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Why is this true?

$$ \int_0^\infty e^{-\frac{1}{2}(b^2+x^2)} I_0(bx) x \,dx = 1 $$

Note that $I_0(x)$ is a modified bessel function of the first kind. The difficulty for me lies in a) translating the bessel function into something basic (the argument is not an integer, hence making it complicated to me), and 2) this doesn't seem to be on integral-table.com.

If it helps, the equation above is identical to: $$ Q(b,0) = 1,$$ where $Q(a,b)$ is a $Q$-Marcum function defined as: $$ Q(a,b) = \int_b^\infty e^{-\frac{1}{2}(a^2+x^2)} I_0(ax) \,dx$$

Thanks.

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    @J.M. I don't understand the edit you did to this question. The $b$'s do not need to be written as $a$'s, (perhaps it makes it clearer, but this is debatable. It is clear already) and there should be a $x$ in the integral. Everything the OP wrote down is correct.2011-04-22
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    @Eric: Yes, I slipped up on the $x$; on the other hand, the standard name for the function in question is "Marcum Q", not "Q-Marcum". Anyway...2011-04-24

2 Answers 2

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Lets start from the definition of the modified Bessel Function. Recall

$$I_{0}(z)=\sum_{n=0}^{\infty}\frac{1}{4^{n}}\frac{z^{2n}}{n!n!}.$$ So our integral is

$$\int_{0}^{\infty}e^{-\frac{1}{2}\left(x^{2}+b^{2}\right)}x\sum_{n=0}^{\infty}\frac{1}{4^{n}}\frac{b^{2n}x^{2n}}{n!n!}dx.$$ Switching the order of sumation and integration we get

$$e^{-\frac{1}{2}b^{2}}\sum_{n=0}^{\infty}\frac{b^{2n}}{4^{n}n!n!}\int_{0}^{\infty}e^{-\frac{1}{2}x^{2}}x^{2n+1}dx.$$ Now, let $u=\frac{1}{2}x^{2}$ and $du=xdx$, to see that $$\int_{0}^{\infty}e^{-\frac{1}{2}x^{2}}x^{2n+1}dx=2^{n}\int_{0}^{\infty}e^{-u}u^{n}du=2^{n}\Gamma(n+1)=2^{n}n!.$$ Hence

$$e^{-\frac{1}{2}b^{2}}\sum_{n=0}^{\infty}\frac{b^{2n}}{4^{n}n!n!}\int_{0}^{\infty}e^{-\frac{1}{2}x^{2}}x^{2n+1}dx=e^{-\frac{1}{2}b^{2}}\sum_{n=0}^{\infty}\frac{b^{2n}}{4^{n}n!n!}\left(2^nn!\right)$$

$$=e^{-\frac{1}{2}b^{2}}\sum_{n=0}^{\infty}\frac{\left(\frac{b^{2}}{2}\right)^{n}}{n!}=e^{-\frac{1}{2}b^{2}}e^{\frac{1}{2}b^{2}}=1$$

as desired.

Notice that the exact same argument shows $$\int_0^\infty e^{\left(\frac{1}{2}b^2-\frac{1}{2}x^2\right)}J_0(bx)xdx=1$$ where $J_0(x)$ is the Bessel Function.

Hope that helps,

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    @Eric: Bravo. Seems like the definitive answer to me.2011-04-03
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    @Eric I don't see the extra $x$ in the original question, am I blind, or is something going on here? Did the original question have a typo? If that is the case, we should edit it.2011-04-22
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    @Glen Wheeler, @J.M: It should be there, and was originally there when I posted this answer. Looking at the edit history, for some reason J.M. removed it? I am going to put it back, as without that extra $x$ you would recover $\Gamma(n+\frac{1}{2})$ rather than $\Gamma(n+1)$, making it so the factorials do not cancel as nicely.2011-04-22
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    @Eric great derivation, this is very useful. One question to help clarify my own understanding: how did you get rid of the $1/4^n$ go in the last step?2011-04-22
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    @Nick: Thanks. I edited that line to make it a bit clearer, but what happened was $2^n n!$ came from the Gamma-Function-like integral, and then I wrote $$\frac{b^{2n}}{2^n}=\left(\frac{b^2}{2}\right)^n$$ so that it was clearer that we had the power series for $e^{b^2/2}$.2011-04-22
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    @Eric I recently learned that only the first "ping" in a comment actually functions, so this means that JM was not notified. I'm not sure if there is any solution to this problem other than posting multiple comments. A pretty poor solution however.2011-04-23
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    @J.M Just like this ;). Eric: Thanks for the clarification on my query.2011-04-23
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    @Glen and Eric: I slipped up in deleting that factor; sorry for any confusion/inconvenience.2011-04-24
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See also the Boros-Moll article in http://www.mat.utfsm.cl/scientia/vol12.html , equation (5.8).