Inverse Laplace transform seems to be always vanishing but it couldn't! Let's consider $x\in (0,1)$ and the distribution $p(x)=\lambda x^\lambda$, $\lambda>0$. I would like to find the pdf of the sum. The characteristic function of the $N$ sum reads:
\begin{equation}
\phi_{S_N}=(\lambda s^{-\lambda}[\Gamma(\lambda)-\Gamma(\lambda,s)])^N
\end{equation} 
To compute the inverse Laplace transform I set $\lambda N\in \mathcal N$ (i.e if $\lambda=0.1$ I use only $N=10,20,..$) and I use the Residue theorem. But this give me ALWAYS ZERO since the only pole is in $s=0$ and $\Gamma(\lambda,0)=\Gamma(\lambda)$. Is this CORRECT? I have also decomposed each term using the Binomial theorem and the result is always vanishing. Since this defines a pdf it couldn't be vanishing everywhere! Someone see the error ?
Cheers
 A: The problem is that you are ignoring the effect of the branch point at $s=0$ due to the $s^{-\lambda}$ behavior.  This is true despite the pole.  To illustrate, let's expand out the binomial and look at the first term, the constant. The ILT in this case is
$$\frac{\Gamma(\lambda)^N}{i 2 \pi} \int_{c-i \infty}^{c+ i \infty} ds \, s^{-\lambda} e^{s t} $$
To evaluate this, we consider a Bromwich contour $C$ to the left of the line $\operatorname{Re}{z}=c$.  However, we take the negative real axis to be a branch cut, so we actually have separate circular arcs of radius $R$, divided by separate traversals above and below the negative real axis and a circular arc about the origin of radius $\epsilon$.  The integral about this contour,
$$\oint_C dz \, z^{-\lambda} e^{t z}$$
is equal to
$$\int_{c-i R}^{c+i R} ds \, s^{-\lambda} e^{s t} + i R^{1-\lambda} \int_{\pi/2}^{\pi} d\theta \, e^{i (1-\lambda) \theta} e^{t R e^{i \theta}} \\ + e^{i \pi} \int_R^{\epsilon} dx \, e^{-i \lambda \pi}  x^{-\lambda} e^{-t x}+i \epsilon^{1-\lambda} \int_{\pi}^{-\pi} d\phi \, e^{i (1-\lambda)\phi} e^{t \epsilon e^{i \phi}}\\+ e^{-i \pi} \int_{\epsilon}^R dx \, e^{i \lambda \pi} x^{-\lambda} e^{-t x}+i R^{1-\lambda} \int_{\pi}^{3 \pi/2} d\theta \, e^{i (1-\lambda) \theta} e^{t R e^{i \theta}}$$
We can show that the second and sixth integrals vanish as $R \to \infty$.  For example, the magnitude of the second integral is bounded by
$$R^{1-\lambda} \int_0^{\pi/2} d\theta \, e^{-t R \sin{\theta}} \le R^{1-\lambda} \int_0^{\pi/2} d\theta \, e^{-2 t R \theta/\pi} \le \frac{\pi}{2 t R^{\lambda}}$$
which vanishes because $\lambda \gt 0$.
We can also see that the fourth integral vanishes as $\epsilon \to 0$ because $\lambda \lt 1$.
In these limits, then, we are left with, as the contour integral,
$$\int_{c-i \infty}^{c+i \infty} ds \, s^{-\lambda} e^{s t} + \left ( e^{-i \lambda \pi} - e^{i \lambda \pi} \right ) \int_0^{\infty} dx \, x^{-\lambda} e^{-t x} $$
By Cauchy's theorem, the contour integral is zero because there are no poles inside the contour $C$.  Thus, we may now write down the ILT:
$$\frac{\Gamma(\lambda)^N}{i 2 \pi}\int_{c-i \infty}^{c+i \infty} ds \, s^{-\lambda} e^{s t} = \frac{t^{-(1-\lambda)}}{\pi} \Gamma(\lambda)^N \Gamma(1-\lambda) \sin{\pi \lambda} = \Gamma(\lambda)^{N-1} t^{-(1-\lambda)}$$
Of course, in reality, you have to contend with all of the other terms involving the upper incomplete Gamma, or the binomial taken as a whole.  But what I did should outline for you the approach necessary to compute the ILT here.
