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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

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


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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. – Eric Naslund Apr 22 '11 at 19:30
@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... – J. M. Apr 24 '11 at 13:24
up vote 15 down vote accepted

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



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. – Did Apr 3 '11 at 21:31
@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. – Glen Wheeler Apr 22 '11 at 12:50
@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. – Eric Naslund Apr 22 '11 at 18:14
@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? – Nick Apr 22 '11 at 18:53
@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 $e^{b^2/2}$. – Eric Naslund Apr 22 '11 at 19:21

See also the Boros-Moll article in , equation (5.8).

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