bookmark_borderRamanujan’s Master Theorem: Part I

Ramanujan’s master theorem (RMT) is powerful tool that provides the Mellin transform of an analytic function using it’s power series around zero.

Theorem (RMT): If the complex-valued function f(z) has the following expansion f(z) = \sum_{n=0}^\infty \frac{\varphi(n)}{n!}(-z)^n then it’s Mellin transform is given by \int_0^\infty x^{s-1} f(x)\; dx = \Gamma(s) \varphi(-s)

Sometimes RMT provides short proofs for integrals that appear to be very difficult. One such example is the following: \int_0^\infty x^{\sigma-1} J_\nu (x) \; dx = \frac{2^{\sigma-1}\Gamma\left(\frac{\sigma+\nu}{2}\right)}{\Gamma\left(\frac{2-\sigma+\nu}{2} \right)} \; ,\quad -\nu < \sigma < \frac{3}{2} \quad (1) where J_\nu (x) is the Bessel function of the first kind.

Indeed, J_\nu(x) has the following series expansion around x=0: J_\nu(x) =\sum_{n=0}^\infty \frac{(-1)^n }{n! \Gamma(n+\nu+1)}\left(\frac{x}{2} \right)^{\nu+2n} Hence, we may apply the RMT to the function x^{-\nu/2}J_\nu(2\sqrt{x}) to obtain: \int_0^\infty x^{s-\nu/2-1} J_\nu(2\sqrt{x}) dx=\frac{\Gamma(s)}{\Gamma(\nu+1-s)} Now, one can perform the substitutions x=t^2/4 and s=\frac{\sigma+\nu}{2} to transform the above integral into the integral of equation (1).

It turns out that a similar approach also works for the product J_\nu (x) J_\mu(x). First, we need to compute it’s series expansion. This can be done by multiplying the individual series expansions and rearranging/grouping certain terms.

\begin{aligned} J_{\nu}(x)J_{\mu}(x) &= \left(\sum_{n=0}^\infty \frac{(-1)^n }{n! \Gamma(n+\nu+1)}\left(\frac{x}{2} \right)^{\nu+2n}\right) \left(\sum_{m=0}^\infty \frac{(-1)^m }{m! \Gamma(m+\nu+1)}\left(\frac{x}{2} \right)^{\nu+2m} \right) \\ &= \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{(-1)^{n+m} x^{\nu+\mu+2(m+n)}}{2^{\nu+\mu+2(m+n)}} \cdot \frac{1}{m! n!\Gamma(n+\nu+1)\Gamma(m+\mu+1)} \\ &= \sum_{r=0}^\infty \frac{(-1)^{r} x^{\nu+\mu+2r}}{2^{\nu+\mu+2r}}\sum\limits_{\substack{0\leq m,n \leq r \\ m+n=r}}\frac{1}{m! n!\Gamma(n+\nu+1)\Gamma(m+\mu+1)} \\ &= \sum_{r=0}^\infty \frac{(-1)^{r} x^{\nu+\mu+2r}}{2^{\nu+\mu+2r} r! \Gamma(r+\nu+1)\Gamma(r+\mu+1)}\sum\limits_{\substack{0\leq m,n \leq r \\ m+n=r}}\binom{r}{m,n} \left(\prod_{i=m+1}^r (\nu+i)\right) \left(\prod_{j=n+1}^r (\mu+j)\right) \\ &= \sum_{r=0}^\infty \frac{(-1)^{r} x^{\nu+\mu+2r}}{2^{\nu+\mu+2r} r! \Gamma(r+\nu+1)\Gamma(r+\mu+1)}\sum _{n=0}^r\binom{r}{n}(\nu +r)^{\underline{n}}(\mu +r)^{\underline{r-n}} \\ &= \sum_{r=0}^\infty \frac{(-1)^{r} x^{\nu+\mu+2r}}{2^{\nu+\mu+2r} r! \Gamma(r+\nu+1)\Gamma(r+\mu+1)} (\nu +\mu+2r)^{\underline{r}} \\ &= \sum_{r=0}^\infty \frac{(-1)^{r} \Gamma(\nu+\mu+2r+1)}{ r! \Gamma(r+\nu+1)\Gamma(r+\mu+1)\Gamma(\mu+\nu+r+1)} \left( \frac{x}{2}\right)^{\nu+\mu+2r} \quad \; \quad (2) \end{aligned}

To simplify the inner summation, we used the binomial theorem for falling factorials.

Therefore, applying RMT to the function x^{-\frac{\nu+\mu}{2}}J_{\nu}(2\sqrt{x})J_{\mu}(2\sqrt{x}), gives us: \int_0^\infty x^{s-\frac{\nu+\mu}{2}-1} J_{\nu}(2\sqrt{x})J_{\mu}(2\sqrt{x}) \; dx = \frac{\Gamma(s)\Gamma(\nu+\mu+1-2s)}{\Gamma(\nu+1-s)\Gamma(\mu+1-s)\Gamma(\nu+\mu+1-s)} Finally, make the substitutions x=t^2/4 and s=\frac{\sigma+\nu+\mu}{2} in the above integral to get: \boxed{ \int_0^\infty x^{\sigma-1}J_\nu(x)J_\mu(x)\; dx = \frac{2^{\sigma-1} \Gamma\left(\frac{\nu+\mu+\sigma}{2} \right)\Gamma\left(1-\sigma\right)}{\Gamma\left(\frac{\nu-\mu-\sigma}{2} +1\right)\Gamma\left(\frac{\mu-\nu-\sigma}{2} +1\right)\Gamma\left(\frac{\nu+\mu-\sigma}{2} + 1\right)} } \quad \;\quad (3) The above equation holds provided that the condition -(\nu+\mu)< \sigma < 1 is satisfied.