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Here I have the integral $$\int_0^{\infty}x^{\ell-1}K_m(x)K_n(\frac bx)\,dx$$ which is the integral of the multiplication of two modified bessel function of the second kind. I find that this integral is expressed as as a series of the form $$\sum_{v=0}^{\infty}c_vb^{\rho+v}$$ in the reference, i.e., "An Infinite Integral Involving a Product of two Modified Bessel functions of the Second Kind" by T. M. Macrobert. which is used to express the integral as generalized hypergeometric function. I do not know why the integral can be expressed as this series. Can you give me some advice?

bluestool
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Michael.Andy
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  • Could you give the "reference" you use? Expression for $c_\nu$ and $\rho$ and/or the result as a generalized hypergeometric function may also be useful. – Paul Enta Jun 14 '19 at 08:12
  • Thanks for your advice! The reference I use is "AN INFINITE INTEGRAL INVOLVING A PRODUCT OF TWO MODIFIED BESSEL FUNCTIONS OF THE SECOND KIND" by T. M. Macrobert. – Michael.Andy Jun 14 '19 at 14:05
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    Their notation $\sum_{n, -n} f(n)$ means $f(n) + f(-n)$, so the result is a sum of four hypergeometric functions. This is solvable with the general method described here. The Mellin transform of $K_m$ is known, and the convolution evaluates to $$2^{l - 3} G_{0, 4}^{4, 0} {\left( \frac {b^2} {16} \middle| {\frac {l - m} 2, \frac {l + m} 2, -\frac n 2, \frac n 2 \atop -} \right)}.$$ When all poles are simple, this expands into a sum of ${_0 F_3}$ functions by Slater's theorem. – Maxim Jun 14 '19 at 16:37
  • @Maxim: Thank you! I follow the steps that you provide in the link and I have a problem about the variable $\omega$. When using mellin convolution, $\omega$ is the variable of integration, if $\omega$ is equal to a constant, how to integrate it? – Michael.Andy Jun 22 '19 at 08:48
  • Let's make it explicit that $\mathcal M$ takes a function and returns a function and that $f * g$ is also a function: $$\mathcal M\omega \mapsto (f * g)(\omega) = \mathcal Mf \mathcal Mg, \ (f * g)(\omega) = \mathcal M^{-1}p \mapsto \mathcal M[f](p) \mathcal M[g](p).$$ – Maxim Jun 22 '19 at 13:04
  • @Maxim: Thank you! Here is what I have done.$$f(x)=\frac{1}{(x^{n_1}+a_1)^{m_1}}$$ $$g(\frac{\omega}{x})=\frac{\omega^{-\sigma} x^{\sigma}}{(\omega^{-n_2}x^{n_2}+a_2)^{m_2}}$$ Follow the inverse Mellin Convolution $$(f*g)(\omega)=\mathcal{M}{-1}p\mapsto \mathcal{M}[f](p)\mathcal{M}[g](p)$$ The function of $\omega$ is $$(\int{0}^{\infty}\frac{1}{(x^{n_1}+a_1)^{m_1}} \frac{\omega^{-\sigma}x^{-(k_1+k_2)}}{(\omega^{-n_2}x^{n_2}+a_2)^{m_2}} dx)(\omega) =(\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} \mathcal{M}f \mathcal{M}g\omega^{-p} dp)(\omega) $$Set $\omega=1$. – Michael.Andy Jun 27 '19 at 15:41
  • That is correct, except that the integrand on the lhs in the last formula should contain $x^{\sigma - 1} = x^{k_1 + k_2}$. – Maxim Jun 27 '19 at 21:22
  • @Maxim: Thanks for pointing that error out ! – Michael.Andy Jun 28 '19 at 01:50
  • @Maxim: Following the general method, I obtain $$ 2^{l-3} G_{0, 4}^{4, 0} {\left( \frac{b^2}{16} \middle|{ - \atop \frac{l-m}{2},\frac{l+m}{2}, - \frac{n}{2},\frac{n}{2} } \right)}.$$ which is different form yours. – Michael.Andy Jun 30 '19 at 03:03
