Let $f(t)=1-t^2$ , for $|t|<1$ and $0$ elsewhere.

Compute the Fourier transform of $f(t)$ and use the result to find the value of the integral $$ \int_{-\infty}^{\infty}\frac{\sin t-t \cos t}{t^3}dt $$


So the Fourier transform is pretty easy and I got $\hat{f}(\omega)= 4 \frac{\sin \omega- \omega \cos \omega}{\omega^3}$. How do I use this to compute the integral? Since there is an obvious connection between these, can I use the inverse theorem or Parseval's formula?


I solved it myself, here is my solution if anyone else counter the same problem.

So I used Parseval's formula

$$ \int_{-\infty}^{\infty}f(x)\overline{g(x)}dx = \frac{1}{2\pi}\int_{-\infty}^{\infty}\hat{f}(\omega) \overline{\hat{g}(\omega)}d\omega $$ and then I just took $\overline{\hat{g}(\omega)} = 1$ with $g(t) = \delta(t)$, combining this I simply got

$$ \int_{-\infty}^{\infty} \frac{\sin t- t \sin t}{t^3}dt = \frac{2 \pi}{4} \int_{-1}^{1}(1-t^2)\delta(t)dt = \frac{\pi}{2}\,f(0) = \frac{\pi}{2} $$

  • $\begingroup$ Why your bounds in the last integral turned into $\pm1$ ? $\endgroup$ – Nicolas Jan 1 '16 at 22:02
  • $\begingroup$ I saw now that my problem was written all wrong -.-. Look how $f(t)$ is defined now $\endgroup$ – user269620 Jan 1 '16 at 22:03
  • $\begingroup$ You then have a typo in your question. **Corrected now ! But has you said, only the value at $0$ does matter. $\endgroup$ – Nicolas Jan 1 '16 at 22:05
  • $\begingroup$ yes I saw it too, I'm sorry :) $\endgroup$ – user269620 Jan 1 '16 at 22:06
  • $\begingroup$ No problem, good answer! $\endgroup$ – Nicolas Jan 1 '16 at 22:06

If $$ \hat{f}(\omega) = \int_{-\infty}^\infty f(t)e^{-i\omega t}dt$$ is the Fourier Transform of $f$, then we have the inverse Fourier Transform $$ f(t) = \frac{1}{2\pi} \int_{-\infty}^\infty \hat{f}(\omega)e^{i\omega t}d\omega. $$ Then you get the integral by evaluating the inverse transform at $t=0$.

  • $\begingroup$ That would work too, thanks for pointing this out :) $\endgroup$ – user269620 Jan 1 '16 at 22:07

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