1
$\begingroup$

I want to compute the following integral

$$- \frac{1}{M(\lambda_1-\lambda_2)}\int\limits_{-\infty}^t(e^{\lambda_1(t-t')}-e^{\lambda_2(t-t')})(\beta\omega A\sin\omega t' +g)\;dt'$$

Here the integral that's not trivial is

$$\int_{-\infty}^t (e^{\lambda_1(t-t')}-e^{\lambda_2(t-t')}) \sin\omega t'\; dt'.$$

If I use $\sin \omega t' = \frac{1}{2i}(e^{i\omega t'}-e^{-j\omega t'})$, then I get that the last integral is:

$$\left[\frac{e^{\lambda_1t + (i\omega - \lambda_1)t'}}{i\omega-\lambda_1}+\frac{e^{\lambda_1t - (i\omega + \lambda_1)t'}}{i\omega+\lambda_1}+\frac{e^{\lambda_2t + (i\omega - \lambda_2)t'}}{-i\omega+\lambda_2}-\frac{e^{\lambda_2t - (i\omega + \lambda_2)t'}}{i\omega+\lambda_2}\right]^t_{-\infty}\tag{1}$$ which diverges.

My question is. Can I assume that the integral is the imaginary part of whatever comes out if I use $\sin\omega t' = Im(e^{i\omega t'})$? I mean, I will mix this exponential with the others and I'm not sure if after I do some algebra and take the imaginary part I will get a true result. Also, why does the expression in (1) diverge (I don't see that I made any mistake)?

I appreciate your help.

$\endgroup$
  • $\begingroup$ Are $\lambda_1$ and $\lambda_2$ real and of positive value? $\endgroup$ – Mark Viola Apr 14 '15 at 2:37
  • $\begingroup$ @Dr.MV I'm very sorry about the variable of integration. I thought I wrote $t'$, because this is the correct one. Also $\lambda_1,\lambda_2\in\mathbb C$. So they can be real and take any positive or negative value. $\endgroup$ – Vladimir Vargas Apr 14 '15 at 2:41
  • $\begingroup$ The integral diverges if the real part of either $\lambda_1$ or $\lambda_2$ is positive. $\endgroup$ – Mark Viola Apr 14 '15 at 2:46
1
$\begingroup$

We must assume that $\text{Re}\{\lambda_n\}<0$, ($n=1,2$) in order to render the integral convergent. Then, we can write the integral of interest as

$\begin{align} \int_{-\infty}^t \left( e^{\lambda_1(t-t')}-e^{\lambda_2(t-t')} \right)\sin(\omega t')\,\,dt'&=\text{Im}\left\{\int_{-\infty}^t \left( e^{\lambda_1(t-t')}-e^{\lambda_2(t-t')} \right)e^{i\omega t'}\,\,dt'\right\}\\\\ &=\text{Im}\left\{e^{\lambda_1 t}\int_{-\infty}^t e^{-(\lambda_1-i\omega)t'}\,\,dt'-e^{\lambda_2 t}\int_{-\infty}^te^{-(\lambda_2-i\omega)t'}\,\,dt'\right\}\\\\ &=\text{Im}\left\{e^{i\omega t}\left(\frac{1}{\lambda_2-i\omega}-\frac{1}{\lambda_1-i\omega}\right)\right\} \end{align}$

Note that we have

$$\text{Im}\left\{e^{i\omega t}\frac{1}{\lambda_n-i\omega}\right\}= \text{Im}\left\{e^{i\omega t}\frac{\lambda_n^*+i\omega}{|\lambda_2-i\omega|^2} \right\}$$

where $n=1,2$, and $z^*$ denotes the complex conjugate of $z$. Then,

$$\text{Im}\left\{e^{i\omega t}\frac{1}{\lambda_n-i\omega}\right\}= \frac{(\omega-\lambda_{ni})\cos (\omega t)+\lambda_{nr} \sin(\omega t)}{|\lambda_n-i\omega|^2} $$

where $z_r$ and $z_i$ designate the real and imaginary parts of $z$, respectively.

$\endgroup$

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.