  • You're absolutely right, I put the dash in the wrong place, so what I wrote is just an undefined thing. The lower indices $(0, 4)$ mean there are four $b_k$ coefficients. The $b_k$ coefficients go into the lower row. Your result is correct. – Maxim Jun 30 '19 at 04:07
  • @Maxim: Hello Maxim! I encounter a problem when I apply the above meijerG form to calculate the intergral, which is when $b$ is big enough, the calculation results of the above meijerG form diverge. For comparison, I apply the numerical integration method to calculate and the results show that the intergral converge to zero. In addition, the calculation time of the meijerG form is 15 times more than the numerical integration method. I do not understand why the results of meijerG form diverge. – Michael.Andy Aug 16 '19 at 12:29
  • The G-function decays subexponentially: $$2^{l - 3} G_{0, 4}^{4, 0} {\left(\frac {b^2} {16} \middle| {- \atop \frac {l - m} 2, \frac {l + m} 2, -\frac n 2, \frac n 2} \right)} \sim \frac {\pi^{3/2}} 2 b^{(2 l - 3)/4} e^{-2 \sqrt b}, \quad b \to \infty.$$ For numerical evaluation, does it help if you use the ${_0 F_3}$ form? What are the specific parameter values you're interested in? Perhaps this asymptotic expansion already gives enough precision. – Maxim Aug 16 '19 at 14:48
  • @Maxim: I have tried the $ {0}F{3}$ form, it's results also diverge. The formula I evaluate is $$s I^m \int_{0}^{\infty} t^{a-1} K_{b}(t) K_{c}(\frac{\sqrt{hI}}{t}),dt$$ For numerical integration, I apply the quadgk function of matlab to integrate $t$ at every given I. As for MeijerG form, the evaluated formula is $$s I^m 2^{a-3} G_{0, 4}^{4, 0} {\left( \frac{hI}{16}\middle| {- \atop \frac{a-b}{2}, \frac{a+b}{2}, -\frac{c}{2}, \frac{c}{2} } \right)}$$ The corresponding ${0}F{3}$ form is $$\sum_{c,-c} 2^{a-2c-3} \Gamma(-c) \Gamma((a+b-c)/2) \Gamma((a-b-c)/2) \sqrt{hI}^{,c} ...$$ – Michael.Andy Aug 18 '19 at 06:58
  • @Maxim: $$... {0}F{3}(; 1+c, 1-\frac{a+b-c}{2}, 1-\frac{a-b-c}{2}; \frac{hI}{16}) +\sum_{b,-b} 2^{-a-2b-3} \Gamma(-b) \Gamma(\frac{-a-b+c}{2})\Gamma(\frac{-a-b-c}{2}) \sqrt{hI}^{a+b},{0}F{3}(; 1+b, 1-\frac{a+b-c}{2}, 1+\frac{a+b+c}{2};\frac{hI}{16}) $$ The parameter values are $s=2.5469^{77}, m=4.1101, a=-0.8212, b=7.3434, c = 8.1800, h=7.0965^{17}.$ Moreover, I assume $E = 1.8207^{-15}$. In this case, when parameter I ranges from $10^{-5}E$ to $5E$, the results of the three methods are the same. However, when parameter I ranges from $10^{-5}E$ to $500E$, the results of the MeijerG form – Michael.Andy Aug 18 '19 at 07:12
  • @Maxim: and the ${0}F{3}$ form diverge at higher I, while the numerical integration remains stable. Modifications: The second ${0}F{3}$ 's parameters are ${0}F{3}(;1+b, 1+\frac{a+b-c}{2}, 1+\frac{a+b+c}{2}; \frac{hI}{16})$. – Michael.Andy Aug 18 '19 at 07:27
  • This is a numerical issue. You have something like $$G_{0, 4}^{4, 0} {\left(1000 \middle| {- \atop -\frac {409} {100}, -\frac {408} {100}, \frac {326} {100}, \frac {409} {100}} \right)}.$$ The issue is more clear if you consider the sum of ${_0 F_3}$ functions: you have four terms which have large magnitudes and sum up to a small number. All significant digits cancel out. You have to use variable-precision arithmetic or a different method to evaluate the G-function (perhaps as the integral over $(\gamma - i \infty, \gamma + i \infty)$, where the gamma functions decay exponentially). – Maxim Aug 18 '19 at 13:12
  • @Maxim: Thank you Maxim! I have used variable-precision arithmetic and the results become stable. – Michael.Andy Aug 19 '19 at 14:27

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