Let $ X_1 $ and $ X_2 $ be iid with pdf $$ f(x)=\lambda e^{-\lambda x}, \quad \lambda>0, \quad x \in (0, \infty)$$ Find the density function of $Z = X_1+X_2$.
I tried to solve it using this formula: $$ f_Z \left( z \right) = \int_{-\infty}^{\infty} f_X \left( z - y \right) f_Y \left( y \right) d y $$
$$ \lambda^2 \int_{-\infty}^{\infty} e^{-\lambda (z-w)} \cdot e^{-\lambda w} dw= \lambda^2 \int_{-\infty}^{\infty} e^{-\lambda z} dw$$ but the integral diverges...
I tried the second way: $$ F(x) = 1-e^{-\lambda z} \quad Z=X_1+X_2 = 2X_1$$
$ F_{2X_1}(z)=P(2X_1 \le z) = P(X_1 \le \frac{t}{2})=1-e^{\frac{-\lambda}{2} z} \\ $
$ f_Z(z)=\frac{-\lambda}{2} e ^{\frac{-\lambda}{2} z} $ Is it ok?
Which way is good?
Specifically, you should be guided by the general result $X_{1}+X_{2} \sim Gamma(2, \lambda)$. In particular, using convolution you should integrate over $(-\infty, z]$, because, as @Gaffney said, the pdf is $0$ for negative values. Hence,
\begin{align} f_Z \left( z \right) &= \int_{-\infty}^{z} f_X \left( z - y \right) f_Y \left( y \right) \mathrm{d} y\\ &= \int_{-\infty}^{z} \lambda e^{-\lambda\left( z - y \right)} \lambda e^{-\lambda y} \mathrm{d} y \\ &=\lambda ^2 e^{-\lambda z}z, \quad z\ge 0 . \end{align} Which is pdf of $Erlang(2,\lambda)$. Another possible way to show it is by using MGF of the exponential distribution.