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I need your kind help in the computation of the small-$q$ expansion of the following definite integral

$\displaystyle\int_0^\infty\frac{dp}{p \left(p^2+q^2\right) \Big[(a/p)^{\eta }+1\Big]^{1/2} \Big[a^\eta/(p^2+q^2)^{\eta/2}+1\Big]^{1/2}}$,

where $2>\eta>0$, $a>0$, and $q\ll a$.

I can show that for $\eta=2$ the integral behaves as $\displaystyle \frac{2}{a^2}\log\left(\frac{a}{q}\right)$. However, I am not able to find its small-$q$ asymptote for other values of $\eta$.

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The main contribution to the integral comes from the vicinity of $p=0$. To get the leading term, it's sufficient to replace the integrand with a simpler one that has the same behavior for small $p$ and small $q$: $$\int_0^\infty \frac 1 {p \left( p^2+q^2 \right)} \left( \left( a/p \right) ^ \eta + 1 \right) ^ {-1/2} \left( a ^ \eta / \left( p^2+q^2 \right) ^ {\eta/2} + 1 \right) ^ {-1/2} dp \sim\\ \int_0^\infty \frac 1 {p \left( p^2+q^2 \right)} \left( \left( a/p \right) ^ \eta \right) ^ {-1/2} \left( a ^ \eta / \left( p^2+q^2 \right) ^ {\eta/2} \right) ^ {-1/2} dp =\\ \frac {\Gamma \left( \frac \eta 4 \right) \Gamma \left( \frac 1 2 - \frac \eta 4 \right)} {2 ^ {\eta/2 + 1} \sqrt \pi a ^ \eta q ^ {2 - \eta}}.$$

The leading term for $\eta=2$ can be obtained in the same manner; it'll be $-\ln(q)/a^2$ (you have an extra factor of 2 in your result).

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  • $\begingroup$ By using such a procedure, can we be sure about the prefactor? Or you just provide the scaling? $\endgroup$ – yarchik Mar 3 '18 at 19:36
  • $\begingroup$ Those are the leading terms for $0<\eta<2$ and for $\eta=2$, asymptotically equivalent to your integral. It's always necessary to estimate the remainder; for $0<\eta<1$, the remainder is $O\left( 1 / q ^ {2 - 2\eta} \right)$, and similarly for other cases. $\endgroup$ – Maxim Mar 3 '18 at 21:03

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