Convolution of two independent exponential random variables

Let $$X\sim \text{Exponential}(\frac{1}{\lambda})$$ and let $$Y \sim \text{Exponential}(\frac{1}{\mu})$$. Let $$Z = X+Y$$. I want to find the pdf of $$Z$$. I start by using the convolution $$f_Z(z) = \int_{-\infty}^{\infty} f_X(z-t)f_Y(t)dt$$

$$f_Z(z) = \int_{0}^{\infty} \frac{1}{\lambda}e^{-\frac{z-t}{\lambda}}\cdot \frac{1}{\mu}e^{-\frac{t}{\mu}}dt = \frac{e^{-\frac{z}{\lambda}}}{\lambda\mu}\int_{0}^{\infty}e^{t(\frac{1}{\lambda}-\frac{1}{\mu})}dt$$

But the final integral I have is undefined. I assume I have made an error somewhere. Any help is appreciated.

• Asked here many times before. Here are some relevant threads: math.stackexchange.com/questions/655302/…, math.stackexchange.com/questions/635443/…, math.stackexchange.com/questions/2018282/… – StubbornAtom Oct 6 '18 at 9:55
• I had already seen the first two links you gave. The first one the parameters are the same. In the second one, there is some notation that I am unfamiliar with, and the question is several years old, so I didn't think I'd get a response posting on it. I hadn't seen the third link. From reading that, it seems my limit should be up to $z$ rather than $\infty$, is that correct? – mrnovice Oct 6 '18 at 10:06
• Consider the support of the densities. $$z-t>0,t>0\implies 0<t<z$$ – StubbornAtom Oct 6 '18 at 10:09
• Ah thanks, that solved my problem. Feel free to submit it as an answer if you want. – mrnovice Oct 6 '18 at 10:11