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Let $f(x)$ function define by

$$ f(x)=x^m e^{-bx}K_{n+1}(ax)^{'} $$ where $K_v(⋅)$ is the $v$-th order modified Bessel function of the second kind and $L^{'}(x)$ is the derivation of function $L(x)$.

I would like to compute the following integral integral

$$ I=\int_{0}^{\infty}f(x)dx=\int_{0}^{\infty}x^m e^{-bx} K_{n+1}(ax)^{'}dx. $$

So my question is what is the derivation of $K_{n+1}(ax)$.

I found the following formula to evaluate the integral $$ \int_{0}^{\infty}x^{\mu-1}e^{-\alpha x}K_{v}(bx)dx= $$ $$\frac{\sqrt\pi (2b)^v}{(\alpha+b)^{\mu+v}} \frac{\Gamma(\mu+v)\Gamma(\mu-v)}{\Gamma(\mu+\frac{1}{2})} F\left( \mu+v,v+\frac{1}{2},\mu+\frac{1}{2};\frac{\alpha-b}{\alpha+b} \right) $$ where $F(a,b,c;z)$ is is the generalized hypergeometric function.

So what is the final result of Integral $I$.

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  • $\begingroup$ There is some recursion for the derivatives of the $K_n$ though I have forgotten it. Check mathworld or Abramowitz and Stegun. $\endgroup$ – Ian Jan 23 at 17:54
  • $\begingroup$ I would like just the exact expression of $K_{n+1}(ax)^{'}$, just the derivation. $\endgroup$ – Monir Jan 23 at 17:57

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