^{+\infty }{\frac{\sin (\lambda x)}{{{({{x}^{2}}+1)}^{2}}}}\,dx
$$
as a result
$$\lambda I(\lambda )=2\int_{-\infty }^{\infty }{\frac{x\sin \lambda x}{{{({{x}^{2}}+1)}^{2}}}\,}dx \,.\quad(1)$$
By differentiate with respect $\lambda$ to get
$$\lambda \frac{dI}{d\lambda }+I(\lambda )=2\int_{-\infty }^{\infty }{\frac{{{x}^{2}}\cos \lambda x}{{{({{x}^{2}}+1)}^{2}}}\,}dx=\underbrace{2\int_{-\infty }^{\infty }{\frac{\cos \lambda x}{{{x}^{2}}+1}\,}dx}_{2I(\lambda )}-2\int_{-\infty }^{\infty }{\frac{\cos \lambda x}{{{({{x}^{2}}+1)}^{2}}}\,}dx$$
therefore
$$\lambda \frac{dI}{d\lambda }-I(\lambda )=-2\int_{-\infty }^{\infty }{\frac{\cos \lambda x}{{{({{x}^{2}}+1)}^{2}}}\,}dx$$
and 
$$\lambda \frac{{{d}^{2}}I}{d{{\lambda }^{2}}}=2\int_{-\infty }^{\infty }{\frac{x\sin \lambda x}{{{({{x}^{2}}+1)}^{2}}}\,}dx.\quad(2)$$
$(1)$ and $(2)$
$$\frac{{{d}^{2}}I(\lambda)}{d{{\lambda }^{2}}}- I(\lambda )=0$$
thus
$$I(\lambda)=c_1e^{\lambda}+c_2e^{-\lambda}$$
on the other hand
\begin{align}
  & I(0)={{c}_{1}}+{{c}_{2}}=\int_{-\infty }^{+\infty }{\frac{1}{{{x}^{2}}+1}}\,dx=\pi \,\,\,\,\Rightarrow \,\,{{c}_{1}}+{{c}_{2}}=\pi \, \\ 
 & I(\lambda )=\frac{2}{\lambda }\int_{-\infty }^{+\infty }{\frac{x\sin \lambda x}{{{({{x}^{2}}+1)}^{2}}}\,\,}dx\,\,\,\Rightarrow \,\,\underset{\lambda \to \infty }{\mathop{\lim }}\,I(\lambda )=0\,\,\,\Rightarrow \,{{c}_{1}}=0 \\ 
\end{align}
then
$$I(\lambda )=\pi {{e}^{-\lambda }}$$
set $\lambda=1$, we have

$$\cos (a-b)\int_{-\infty}^{\infty}{\cos x\over x^2+1}dx=\frac{\pi}{e}\cos (a-b)$$

Now set 
$$J(\lambda)=\int_{-\infty}^{\infty}{\sin(\lambda x)\over x^2+1}dx$$
WE repeat this produce,to get
$$J(\lambda)=c_1e^{\lambda}+c_2e^{-\lambda}$$
and
\begin{align}
  & J(0)={{c}_{1}}+{{c}_{2}}=0 \\ 
 & \underset{\lambda \to \infty }{\mathop{\lim }}\,J(\lambda )=0\Rightarrow \,{{c}_{1}}=0 \\ 
\end{align}
i.e. $J(\lambda)=0$ thus

$$\int_{-\infty}^{\infty}{\cos(x+a)\over(x+b)^2+1}dx=\frac{\pi}{e}\cos(a-b)$$