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I need to find $\int\limits_0^t e^{\alpha t}\sin(\omega t)\,\mathrm dt$. I'd like to know whether it can be brought into some closed form expression. Please suggest me some hints to solve this.

J. M. ain't a mathematician
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Rajesh D
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    this should help: http://math.stackexchange.com/questions/19796/name-of-this-identity-int-e-alpha-x-cos-beta-x-frace-alpha-x-alp –  Apr 23 '11 at 03:55

3 Answers3

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This is the imaginary part of a similar integral that is easy to compute.

Carl Brannen
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  • @Carl: Is the solution $e^{\alpha t}sin(\omega t) + \frac{\omega}{\alpha^2 + \omega^2}$ correct ? – Rajesh D Apr 23 '11 at 04:08
  • I get the $\omega/(\alpha^2+\omega^2)$ part, but not the other part. Instead of it, I'm getting stuff like $\omega\cos(\omega t)/(\alpha^2+\omega^2)$ and similar for $\alpha \sin(\omega t)$, with various minus signs. – Carl Brannen Apr 23 '11 at 04:15
  • My instinct is to do it as the imaginary part of the complex integral $\int \exp(\alpha t + i\omega t);dt$, but I see (from the Chandru1 hint) that you can do it by messing around with integration by parts. That seems messier, but will get the same answer given no errors in the algebra. – Carl Brannen Apr 23 '11 at 04:18
  • @Carl: I am getting it as $e^{\alpha t} \frac{\alpha\sin(\omega t) - \omega \cos(\omega t) + \omega}{\alpha^2 + \omega^2}$ – Rajesh D Apr 23 '11 at 04:34
  • @Rajesh D; That is almost what I got, but I also have an $\exp(\alpha t)$. Note that it is zero when $t=0$, as required. When you get the right answer, you can differentiate it to see if you get the thing under the integral in the original question. – Carl Brannen Apr 23 '11 at 04:38
  • @Carl Brannen : We should consider only the imaginary part of the integral.Is this correct ? please let me know if that is the procedure you have done. – Rajesh D Apr 23 '11 at 04:40
  • Yes, after integrating, you take the imaginary part. And I think you did that correctly, but in the algebra, left off the $\exp(\alpha t)$. Oh, now I see it's in there. So never mind! – Carl Brannen Apr 23 '11 at 04:44
  • @Rajesh; I seem to get $e^{\alpha t} \frac{\alpha\sin(\omega t) - \omega \cos(\omega t)}{\alpha^2 + \omega^2} + \frac{\omega}{\alpha^2+\omega^2}$. This is not quite the same. – Carl Brannen Apr 23 '11 at 04:50
  • @Carl : Now I am getting the same, thank you. – Rajesh D Apr 23 '11 at 05:18
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Integration by parts may help.

JonnyQuiznos
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I would suggest to use $\sin \omega t = \frac{1}{2 i}\left(e^{i\omega t} - e^{-i\omega t}\right)$.

Peter
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