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Is there a closed form expression for an integral of the modified Bessel function of the first kind zero order including the following?

$\int_0^\infty x^a e^{-bx^2} I_0(cx)\ x \,dx$

where a is positive integer, $b$ and $c$ are positive real.

Please also clarify if integrals including such expressions can be of closed form, e.g., for specific $a$,$b$, or $c$, or if they can only be numerically evaluated for any value included. If this can be only numerically evaluated, is there any closed formed formula for approximating this expression?

Thanks in advance for both your time and patience

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Prudnikov-Brychkov-Marychev book (Vol. 2) contains the following formula:

$$ \int_{0}^{\infty}x^{\alpha-1}e^{-px^2}I_{\nu}(cx)dx=2^{-\nu-1}c^{\nu}p^{-\frac{\alpha+\nu}{2}}\frac{\Gamma(\frac{\alpha+\nu}{2})}{\Gamma(\nu+1)}\, _1F_1\left(\frac{\alpha+\nu}{2},\nu+1,\frac{c^2}{4p}\right).$$ Here $\mathrm{Re}\,p>0$, $\mathrm{Re}(\alpha+\nu)>0$, $-\pi<\arg c<\pi$, and $_1F_1$ denotes confluent hypergeometric function. For $\nu=0$ this gives $$ \int_{0}^{\infty}x^{\alpha-1}e^{-px^2}I_{0}(cx)dx=\frac12p^{-\frac{\alpha}{2}}\Gamma(\alpha/2)\; _1F_1\left(\frac{\alpha}{2},1,\frac{c^2}{4p}\right).$$ Further, for integer $\alpha$ (evens are better than odds) $_1F_1$ simplifies. For example, $$ _1F_1(1/2,1,z)=e^{z/2}I_0(z/2),\qquad _1F_1(1,1,z)=e^z.$$ Similar answers for greater integer values of $\alpha$ can be obtained using recursion formulas for $_1F_1$ w.r.t. parameters.

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  • $\begingroup$ Thank you O.L. for the answer. Cheers $\endgroup$ – dioxen Apr 25 '13 at 11:36

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