Computing $\int_0^1dx\:I_0(\alpha \sqrt{1-x})e^{\beta x}$

Question

Let $\alpha,\:\beta$ be positive constants and $I_0(x)$ the modified Bessel function. Any idea about how to calculate the following integral ?

$$ \begin{equation} f(\alpha,\beta)=\int_0^1dx\:I_0(\alpha \sqrt{1-x})\:e^{\beta x}. \end{equation} $$

I've tried expanding the exponential but then the coefficients are given by some $_p F_q$ function whose summation is perhaps more difficult than the original problem.

Answer 1

We have \begin{align*} \int_0^1 {I_0 (\alpha \sqrt {1 - x} )e^{\beta x} dx} & = 2e^\beta \int_0^1 {I_0 (\alpha t)e^{ - \beta t^2 } tdt} \\ & = 2e^\beta \sum\limits_{n = 0}^\infty {\frac{1}{{n!^2 }}\left( {\frac{\alpha }{2}} \right)^{2n} \int_0^1 {e^{ - \beta t^2 } t^{2n + 1} dt} } \\ & = \frac{{e^\beta }}{\beta }\sum\limits_{n = 0}^\infty {\frac{{P(n + 1,\beta )}}{{n!}}\left( {\frac{\alpha }{{2\sqrt \beta }}} \right)^{2n} } , \end{align*} where $P$ is the normalised lower incomplete gamma function. It may be written in terms of the Marcum $Q$-function as $$ \frac{1}{\beta }\exp \left( {\frac{{\alpha ^2 }}{{4\beta }} + \beta } \right)\left( {1 - Q_1 \!\left( {\frac{\alpha }{{\sqrt {2\beta } }},\sqrt {2\beta } } \right)} \right). $$

Answer 2

Note change variables $y=\alpha\sqrt{1-x}$ to get $$ f(\alpha,\beta) = \frac{2e^\beta}{\alpha^2}\int_0^\alpha I_0(y) \exp\left(\frac{-\beta}{\alpha^2}y^2\right)y\;dy $$ So evaluating this for all $\alpha,\beta$ means finding the indefinite integral $$ \int I_0(y) e^{-cy^2}y\;dy $$

This is not listed in

Gradshteyn, I. S.; Ryzhik, I. M.; Zwillinger, Daniel (ed.); Moll, Victor (ed.), Table of integrals, series, and products. Translated from the Russian. Translation edited and with a preface by Victor Moll and Daniel Zwillinger, Amsterdam: Elsevier/Academic Press (ISBN 978-0-12-384933-5/hbk; 978-0-12-384934-2/ebook). xlv, 1133 p. (2015). ZBL1300.65001.

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On the other hand, there is a definite integral: $$ \int_0^\infty x e^{-cx^2} I_0(x)\;dx = \frac{e^{1/(4c)}}{2c},\quad c>0 $$ which becomes $$ \int_{-\infty}^1 e^{\beta x}I_0(\alpha\sqrt{1-x}) = \frac{\exp\left(\beta - \frac{\alpha^2}{4\beta}\right)}{\beta} $$

Answer 3

$$I_0(\alpha \sqrt{1-x})=\sum_{n=0}^\infty (-1)^n \frac{ \alpha ^{2 n} }{4^n(n!)^2}(x-1)^n$$

$$\int (x-1)^n e^{\beta x}\,dx=e^\beta\int y^n e^{\beta y} \,dy=\frac {e^\beta }{\beta^{n+1} }\int t^n \,e^t\,dt$$ Now, the incomplete gamma function.

Answer 4

It seems a closed form exists using the following definition from Wikipedia of the Marcum Q function used in many distributions. It is a bit obscure, but works. You will find the rest of the needed information to understand this result in the link:

$$\mathrm{Q_m(a,b)=1-e^{-\frac{a^2}2}\sum_{n=0}^\infty\left(\frac{a^2}{2}\right)^n\frac{P\left(m+n,\frac{b^2}2\right)}{n!}}$$

This needs to be equal to @Gary’s solution

$$\mathrm{A\,Q_x(y,z)+B=\frac{e^β}β\sum_{n=0}^\infty \frac{P(n+1,β)}{n!}\left(\frac{α}{2\sqrtβ}\right)^{2n}}$$