Integral calculation by using Mellin Transform I want to use the Mellin Transform (MT) to calculate the integral:
$\int_0^{1 } \exp(-2\rho^2) J_0(\pi  \rho  r)\rho \, d\rho$
in which $r>=0$ and real.
I have calculated it by numerical methods. However, I want to use MT to obtain a closed form. I have applied the convolution property in MT, and the equivalence between Laplace Transform and finite MT.
I get this Mellin-Barnes integral for my integral:
$\int_{0}^{1}g(r\rho)h(\rho)d\rho=\frac{1}{2\pi i}\oint_{\delta-i\infty}^{\delta+i\infty}g(z)h(1-z)r^{-z}dz\\=
\frac{1}{2\pi i}\oint_{\delta-i\infty}^{\delta+i\infty}\frac{\sqrt{\pi } e^{\frac{(z-1)^2}{8 }} \text{erfc}\left(-\frac{z-1}{2 \sqrt{2 }}\right)}{2 \sqrt{2 }}\frac{2^z \pi ^{-z-1} \Gamma \left(\frac{z+1}{2}\right)}{\Gamma \left(\frac{1}{2}-\frac{z}{2}\right)}r^{-z-1}dz$
since:
$h(z)=\int_0^{1 } \exp(-2\rho^2)\rho^{z-1} \, d\rho=\frac{\sqrt{\pi } e^{\frac{(z-1)^2}{8 }} \text{Erfc}\left(-\frac{z-1}{2 \sqrt{2 }}\right)}{2 \sqrt{2 }}\qquad g(z)=\frac{1}{r}\int_0^{\infty}J_0(\pi  \rho )\rho^{z} \, d\rho=\frac{2^z \pi ^{-z-1} \Gamma \left(\frac{z+1}{2}\right)}{r\,\Gamma \left(\frac{1}{2}-\frac{z}{2}\right)}$
However, I do not get an agreement with the numerical result after calculating the residues ($z=-(2n+1),\;n\geq0 $).
Any help and/or check will be appreciated.
 A: I have a summation result via Mellin Transform, introduce a parameter $a$ to give 
$$
I(a,r) = \int_0^{1 } \exp(-2a\rho^2) J_0(\pi  \rho  r)\rho \, d\rho
$$
take the Mellin transform of both sides changing variables from $a \to s$
$$
\mathcal{M}_{a\to s}[I(a,r)] = \int_0^{1 } \int_0^\infty a^{s-1}\exp(-2a\rho^2)\;da \; J_0(\pi  \rho  r)\rho \, d\rho
$$
$$
\mathcal{M}_{a\to s}[I(a,r)] = \frac{\Gamma(s)}{2^s}\int_0^{1 }\frac{J_0(\pi  \rho  r)\rho}{\rho^{2s}} \, d\rho
$$
this gives a hypergeometric function according to Mathematica
$$
\mathcal{M}_{a\to s}[I(a,r)] = \frac{\Gamma(s)}{2^s(2-2s)}\;_1F_2\left(1-s;1,2-s;\frac{-\pi^2r^2}{4}\right)
$$
using the Ramanujan Master theorem if the Mellin transform of $I(a,r)=\Gamma(s)\phi(-s)$ then
$$
I(a,r)=\sum_{k=0}^\infty \phi(k)\frac{(-a)^k}{k!}
$$
so we can write the orginal integral as a series expansion
$$
I(a,r)=\sum_{s=0}^\infty \frac{2^s}{(2+2s)}\;_1F_2\left(1+s;1,2+s;\frac{-\pi^2r^2}{4}\right)\frac{(-a)^s}{s!}
$$
and by setting $a=1$ we have a series representation for the integral you were originally interested in
$$
\int_0^{1 } \exp(-2\rho^2) J_0(\pi  \rho  r)\rho \, d\rho=\sum_{s=0}^\infty \frac{(-1)^s 2^s}{(2+2s)s!}\;_1F_2\left(1+s;1,2+s;\frac{-\pi^2r^2}{4}\right)
$$
this seems to work numerically for a few values of $r$ at least. The hypergeometric function appears to reduce to sums of Bessel-J functions divided by powers of r.
