How do I prove $\int_{-\infty}^{\infty}{\cos(x+a)\over (x+b)^2+1}dx={\pi\over e}{\cos(a-b)}$?

Question

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 $\cos(a-b+i)$?

Answer 1

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

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

Answer 2

With residues at hand you are well equipped to evaluate these improper integrals.

But before doing so I'd propose to transform: Shift $x$ to get rid of $b$ in the denominator, $(1)$ then reads $$\int_{-\infty}^\infty{\sin(x+a)\over (x+b)^2+1}\,dx\; =\; \int_{-\infty}^\infty{\sin(x+a-b)\over x^2+1}\,dx\:,$$ and apply trigonometric addition formulae to the numerator to obtain $$ =\: \sin(a-b)\int_{-\infty}^\infty{\cos x\over x^2+1}\,dx\; +\; \cos(a-b)\int_{-\infty}^\infty{\sin x\over x^2+1}\,dx\:,$$ with a vanishing second integral—note that its integrand is an odd function.
Treating $(2)$ along the same lines yields $$\cos(a-b)\int_{-\infty}^\infty{\cos x\over x^2+1}\,dx\; .$$ The remaining integral to be evaluated $$\int_\mathbb{R}{\cos x\over x^2+1}\,dx$$ is well-known, cf. Jack D'Aurizio's answer, it is a showcase for the calculus of residues.

Remark: The method extends to the more general case where an even power $x^{2n}$ is present in the denominator, instead of $x^2$.

Answer 3

You may compute both integrals at once by computing $$ \int_{-\infty}^{+\infty}\frac{e^{i(x+a)}}{(x+b)^2+1}\,dx $$ (i.e. the rescaled Fourier transform of $\frac{1}{x^2+1}$) then considering the real/imaginary part.

Since the CF of the Laplace distribution is well-known and easy to compute, the claim is trivial by Fourier inversion. Using residues:

$$\int_{-\infty}^{+\infty}\frac{e^{i\xi x}}{x^2+1}\,dx = 2\pi i\cdot\text{Res}\left(\frac{e^{i\xi x}}{x^2+1},x=\pm i\right) = \pi e^{-|\xi|} $$ by simply considering a semicircular contour in the upper or lower half-plane, according to the sign of $\xi\in\mathbb{R}$.

Answer 4

Using the Euler identity: $e^{xi}=\cos x+\sin x$, we have $$ \begin{aligned} I & =\int_{-\infty}^{\infty} \frac{\cos (x+a)}{(x+b)^2+1} d x \\ & =\Re \int_{-\infty}^{\infty} \frac{e^{(x+a) i}}{(x+b)^2+1} d x\\&= \Re\left(e^{a i} \int_{-\infty}^{\infty} \frac{e^{x i}}{(x+b)^2+1} d x\right) \end{aligned} $$ Letting $y=x+b$ transforms the integral into $$ I=\int_{-\infty}^{\infty} \frac{e^{(y-b) i}}{y^2+1} d x=e^{-b i} \int_{-\infty}^{\infty} \frac{e^{yi}}{y^2+1} d y $$ By the contour integration of the upper unit semi-circle in anti-clockwise direction, we get the result $$ \displaystyle \int_{-\infty}^{\infty} \frac{e^{y i}}{y^2+1} d y =2 \pi i \lim _{z \rightarrow i}(z-i) \cdot \frac{e^{z i}}{z^2+1}=\dfrac{\pi}{e}$$

Hence

$$ \begin{aligned} I & =\Re\left(\frac{e^{a i} \cdot e^{-b i} \pi}{e}\right) \\ & =\frac{\pi}{e} \Re\left(e^{(a-b) i}\right)\\&=\frac{\pi}{e} \cos(a-b) \end{aligned} $$ On the other hand, we take the imaginary part in the second step and get

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