Laplace transform calculation Please can someone help me? I have tried to find the Laplace transform of the form:
$$\int_{0}^{+\infty} (v+1)^{\nu}(2v+1)^{k}v^{\alpha} \exp(-pv), \mbox{ where }\alpha,\nu, k \mbox{ are integers }$$
I have searched on the two Books "Erdelyi. Tables of integral transforms, Vol I, 1954" and "Oberhettinger, L. Badii, Tables of Laplace transforms, 1973. But i didn't found this type of integral transform is there any reference or  method in order to compute this formula above.
Thanks and best regards.
 A: For $n,k\in\mathbb{N}$ and $\alpha\geq 0$ we can use the binomial expansion,
$$(v+1)^{\nu}(2v+1)^{k}v^{\alpha}=\sum_{s=0}^{\nu+k}\sum_{q=0}^{s}{{\nu}\choose{q}} {{k}\choose{s-q}}2^{s-q}v^{s+\alpha} $$
$$\Rightarrow \int_{0}^{\infty} (v+1)^{\nu}(2v+1)^{k}v^{\alpha} e^{-pv}\,dv=\sum_{s=0}^{\nu+k}\sum_{q=0}^{s}{{\nu}\choose{q}} {{k}\choose{s-q}}2^{s-q}\frac{(s+\alpha)!}{p^{s+\alpha+1}}$$
$$\qquad=\sum_{s=0}^{\nu+k}2^s \binom{k}{s} \, _2F_1\left(-\nu,-s;k-s+1;\tfrac{1}{2}\right)\frac{(s+\alpha)!}{p^{s+\alpha+1}}.$$
A: Another way to look at this integral is by considering $$F(q,r,p) = \int_{0}^{\infty}\exp\left[-q(v+1)-r(2v+1)-pv\right]\,dv = \frac{\exp(-q-r)}{p+q+2r}$$ for $p+q+2r>0$. If $\nu\geq 0$, $k\geq 0$, and $\alpha\geq 0$, then your integral can be represented as
$$(-1)^{\nu+k+\alpha}\left.\frac{\partial^{\nu+k+\alpha}F}{\partial q^{\nu}\,\partial r^{k}\,\partial p^{\alpha}}\right|_{q\,=\,r\,=\,0}.$$
For negative values of $\nu$ or $k$, integration should be used instead of differentiation.
Of course, it depends on what you're looking for, so this method may not give anything better than the previous answer.
