Okay, I am going to be very specific. I have the following integral $$\int_{-1}^1 \mathop{dx}\frac{x^{n-2m}(a^2+x^2)^{(k-2)n/2+(3-k)m/2}}{(c_1 +c_2(a^2+x^2)^{k/2}+c_3(a^2+x^2)^{k/2-1}x^2)^{n-m+1/2}}\left(c_4+c_5\frac{x^2}{a^2+x^2}\right)^{n-2m}\left(c_6\frac{x^2}{a^2+x^2}+c_7\right)^m.$$ where all $c_i$ are real constants, $n$ and $m$ are non-negative integers such that $n\geq 2m$. $k$ is a real number which could be both positive or negative.

I am interested in the scaling of the above integral when $a\ll 1$. I believe the scaling will be different when $k>0$ in comparison to when $k<0$.

My approach is the following

As $a\ll1$, $a^2$ is negligible in comparison to $x^2$, except in the region $x\sim 0$. But, in that region, our integrand contains $x^{n-2m}$, and therefore, the contribution of $a$ is negligible even in the region $x\sim 0$. Therefore, my integral should be of the following from $$ \mathcal{I}\sim \int_{-1}^1 \mathop{dx}\frac{x^{n-2m}x^{(k-2)n+(3-k)m}}{(c_1 +(c_2+c_3)x^{k})^{n-m+1/2}}\left(c_4+c_5\right)^{n-2m}\left(c_6+c_7\right)^m.$$ It is quite clear from the above equation that the integral is independent of $a$. Therefore the scaling of the integral is $$\mathcal{I}\sim a^0.$$ The unfortunate part of my answer is that the information about $k$ and $a$ is completely lost in my crude approximation. I do not want this to happen. Do you guys have any idea how to improve this approximation? I am solely interested in the dependence of integral on $a$.

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    $\begingroup$ I think no one will really care about estimating an integral depending on eleven parameters. Do you really need that? Why? A bit of context would be helpful. $\endgroup$ – Jack D'Aurizio Oct 31 '15 at 22:00
  • $\begingroup$ It came up in my research. I won't need to evaluate the entire integral exactly (of course), and just getting the scaling of this integral on $a$ suffices for me. $\endgroup$ – titanium Oct 31 '15 at 22:02

I write here because this stupid site allows me to give an answer but not a simple comment because I don't have 50 reputations yet. NONSENSE.

Anyway: think about the power:

$$(x^2 + a^2)^k$$

where $k$ is the whole stuff you have as exponent. Now:

$$\left(x^2 \left(1 + \frac{a^2}{x^2}\right)\right)^k = x^{2k}\cdot (1+ X)^k$$

and then use the binomial expansion for small $X$

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  • $\begingroup$ Yeah, but I can't have binomial expansion for $X$, as $X=a/x$, and this is not necessarily small, at least not in the regime when $x\sim 0$. $\endgroup$ – titanium Oct 31 '15 at 21:56
  • $\begingroup$ Uhm.. What is $j$? $\endgroup$ – Von Neumann Oct 31 '15 at 22:04
  • $\begingroup$ Sorry. It was supposed to be $m$. I have corrected that. $\endgroup$ – titanium Oct 31 '15 at 22:06
  • $\begingroup$ Exactly. That's what I was thinking too. I believe my approximation is alright for $k>0$. Even in that case, I would like to get a better approximation, which incorporates $a$ as well. $\endgroup$ – titanium Oct 31 '15 at 22:12
  • $\begingroup$ Oh gosh.. I deleted my last comment -.- Sorry!! Anyway for $k > 0$ I think that $a$ becomes really negligible. But for $k < 0$ there are things that have to be considered.. $\endgroup$ – Von Neumann Oct 31 '15 at 22:13

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