How could i find the pdf of exponential distribution from its characteristics function? I know that the characteristics function of the exponential distribution is as following:
$$ \phi_x(t)  =\frac{\lambda}{(\lambda -it)}$$
Also, I know that the pdf of the exponential distribution is:
$$f_x(x)=\lambda e^{\lambda x}$$
Moreover, I know that the relation ship between the pdf and the characteristics function can be describe as following:
$$ f_x(x)= \int_0^\infty e^{-itx} \phi_x(t) $$
$$ f_x(x)= \int_0^\infty e^{-itx} \frac{\lambda}{(\lambda -it)} $$
However, I can't compute the last equation to find the exactly pdf that i already mentioned before. Could you guys help me to solve this integral. I used wolfram, but without any result.
Thanks .
 A: If you start defining the probability density function as
$$
f(x) = \lambda e^{-\lambda x} \Theta(x),
$$
with $\lambda \in \mathbb{R}$, $\lambda>0$ and $\Theta(x)$ the Heaviside step function since we only want to take care of $x>0$. We can take its characteristic function performing the inverse Fourier transform of it, where
$$
\phi(t) = \frac{1}{\sqrt{2\pi}} \int_\mathbb{R} e^{itx}f(x) \mathrm{d}x = \frac{1}{\sqrt{2\pi}} \int_0^\infty e^{itx} \lambda e^{-\lambda x} \mathrm{d}x = \frac{1}{\sqrt{2\pi}}\frac{\lambda}{\lambda - it},
$$
as written. We can invert the process computing the Fourier transform of the characteristic function $\phi(t)$, where
$$
\begin{aligned}
f(x) &= \frac{1}{\sqrt{2\pi}} \int_\mathbb{R} e^{-ixt}\phi(t) \mathrm{d}t = -\frac{\lambda e^{-\lambda x}}{2}  \left[\mathrm{sgn}(x)\left(\mathrm{sgn}(|\mathrm{Re}(\lambda)|) - 1\right) -2\,\mathrm{sgn}\left(\mathrm{Re}(\lambda)\right)\Theta(x\,\mathrm{sgn}\left(\mathrm{Re}(\lambda)\right))\right] \\
&= \lambda e^{-\lambda x} \Theta(x),
\end{aligned}
$$
since $\mathrm{sgn}\left(\mathrm{Re}(\lambda)\right) = 1$.
