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Is it possible to get the equation below into closed form? I have tried using integration tables but I haven't found anything that matches. Are there any other methods to achieve a closed form expression of a bessel function with two exponentials? If so could someone please advise on how I would go about this? Kind regards.

$$\int^\infty_0 \frac{1}{2}\frac{x}{a}^\frac{b-2}{4}\exp\left(-\frac{a+x}{2}\right)\left(1-\exp\left(-\frac{x}{2}\right)\right)^{n} I_{\frac{b}{2}-1}\left(\sqrt{xa}\right) \mathrm{d}x$$

To clarify a, b and n are all integer values $\ \ge 1 $.

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    $\begingroup$ Looks a bit like a Marcum Q function: en.wikipedia.org/wiki/Marcum_Q-function ; A. H. Nuttall, “Some integrals involving the Q-function,” Technical report, DTIC Document (1972). $\endgroup$ – user25199 Feb 6 '15 at 15:38
  • $\begingroup$ Apologies, I have made a mistake in the original equation. To clarify a, b and n are all integer values $\ \ge 1 $. Thanks $\endgroup$ – James Crawford Feb 6 '15 at 15:48
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    $\begingroup$ Maple fails to find closed form in case $a=b=1,n=2$. It does $n=1$. $\endgroup$ – Gerald Edgar Feb 6 '15 at 16:43
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    $\begingroup$ I deleted my earlier comment, because it contained the wrong references. For n≥1 and integer one could expand the (n−1)th power using binomial formula and then possibly exploit e.g. the formula 6.643(2) in Gradshteyn & Rizhik (7th edition). Btw, after transformation of the integration variable, $y:=\sqrt{a x}$, Mathematica 10 finds solutions for fixed integer $n$. It seems to be essentially a sum of $n$ exponentials. $\endgroup$ – Johannes Trost Feb 7 '15 at 8:18
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Maybe you can accept the following result as a closed form. Transforming the variable of integration in your integral $$ I_{n}(a,b):= \frac{1}{2 a}\ e^{-a/2}\int_{0}^{\infty} dx\ x^{(b-2)/4} e^{-x/2}(1-e^{-x/2})^{n} I_{\frac{b}{2}-1}(\sqrt{a x}) $$ to $y:=\sqrt{a x}$, we get $$ I_{n}(a,b)=e^{-a/2}\ a^{-\frac{b}{4}-\frac{3}{2}} \int_{0}^{\infty}dy\ y^{b/2}\ e^{-y^{2}/(2 a)} I_{\frac{b}{2}-1}(y) \sum_{m=0}^{n} {n \choose m} (-1)^m e^{-m\frac{y^{2}}{2a}} $$ where the $n$th power is expanded using the binomial formula. After interchanging summation and integral use DLMF formula http://dlmf.nist.gov/10.43.E23 to get $$ I_{n}(a,b)=e^{-a/2}\ a^{\frac{b}{4}-\frac{3}{2}}\sum_{m=0}^{n}{n \choose m} (-1)^{m} (m+1)^{-b/2} \exp\left(\frac{a}{2(m+1))}\right). $$

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  • $\begingroup$ This is very helpful, thank you very much! $\endgroup$ – James Crawford Feb 10 '15 at 14:51
  • $\begingroup$ I am having trouble understanding your equation after you substitute for y. I have been getting the following: $$I_{n}(a,b)=e^{-a/2}\ \frac{1}{2} a^{\frac{b}{4}-\frac{3}{2}} \int_{0}^{\infty}dy\ y^{b/2-1}\ e^{-y^{2}/(2 a)} I_{\frac{b}{2}-1}(y) \sum_{m=0}^{n} {n \choose m} (-1)^m e^{-m\frac{y^{2}}{2a}}$$ Is there something I am missing? Kind regards. $\endgroup$ – James Crawford Feb 18 '15 at 15:31
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    $\begingroup$ @JamesCrawford It seems that you have forgotten to substitute $d\ x$ correctly. For $y:=\sqrt{a x}$ you have to substitute $dx=\frac{2}{a} \ y d y$. $\endgroup$ – Johannes Trost Feb 19 '15 at 8:19

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