2
$\begingroup$

How to calculate the following integral using complex analysis? $\int_0^\infty\frac{\cos(5x)}{(1+x^2)^2}\mathrm{d}x$.

So far I have $$\int_0^\infty\frac{\cos(5x)}{(1+x^2)^2}\mathrm{d}x = \int_{-\infty}^\infty\frac{1}{(1+x^2)^2}e^{5ix}\mathrm{d}x$$ Then, $$Res(f(x),i)=\frac{d}{dx}[e^{5ix}]|_i=5ie^{5ix}|_i=2\pi i5ie^{5i(i)}=\frac{-10\pi}{e^5}$$ Then I might have to multiply by 1/2 to get from 0 to infinity only but that gives $\frac{-5\pi}{e^5}$ and the answer should be $\frac{3\pi}{2e^5}$ and I am not sure what I am doing wrong...

$\endgroup$
2
$\begingroup$

$$\int_{0}^{+\infty}\frac{\cos(5x)}{(1+x^2)^2}\,dx = \frac{1}{2}\text{Re}\int_{-\infty}^{+\infty}\frac{e^{5ix}}{(1+x^2)^2}\,dx \tag{1}$$ and $x=i$ is a double pole for $\frac{e^{5ix}}{(1+x^2)^2}$, in particular

$$ \text{Res}\left(\frac{e^{5ix}}{(1+x^2)^2},x=i\right) = \lim_{x\to i}\frac{d}{dx}\left(\frac{e^{5ix}}{(x+i)^2}\right)=-\frac{3i}{2e^5}\tag{2}$$ and $$ \int_{0}^{+\infty}\frac{\cos(5x)}{(1+x^2)^2}\,dx = \text{Re}\left(\frac{(-3i)\cdot(\pi i)}{2e^5}\right)=\color{red}{\frac{3\pi}{2e^5}}.\tag{3}$$

$\endgroup$
  • $\begingroup$ Oh thank you. I am a bit confused because Res$(f(z),z_0)=\frac{1}{(n-1)!} \frac{d^{n-1}}{dz^{n-1}}[(z-z_0)^nf(z)]}$. Wouldn't that give $\frac{d}{dz}[(z-i)^2 \frac{e^{5iz}{(z+i)^2}]$ or something like that? Can you explain how you used the definition to get the Residue? Thanks for your time. $\endgroup$ – MathIsHard Apr 25 '17 at 16:41
  • 1
    $\begingroup$ @Math4Life: I took the definition of the residue at a double pole and simply noticed that $(x^2+1)^2=(x+i)^2(x-i)^2$. $\endgroup$ – Jack D'Aurizio Apr 25 '17 at 16:46
  • $\begingroup$ Oh thank you. I missed that fact. I appreciate the help. $\endgroup$ – MathIsHard Apr 25 '17 at 16:47
1
$\begingroup$

Hint: $$\dfrac{e^{5iz}}{(z-i)^2(z+i)^2}=\dfrac{1}{(z-i)^2}\dfrac{e^{5iz}}{(z+i)^2}$$ so compute $$2\pi i\Big[\dfrac{e^{5iz}}{(z+i)^2}\Big]'_{z=i}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.