2
$\begingroup$

Find the value of $\int_{-\infty}^{\infty} f(x) dx$, where $f(x)=sin(x)/(x^3+x)$.

How do I go about solving this? I have tried to expand the sine part into complex exponentials to try and resemble the definition of the transform but have had no luck. Is there a general method applicable here?

Thank you!

$\endgroup$
  • 2
    $\begingroup$ Side note: If you find $\hat{f}$, then $\int_{-\infty}^{\infty} f(x)\ dx = \hat{f}(0)$. $\endgroup$ – Neal Feb 2 '16 at 13:40
1
$\begingroup$

You can use Parseval's Theorem: If $f(x)$ and $g(x)$ have respective FT's

$$F(k) = \int_{-\infty}^{\infty} dx \, f(x) e^{i k x} $$ $$G(k) = \int_{-\infty}^{\infty} dx \, g(x) e^{i k x} $$

Then

$$\int_{-\infty}^{\infty} dx \, f(x) g^*(x) = \frac1{2 \pi} \int_{-\infty}^{\infty} dk \, F(k) G^*(k) $$

In your case, $f(x) = \frac{\sin{x}}{x}$ and $g(x) = \frac1{1+x^2}$. Thus,

$$\int_{-\infty}^{\infty} dx \, \frac{\sin{x}}{x (1+x^2)} = \frac1{2 \pi} \int_{-1}^1 dk \, \pi \, \pi e^{-|k|} = \pi \int_0^1 dk \, e^{-k} = \pi \left ( 1-\frac1{e} \right )$$

$\endgroup$
1
$\begingroup$

\begin{align} \int_{-\infty}^{\infty}\frac{\sin x}{x^3+x}dx & =\Im\int_{-\infty}^{\infty}\frac{e^{ix}-1}{x^3+x}dx \\ & = \Im\left(\frac{1}{2i} \int_{-\infty}^{\infty}\frac{e^{ix}-1}{x}\left[\frac{1}{x-i}-\frac{1}{x+i}\right]dx\right) \end{align} The integrand on the right decays when $x=z$ is in the upper half plane. So you can close the contour in the upper half plane and use Cauchy's theorem to evaluate at the one residue in the upper half plane. The result is $$ \Im\left(\left.\pi\frac{e^{iz}-1}{z}\right|_{z=i}\right)=\Im\left(\pi\frac{1/e-1}{i}\right)=\pi(1-1/e). $$

$\endgroup$
  • $\begingroup$ This looks very interesting, but how to you get the first equality? $\endgroup$ – plebmatician Feb 3 '16 at 6:10
  • 1
    $\begingroup$ $\Im (e^{ix}-1)=\Im e^{ix}=\sin(x)$. Any real constant $a$ gives $\Im(e^{ix}-a)$, but you want the integral to converge, which means $a=1$ is the correct choice. $\endgroup$ – DisintegratingByParts Feb 3 '16 at 7:24

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.