I am trying to find the PDF of $Z$, where $Z = X_1 \cdot X_2$ such that both $X_1$, $X_2$ are independent and identically distributed random variables with PDF $f_{X_1}(x_1) = \lambda_1\exp(-\lambda_1x_1)$ and $f_{X_2}(x_2) = \lambda_2\exp(-\lambda_2x_2)$. So, using product distribution and Mellin convolution, the PDF of $Z$ can be written as
$f_Z(z) = \int_0^{\infty}\frac{1}{x_2}f_{X_1}(\frac{z}{x_2})f_{X_2}(x_2)\text{d}x_2$----(1)
My query is did I correctly represented the Eq.(1)
Any help in this regard will be highly appreciated.
Without applying directly the convolution but using the definition of CDF you get
$$F_Z(z)=P(Z\leq z)=P(Y\leq z/X)=\int_0^{\infty}\lambda e^{-\lambda x}\left[ \int_0^{z/x}\theta e^{-\theta y} dy \right]dx$$
thus the requested pdf is the derivative of F
$$f_Z(z)=\frac{d}{dz}F_Z(z)=\int_0^{\infty}\lambda e^{-\lambda x}\theta e^{-\theta z/x}\frac{1}{x}dx=\int_0^{\infty}f_X(x)\frac{1}{x}f_Y\left( \frac{z}{x} \right)dx$$
which is exactly your result. To simplify the notation, avoiding indexes, I used $X,Y$ as the two independent exponential rv's and $\lambda,\theta$ as the two parameters