I have a convolution integral where $F(\tau) = F_0$ and $g(t - \tau) = \sin(\omega_n(t-\tau))$ so $$ F_0\int_{t_0}^{\infty}\sin(\omega_n(t-\tau))d\tau $$ which doesn't converge. Can I do the following: $$ Im\Bigg\{F_0\int_{t_0}^{\infty}e^{i\omega_n(t - \tau)}d\tau\Bigg\} = Im\bigg\{\frac{e^{i\omega_n(t-\tau)}}{-i\omega_n}\bigg|_{t_0}^{\infty}\bigg\} = \frac{-\cos(\omega_n(t-t_0))}{\omega_n} $$

  • $\begingroup$ What is $\omega_n$? Whatever issue you had to start with is still there in the second line.. $\endgroup$ – mathematician Oct 30 '14 at 17:57
  • $\begingroup$ @mathematician $\omega_n$ is a constant. $\endgroup$ – dustin Oct 30 '14 at 17:59
  • $\begingroup$ whats $e^{i \infty}$ ? :) $\endgroup$ – tired Oct 30 '14 at 19:17

i would suggest to add a small imaginary part like $e^{i((\omega_n +i \delta)(t-\tau))}$ and take the limit of the $\delta->0$ in the end. This has the advantage that your integral converges for every finite delta and you don't have to deal with expressions like $e^{i\infty}$ which are not well defined. The price you pay is that your integral is still divergent in the above limit and has to be interpreted in the sense of distributions.

So: \begin{align} \Im \left[\lim_{\delta \rightarrow0}\int_{t_0}^{\infty} dt e^{i((\omega_n +i \delta)(t-\tau))}\right]=\Im \left(i\lim_{\delta \rightarrow0}\left[e^{-i((\omega_n +i \delta)(t_0-\tau)}\frac{1}{\omega_n+i\delta}\right]\right) \end{align}

using Euler's identity and $\lim_{\delta \rightarrow0}\frac{1}{\omega_n+i\delta}=-i\pi \delta(\omega_n)+P\frac{1}{\omega_n}$. See here.

you should arrive at a result. Please note that $P$ denotes the Cauchy principal value and $\delta(x)$ is Dirac's Delta function.

Can you take it from here?

Edit: I think the final result will be the one you posted, but this time well justified :)

| cite | improve this answer | |

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.