\nRamanujan’s master theorem (RMT) is powerful tool that provides the Mellin transform of an analytic function using it’s power series around zero. \n<\/p>\n
\nTheorem (RMT):<\/b> If the complex-valued function $f(z)$ has the following expansion\n$$ f(z) = \\sum_{n=0}^\\infty \\frac{\\varphi(n)}{n!}(-z)^n $$\nthen it’s Mellin transform is given by\n$$\\int_0^\\infty x^{s-1} f(x)\\; dx = \\Gamma(s) \\varphi(-s)$$\n<\/p>\n
\nSometimes RMT provides short proofs for integrals that appear to be very difficult. One such example is the following:\n$$\n\\int_0^\\infty x^{\\sigma-1} J_\\nu (x) \\; dx = \\frac{2^{\\sigma-1}\\Gamma\\left(\\frac{\\sigma+\\nu}{2}\\right)}{\\Gamma\\left(\\frac{2-\\sigma+\\nu}{2} \\right)} \\; ,\\quad -\\nu < \\sigma < \\frac{3}{2} \\quad (1)\n$$\nwhere $ J_\\nu (x) $ is the Bessel function of the first kind.\n<\/p>\n\n\n\n
\nIndeed, $J_\\nu(x)$ has the following series expansion around $x=0$:\n$$J_\\nu(x) =\\sum_{n=0}^\\infty \\frac{(-1)^n }{n! \\Gamma(n+\\nu+1)}\\left(\\frac{x}{2} \\right)^{\\nu+2n}$$\nHence, we may apply the RMT to the function $x^{-\\nu\/2}J_\\nu(2\\sqrt{x})$ to obtain:\n$$\\int_0^\\infty x^{s-\\nu\/2-1} J_\\nu(2\\sqrt{x}) dx=\\frac{\\Gamma(s)}{\\Gamma(\\nu+1-s)}$$\nNow, 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).\n<\/p>\n\n\n\n
\nIt 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.\n<\/p>\n
\n$$\n\\begin{aligned}\nJ_{\\nu}(x)J_{\\mu}(x) &= \\left(\\sum_{n=0}^\\infty \\frac{(-1)^n }{n! \\Gamma(n+\\nu+1)}\\left(\\frac{x}{2} \\right)^{\\nu+2n}\\right) \\left(\\sum_{m=0}^\\infty \\frac{(-1)^m }{m! \\Gamma(m+\\nu+1)}\\left(\\frac{x}{2} \\right)^{\\nu+2m} \\right) \\\\\n&= \\sum_{n=0}^\\infty \\sum_{m=0}^\\infty \\frac{(-1)^{n+m} x^{\\nu+\\mu+2(m+n)}}{2^{\\nu+\\mu+2(m+n)}} \\cdot \\frac{1}{m! n!\\Gamma(n+\\nu+1)\\Gamma(m+\\mu+1)} \\\\\n&= \\sum_{r=0}^\\infty \\frac{(-1)^{r} x^{\\nu+\\mu+2r}}{2^{\\nu+\\mu+2r}}\\sum\\limits_{\\substack{0\\leq m,n \\leq r \\\\ m+n=r}}\\frac{1}{m! n!\\Gamma(n+\\nu+1)\\Gamma(m+\\mu+1)} \\\\\n&= \\sum_{r=0}^\\infty \\frac{(-1)^{r} x^{\\nu+\\mu+2r}}{2^{\\nu+\\mu+2r} r! \\Gamma(r+\\nu+1)\\Gamma(r+\\mu+1)}\\sum\\limits_{\\substack{0\\leq m,n \\leq r \\\\ m+n=r}}\\binom{r}{m,n} \\left(\\prod_{i=m+1}^r (\\nu+i)\\right) \\left(\\prod_{j=n+1}^r (\\mu+j)\\right) \\\\\n&= \\sum_{r=0}^\\infty \\frac{(-1)^{r} x^{\\nu+\\mu+2r}}{2^{\\nu+\\mu+2r} r! \\Gamma(r+\\nu+1)\\Gamma(r+\\mu+1)}\\sum _{n=0}^r\\binom{r}{n}(\\nu +r)^{\\underline{n}}(\\mu +r)^{\\underline{r-n}} \\\\\n&= \\sum_{r=0}^\\infty \\frac{(-1)^{r} x^{\\nu+\\mu+2r}}{2^{\\nu+\\mu+2r} r! \\Gamma(r+\\nu+1)\\Gamma(r+\\mu+1)} (\\nu +\\mu+2r)^{\\underline{r}} \\\\\n&= \\sum_{r=0}^\\infty \\frac{(-1)^{r} \\Gamma(\\nu+\\mu+2r+1)}{ r! \\Gamma(r+\\nu+1)\\Gamma(r+\\mu+1)\\Gamma(\\mu+\\nu+r+1)} \\left( \\frac{x}{2}\\right)^{\\nu+\\mu+2r} \\quad \\; \\quad (2)\n\\end{aligned}\n$$\n<\/p>\n
\nTo simplify the inner summation, we used the binomial theorem for falling factorials.\n<\/p>\n\n\n\n
\nTherefore, applying RMT to the function $x^{-\\frac{\\nu+\\mu}{2}}J_{\\nu}(2\\sqrt{x})J_{\\mu}(2\\sqrt{x})$, gives us:\n$$\\int_0^\\infty x^{s-\\frac{\\nu+\\mu}{2}-1} J_{\\nu}(2\\sqrt{x})J_{\\mu}(2\\sqrt{x}) \\; dx = \\frac{\\Gamma(s)\\Gamma(\\nu+\\mu+1-2s)}{\\Gamma(\\nu+1-s)\\Gamma(\\mu+1-s)\\Gamma(\\nu+\\mu+1-s)}$$\nFinally, make the substitutions $x=t^2\/4$ and $s=\\frac{\\sigma+\\nu+\\mu}{2}$ in the above integral to get:\n$$\\boxed{ \\int_0^\\infty x^{\\sigma-1}J_\\nu(x)J_\\mu(x)\\; dx = \\frac{2^{\\sigma-1} \\Gamma\\left(\\frac{\\nu+\\mu+\\sigma}{2} \\right)\\Gamma\\left(1-\\sigma\\right)}{\\Gamma\\left(\\frac{\\nu-\\mu-\\sigma}{2} +1\\right)\\Gamma\\left(\\frac{\\mu-\\nu-\\sigma}{2} +1\\right)\\Gamma\\left(\\frac{\\nu+\\mu-\\sigma}{2} + 1\\right)} } \\quad \\;\\quad (3)$$\nThe above equation holds provided that the condition $-(\\nu+\\mu)< \\sigma < 1$ is satisfied.\n<\/p>\n","protected":false},"excerpt":{"rendered":"
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 … Continue reading