Let $X$ and $Y$ be independent random variables with densities $$f_{X}(x)= \begin{cases} \gamma e^{-\gamma x} & \text{, } x\geq0 \\ 0& \text{, } x<0 \end{cases}$$
$$f_Y(y)= \begin{cases} \mu e^{-\mu x} & \text{, } y\geq0 \\ 0& \text{, } y<0 \end{cases}$$ where $\gamma$ and $\mu$ are positive constants.
a) Evaluate the density of the sum $Z=X+Y$.
b) Repeat in the case where $\mu = \gamma$.
My solution so far:
a)
$$f_Z(z) = \int_0^z f_X(x)f_Y{(z-x)} \, dx=\int_0^z \gamma e^{-\gamma x}\mu e^{-\mu (z-x)}\,dx$$
$$= \gamma\mu \int_{0}^{z}e^{-\gamma x-\mu z+\mu x}\,dx$$ $$= \gamma\mu e^{-\mu z}\int_{0}^{z}e^{x(\mu-\gamma)}\,dx$$ $$= \gamma\mu e^{-\mu z} \left. \left [ \frac{e^{(\mu-\gamma)x}}{\mu-\gamma} \right ]\right|_0^z$$ $$=\frac{\gamma\mu e^{-\mu z}}{\mu-\gamma}\left [ e^{(\mu-\gamma)z}-1 \right ].$$
So, $$f_Z(z)= \begin{cases} \frac{\gamma\mu e^{-\mu z}}{\mu-\gamma}\left [ e^{(\mu-\gamma)z}-1 \right ] & \text{, } z\geq0 \\ 0 & \text{, } z<0 \end{cases} $$
b)
For the case where $\mu = \gamma$, $f_{Y}(y)=\begin{cases} \gamma e^{-\gamma y} & \text{, } y\geq 0 \\ 0& \text{, } y<0 \end{cases}.$
\begin{align} f_Z(z) & = \int_0^z f_X(x)f_Y{(z-x)} \, dx=\int_0^z \gamma e^{-\gamma x} \gamma e^{-\mu (z-x)} \, dx \\[10pt] & = \gamma^2 \int_0^z e^{(-\gamma x+\gamma x-\gamma z)} \, dx \\[10pt] & = \gamma^2 \int_0^z e^{-\gamma z} \, dx \\[10pt] & = \gamma^2 e^{-\gamma z}\int_0^z 1\,dx \\[10pt] & =\gamma^2 e^{-\gamma z}[x]\Big|_0^z \\[10pt] & =\gamma^2 e^{-\gamma z}z. \end{align}
So, $$f_Z(z)= \begin{cases} \gamma^{2}e^{-\gamma z}z & \text{, } z\geq0, \\ 0 & \text{, } z<0. \end{cases}$$
I just wanted to make sure that what I have is correct. Thanks for answering!!
My answers agree with yours (except for the typo where you have $\mu-z$ where you need $\mu-\gamma$.)
Note that both of these densities start with value $0$ at $z=0,$ unlike the exponential density, and that makes sense when you consider that in a tiny interval $[0,\varepsilon],$ the second arrival is not likely to be seen because the first arrival hasn't happened yet.