What is the integral of $e^{-\alpha|t|}e^{ivt}e^{-ipt}$ over $\mathbb{R}$? Given $\alpha, v > 0$, I'm supposed to get to the following result:
$$
g(p) = \int_{-\infty}^{\infty}e^{-\alpha|t|}e^{ivt}e^{-ipt}dt =\frac{2\alpha}{\alpha^{2}+(v-p)^{2}}
$$
I've tried calculating it as:
$$
\int_{-\infty}^{0} e^{((v-p)i+\alpha)t}dt\ +\ \int_{0}^{\infty} e^{((v-p)i-\alpha)t}dt
$$
but I got stuck at
$$
\frac{2\alpha}{\alpha^{2}+(v-p)^{2}}+\lim_{x\to\infty}\frac{2\alpha(e^{(v-p)ix}-e^{-(v-p)ix})}{e^{\alpha x}(\alpha^{2}+(v-p)^{2})}
$$

I also tried calculating the integral as
$$
i\int_{-\infty}^{\infty}\sin((v-p)t)e^{\alpha|t|}dt+\int_{-\infty}^{\infty}\cos((v-p)t)e^{\alpha|t|}dt
$$
but that didn't lead me very far.

This is an exercise I have to solve in the context of a (relatively easy) first semester math course in a Computer Science bachelor's.

Note that: $$I_1 = \int_{-\infty}^{0} e^{((v-p)i+\alpha)t} dt = \lim_{b \to -\infty} \int_{b}^{0} e^{((v-p)i +\alpha)t} dt = \lim_{b \to -\infty} \frac{1}{((v-p)i +\alpha)}[1-e^{((v-p)i +\alpha) b}]  = \frac{1}{((v-p)i +\alpha)}$$
Can you similarly see the next integral as well?
