Let $X_1$ and $X_2$ be two independent exponential random variables with the PDFs $f_{X_1}(x_1)=\lambda_1 \exp(-\lambda_1 x_1)$ and $f_{X_2}(x_2)=\lambda_2 \exp(-\lambda_2 x_2)$.
Let $X= \frac{X_1X_2}{X_1+X_2+a}$ be a r.v. for which I want to derive the CDF, where $a$ is a positive constant.
In other words, I need to calculate $\mathbb{P}\{X <x \}$.
Any ideas or hints?
Edit (attempt): Using a similar approach to that used here, we have $X_1 \in [0, \infty)$ and $X_2 \in [0, \frac{x(x_1+a)}{x_1-x})$. The latter interval results from: $\frac{x_1x_2}{x_1+x_2+a} < x$, thus $x_2(x_1-x)<x(x_1+a)$ which implies $x_2 < \frac{x(x_1+a)}{x_1-x}$; but I am not sure if this is correct since $x_1-x$ can sometimes be negative (?). Based on the above, we can write: \begin{align} \mathbb{P} \{ \frac{X_1X_2}{X_1+X_2+a} < x \} & = \int_{x_1=0}^\infty \int_{x_2=0}^{ \frac{x(x_1+a)}{x_1-x}} \lambda_1 \exp(-\lambda_1 x_1) \lambda_2 \exp(-\lambda_2 x_2) dx_2 dx_1 \\ &= \int_{x_1=0}^\infty \left(1-\exp(-\lambda_2\frac{x(x_1+a)}{x_1-x}) \right) \lambda_1 \exp(-\lambda_1 x_1) dx_1 \\ & = 1-\lambda_1 \int_{x_1=x}^\infty \exp(-\lambda_2\frac{x(x_1+a)}{x_1-x} ) \exp(-\lambda_1 x_1) dx_1 \end{align} In the last equality: for the integral of $\lambda_1 \exp(-\lambda_1 x_1) dx_1$, I suppose that $x_1 \in [0,\infty)$, whereas for the second term I suppose that $x_1 \ge x$ (since $x_2$ should be positive) (?).
Is my approach correct ?