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I have the following integral

$$ 2\int_r^\infty \frac{x g(x)}{\sqrt{x^2-r^2}}\text{d}x $$

where $g$ is a probability distribution (normalized and symmetrical around its only maximum in 0). I'm trying to study its behaviour for $r\to 0$ (at least second order), but I fail to Taylor expand it. With the help of the computer, the result seems to be $1-r^2/2$ regardless of $g$. I've tried changing variable to $x/r$ or deriving under integral sign, but without convincing results. Thanks.

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Did you check Abel transform? en.wikipedia.org/wiki/Abel_transform –  Mathlover Jun 19 '13 at 6:13
    
Curiosly, it seems to be exacly the Abel transform of $g$. However, this does not help me at the moment... –  Bzazz Jun 21 '13 at 17:57
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1 Answer 1

Consider

$$I(r)=\int_r^\infty \frac{x g(x)}{\sqrt{x^2-r^2}}\text{d}x=\int_r^\infty \frac{ g(x)}{\sqrt{1-\frac{r^2}{x^2}}}\text{d}x$$ The next step is a Taylor expansion

$$\frac{ 1}{\sqrt{1-\frac{r^2}{x^2}}}=1+\sum_{k=1}^{\infty}\frac{(2k-1)!!}{(2k)!!}\frac{r^{2k}}{x^{2k}}$$ where $(2k)!!=2^kk!$ and $(2k-1)!!=\frac{(2k)!}{2^kk!}$ are so called double factorials. Thus

$$I(r)=\int_{r}^{\infty}g(x)dx+\sum_{k=1}^{\infty}\frac{(2k-1)!!}{(2k)!!}\int_{r}^{\infty}\frac{r^{2k}}{x^{2k}}g(x)dx$$ Now, we make a change of variables in the second integral. Namely $x=ry$. This yields

$$I(r)=\int_{r}^{\infty}g(x)dx+r\sum_{k=1}^{\infty}\frac{(2k-1)!!}{(2k)!!}\int_{1}^{\infty}\frac{g(ry)}{y^{2k}}dy$$ Next, let $M=\text{max}\;g(x)$;$\;$ $r<x<\infty$ Then, we can evaluate the integral inside the sum

$$\int_{1}^{\infty}\frac{dy}{y^{2k}}=\frac{1}{2k-1}$$ and get the following estimation

$$I(r)<\int_{r}^{\infty}g(x)dx+rM\sum_{k=1}^{\infty}\frac{1}{2k-1}\frac{(2k-1)!!}{(2k)!!}$$ Finally,for $r\to 0$, an answer could be:

$$1\leqslant I(r)< 1+rM\sum_{k=1}^{\infty}\frac{1}{2k-1}\frac{(2k-1)!!}{(2k)!!}$$ The last sum converges!

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