How we can obtain the time-integration of a sine signal?

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

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

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.

• 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? Aug 22 '16 at 17:27
• @AllEs: I know what you mean, but it is consistent. Have a look at my answer to your other related question. Aug 22 '16 at 17:50
• I've visited your other response and I've been convinced. Aug 22 '16 at 19:33