I have the following function:
$$\frac{1}{2\beta} e^{\frac{-|x-a|}{\beta}} $$ where $\beta >0$ and $-\infty< \alpha < \infty$ and similarly $-\infty<x<\infty$
Here's how far I have reached:
$$\int_{-\infty}^{\infty}\frac{1}{2\beta} e^{\frac{-|x-a|}{\beta}}e^{tx}dx$$ so we get:
$$\int_{-\infty}^{\alpha}\frac{1}{2\beta} e^{\frac{x-a}{\beta}}e^{tx}dx +\int_{\alpha}^{\infty}\frac{1}{2\beta} e^{\frac{-(x-a)}{\beta}}e^{tx}dx $$
and now I'm stuck here... I tried using an online calculator to figure the integral out, the first term works; however, the second term seems to always diverge. Any suggestions?
The integral can be calculated without using a calculator. I´m focussing on the second integral.
$$\int\frac{1}{2\beta} e^{\frac{-(x-a)}{\beta}}e^{tx}\, dx=\int\frac{1}{2\beta} e^{\frac{a-x}{\beta}}e^{tx} \, dx =\frac{1}{2\beta}\cdot e^{\frac{a}{\beta}}\cdot \int e^{\frac{-x}{\beta}}\cdot e^{tx} \, dx$$ $$=\frac{1}{2\beta}\cdot e^{\frac{a}{\beta}}\cdot \int e^{\left(t-\frac{1}{\beta}\right)x} \, dx=\frac{1}{2\beta}\cdot e^{\frac{a}{\beta}}\cdot \frac1{\left(t-\frac{1}{\beta}\right)}\cdot e^{\left(t-\frac{1}{\beta}\right)x}+ \color{grey}C$$
Inserting the limits. We assume that $t<\frac{1}{\beta}$ so that $\lim\limits_{x \to \infty}e^{\left(t-\frac{1}{\beta}\right)x}=0$. That means that the value with the upper limit is 0. Subtracting the term with the lower limit $\alpha$:
$$-\frac{1}{2\beta}\cdot e^{\frac{a}{\beta}}\cdot \frac1{\left(t-\frac{1}{\beta}\right)}\cdot e^{\left(t-\frac{1}{\beta}\right)\alpha}=-\frac{1}{2\beta}\cdot \frac1{\left(t-\frac{1}{\beta}\right)}\cdot e^{\alpha t}$$