1
$\begingroup$

could we calculate the below integral by the Fourier series or the Fourier transform properties?

$$\int_{-\infty}^t \sin(\omega_0\tau)d\tau=?$$

$\endgroup$
1
$\begingroup$

No, this improper integral doesn't have a value. Since for the indefinite integral we have

$$\int\sin(\omega_0t)dt=-\frac{1}{\omega_0}\cos(\omega_0t)+C\tag{1}$$

the given definite integral would be

$$\int_{-\infty}^{t}\sin(\omega_0\tau)d\tau=-\frac{1}{\omega_0}\cos(\omega_0t)+\lim_{\tau\rightarrow -\infty}\frac{1}{\omega_0}\cos(\omega_0\tau)\tag{2}$$

but the limit in $(2)$ doesn't exist.

$\endgroup$
  • $\begingroup$ I know your solution. But it is not consistent with the response which is obtained by the time-integration property of the Fourier transform. When we utilize this property of the Fourier transform, the response of the above integral would be (−1/ω0 )*cos(ω0t). Why we have this inconsistency? $\endgroup$ – AllEs Aug 22 '16 at 17:27
  • $\begingroup$ @AllEs: I know what you mean, but it is consistent. Have a look at my answer to your other related question. $\endgroup$ – Matt L. Aug 22 '16 at 17:50
  • $\begingroup$ I've visited your other response and I've been convinced. $\endgroup$ – AllEs Aug 22 '16 at 19:33

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.