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

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


1 Answer 1


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


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.

  • $\begingroup$ 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, 2016 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, 2016 at 17:50
  • $\begingroup$ I've visited your other response and I've been convinced. $\endgroup$
    – AllEs
    Aug 22, 2016 at 19:33

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