Lebesgue density of a sum of i.i.d. exponentially distributed random variables Let $X$ be an exponentially distributed random variable with parameter $\lambda$, i.e., its Lebesgue density is given by
$$
f(x) =
\begin{cases}
\lambda e^{- \lambda x} & \text{if} \quad x \geq 0 \\
0 & \text{if} \quad x < 0
\end{cases}
$$
The characteristic function of $X$ can be then computed through
$$
\varphi(z) = E[e^{izX}] = \int_{0}^{\infty} e^{izx} \lambda e^{ - \lambda x} d x = \lambda \int_{0}^{\infty} \frac{ e^{ (iz - \lambda)x} }{iz - \lambda } d ((iz - \lambda)x) = \lambda \frac{ e^{ (iz - \lambda) x} }{i z - \lambda} = \frac{\lambda}{\lambda  - iz }
$$
If we further take $n$ i.i.d. copies of $X$, then the characterisitic function of their sum is given by
$$
\frac{ \lambda^n}{ {( \lambda - i z )}^n} \tag{1}
$$
Now, consider a somewhat reverse problem, namely, given the characteristic function in $(1)$, can one come up with a Lebesgue density? I would be interested in an explicit computation. Can this be helpful?
 A: We can compute the density of $S_n=\sum_{i=1}^n X_i$ directly by convolution. Assume that for some $n\geqslant 1$ that $$f_{S_n}(x) = \frac{\lambda(\lambda x)^{n-1}}{(n-1)!} e^{-\lambda x}. $$
Then
\begin{align}
f_{S_{n+1}}(x) &= f_{S_{n}}\star f_{X_{n+1}}(x)\\
&= \int_0^x \frac{\lambda(\lambda y)^{n-1}}{(n-1)!} e^{-\lambda y} \lambda e^{-\lambda(x-y)}\ \mathsf dy\\
&= \frac{\lambda e^{-\lambda x}}{(n-1)!}\int_0^x\lambda(\lambda y)^{n-1}\ \mathsf dy\\
&= \frac{\lambda e^{-\lambda x}}{(n-1)!}\left[\frac{(\lambda y)^n}n \right]_0^x\\
&= \frac{\lambda (\lambda x)^n}{n!}e^{-\lambda x}.
\end{align}
By induction, we see that $f_{S_n} = \frac{\lambda(\lambda x)^{n-1}}{(n-1)!} e^{-\lambda x}$ for all positive integers $n$.
A: I doubt this is possible in general.
Here is a useful result that may work sometimes.
Let $X: \Omega \to \mathbb{R}^k$ be a random vector with characteristic function $\phi$ such that $$\int_{\mathbb{R}^k} |\phi(t)| dt < \infty$$
Then $\mathbb{P}_X$ is absolutely continuous w.r.t. Lebesgue measure with density function
$$x \mapsto \frac{1}{(2 \pi)^k} \int_{\mathbb{R}^k} \exp(-i \langle t,x \rangle) \phi(t)dt $$
