Help in computing this definite integral I want to compute the definite integral $$\int\limits_{-\infty}^{\infty}e^{\frac{-|x-t_1|}{\alpha}}e^{\frac{-|x-t_2|}{\alpha}}dx$$ $\alpha \in \mathbb{R},\alpha >0$
I tried Wolfram Alpha, but it shows answer as $\infty$, but I feel answer should be a finite positive number.
I computed taking $t_1 = t_2 = 0$, and it comes down to $$\int\limits_{-\infty}^{\infty}e^{\frac{-2|x|}{\alpha}}dx = \alpha$$ and I verified it is $\alpha$ using wolfram. So I feel its likely that the integral is finite also for the case where both $t_1$ and $t_2$ are not zeros.
I'd appreciate If I can get an answer for the integral in terms of $\alpha,t_1,t_2$.
 A: Since the absolute value function is defined piecewise, the natural approach is to split the integral up into intervals corresponding to that piecewise definition. Without loss of generality assume $t_1\le t_2$; then
\begin{align*}
\int_{-\infty}^{\infty} & e^{-|x-t_1|/\alpha}e^{-|x-t_2|/\alpha} \,dx \\
&= \int_{-\infty}^{t_1}e^{-|x-t_1|/\alpha}e^{-|x-t_2|/\alpha} \,dx + \int_{t_1}^{t_2}e^{-|x-t_1|/\alpha}e^{-|x-t_2|/\alpha} \,dx + \int_{t_2}^{\infty}e^{-|x-t_1|/\alpha}e^{-|x-t_2|/\alpha} \,dx \\
&= \int_{-\infty}^{t_1}e^{(x-t_1)/\alpha}e^{(x-t_2)/\alpha} \,dx + \int_{t_1}^{t_2}e^{(t_1-x)/\alpha}e^{(x-t_2)/\alpha} \,dx + \int_{t_2}^{\infty}e^{(t_1-x)/\alpha}e^{t_2-x)/\alpha} \,dx \\
&= e^{-(t_1+t_2)/\alpha} \int_{-\infty}^{t_1}e^{2x/\alpha} \,dx + e^{(t_1-t_2)/\alpha}\int_{t_1}^{t_2}1 \,dx + e^{(t_1+t_2)/\alpha} \int_{t_2}^{\infty}e^{-2x/\alpha} \,dx \\
&= e^{-(t_1+t_2)/\alpha} \frac\alpha2 e^{2x/\alpha}\bigg|_{-\infty}^{t_1} + e^{(t_1-t_2)/\alpha}(t_2-t_1) - e^{(t_1+t_2)/\alpha} \frac\alpha2  e^{-2x/\alpha} \bigg|_{t_2}^{\infty} \\
&= e^{-(t_1+t_2)/\alpha} \frac\alpha2 e^{2t_1/\alpha} + e^{(t_1-t_2)/\alpha}(t_2-t_1) + e^{(t_1+t_2)/\alpha} \frac\alpha2  e^{-2t_2/\alpha} \\
&= e^{(t_1-t_2)/\alpha}(\alpha+t_2-t_1).
\end{align*}
A: Suppose $t_1<t_2.$ Then
$$
-|x - t_1| - |x-t_2| = \begin{cases} 2x - t_1 - t_2 & \text{if } x < t_1 < t_2, \\ t_1-t_2 & \text{if } t_1 \le x \le t_2, \\ t_1+t_2-2x & \text{if } t_1<t_2<x. \end{cases}
$$
Integrate over the three intervals separately and add them up.
