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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.

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  • $\begingroup$ Further thinking about where the error can be, I guess I applied wrongly the equivalence between finite MT and Laplace Transform... $\endgroup$ – José Antonio Díaz Navas Jan 24 '16 at 18:31
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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.

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  • $\begingroup$ Nice solution. I am wondering what about the numerical stability depending on $a$ value... $\endgroup$ – José Antonio Díaz Navas Oct 18 '17 at 18:01
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    $\begingroup$ For your original integral it shouldn't matter as it's temporary, but if you want to change the exponent it may only work with $a \le 1$ for the second to last equation to converge. I'm certain on that point however I'll see if I have time to check in a bit. $\endgroup$ – Benedict W. J. Irwin Oct 19 '17 at 10:14
  • $\begingroup$ Yes, you are right. Your solution gives good results for $a\leq1$,as I have compared it to my analytical solution by means of integration by parts. Sorry about my math ignorance, but where is the limitation in $a$ value when doing the Mellin Transform? $\endgroup$ – José Antonio Díaz Navas Oct 26 '17 at 10:48

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