Integrate $\int_{-\infty}^{\infty} \frac{\sin (5x)}{1+(x-\frac{\pi}2)^2}dx$ with a complex integral theorem

Question

Can someone help me to integrate

$$\int_{-\infty}^{\infty} \frac{\sin (5x)}{1+(x-\frac{\pi}2)^2}dx$$

with a complex integral theorem?

Answer 1

Consider the complex integral $$\int_{C_R} \frac{\exp(5iz)}{1+(z-\frac{\pi}2)^2}dz.$$ where $C_R$ is a semicircle in the upper-half plane with diameter $[-R,R]$.

It turns out that $$\int_{-\infty}^{\infty} \frac{\sin (5x)}{1+(x-\frac{\pi}2)^2}dx= \mbox{Im}\left(2\pi i\cdot\mbox{Res}\left(\frac{\exp(5iz)}{1+(z-\frac{\pi}2)^2};i+\frac{\pi}{2}\right)\right)\\ =\mbox{Im}\left(2\pi i\cdot \frac{\exp(5i(i+\frac{\pi}{2})}{2i}\right) =\frac{\pi}{e^5}.$$

Answer 2

Hints:

Define

$$f(z)=\frac{e^{5iz}}{1+\left(z-\frac\pi2\right)^2}$$

Observe this function one two singularities, which are simple poles, at

$$z_{1,2}=\frac\pi2\pm i$$

of which only one is on the upper half plane. Choose "the usuall" contour

$$C_R:=[-R,R]\cup\gamma_R\;,\;\;\text{with}\;\;\gamma_R:=\left\{\,Re^{it}\;/\;0<t<\pi\,\right\}\;,\;\;R\in\Bbb R^+$$

And $\;R\;$ big enough so that the half disk above contains in its interior the pole $\;\frac\pi2+i\;$ .

Check the residue of the function at the above pole is

$$\lim_{z\to\frac\pi2+i}\frac{e^{5iz}}{2\left(z-\frac\pi2\right)}=\frac{e^{5i(\pi/2+i)}}{-2i}=\frac{e^{-5}}{2i}$$

Either use the M-L estimmation lemma or Jordan's lemma to show

$$\lim_{R\to\infty}\int_{\gamma_R}f(z)\,dz=0$$

and now use Cauchy's Residues Theorem, while passing to the limit $\;R\to\infty\;$ , to get:

$$2\pi i\frac{e^{-5}}{2i}=\lim_{R\to\infty}\oint_{C_R}f(z)dz=\int_{-\infty}^\infty\frac{e^{5ix}}{1+\left(x-\frac\pi2\right)^2}dx$$

Finally, separate real and imaginary parts to get the final result: $\;\color{red}{\cfrac\pi{e^5}}\;$

Answer 3

$$ I = \int_{-\infty}^{+\infty}\frac{\cos(5x)}{1+x^2}\,dx = \text{Re}\int_{-\infty}^{+\infty}\frac{e^{5ix}}{1+x^2}\,dx = \text{Re}\left(2\pi i\cdot\text{Res}\left(\frac{e^{5ix}}{1+x^2},x=i\right)\right)$$ and since $$ \text{Res}\left(\frac{e^{5ix}}{1+x^2},x=i\right) = \lim_{x\to i}\frac{e^{5ix}}{x+i}=-\frac{i}{2e^5} $$ it follows that $$ I = \color{red}{\frac{\pi}{e^5}}.$$