# Integral using complex numbers shortcut

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.

• Are $\lambda_1$ and $\lambda_2$ real and of positive value? – Mark Viola Apr 14 '15 at 2:37
• @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. – Vladimir Vargas Apr 14 '15 at 2:41
• The integral diverges if the real part of either $\lambda_1$ or $\lambda_2$ is positive. – Mark Viola Apr 14 '15 at 2:46

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.