I have the integral of the following:
$$I(g) = g \int_{-\pi}^{\pi} dk \sqrt{1+g^{-2} + 2g^{-1} \cos(k)},$$
whose second derivative ($\partial_g^2 I(g)$), I know diverges as $g \rightarrow 1$. How do I go about extracting the power law of this divergence, and what is the general bag of tricks to approximate integrals in certain limits like this?
Thanks.
Due to the periodicity of the integrand $$I(g)=g \int_{-\pi}^{\pi} \sqrt{1+g^{-2} + 2g^{-1} \cos(k)}\,dk=\int_0^{2\pi}\sqrt{1+g^2 + 2g \cos(k)}\,dk$$ Then $$J(g)=\frac{d^2}{dg^2}I(g)=\int_0^{2\pi}\frac{dk}{\sqrt{1+g^2 + 2g \cos(k)}}-\int_0^{2\pi}\frac{(g+\cos k)^2}{\big(1+g^2 + 2g \cos(k)\big)^\frac32}dk=I_1+I_2$$ Second integral is regular at $g=1$, so we are focusing on the first one. $$I_1=\frac{d^2}{dg^2}I(g)=\int_0^{2\pi}\frac{dk}{\sqrt{(1-g)^2 + 2g (1+\cos(k))}}\overset{x=\frac k2}{=}4\int_0^{\pi/2}\frac{dx}{\sqrt{(1-g)^2+4g\cos^2x}}$$ $$\overset{t=\tan x}{=}4\int_0^\infty\frac{dt}{\sqrt{1+t^2}\sqrt{(1-g)^2(1+t^2)+4gt^2}}$$ $$\overset{t\to\frac1t}{=}\frac4{1+g}\int_0^\infty\frac{dt}{\sqrt{1+t^2}\sqrt{t^2+\big(\frac{1-g}{1+g}\big)^2}}$$ $$\overset{IBP}{=}\frac4{1+g}\frac{\ln\big(t+\sqrt{\big(\frac{1-g}{1+g}\big)^2+t^2}\,\big)}{\sqrt{1+t^2}}\,\bigg|_{t=0}^\infty+\frac4{1+g}\int_0^\infty\frac{\ln\big(t+\sqrt{\big(\frac{1-g}{1+g}\big)^2+t^2}\big)}{(1+t^2)^\frac32}\,t\,dt$$ Second integral is regular at $g=1$; the first term gives us the desired singularity: $$J(g)=\frac{d^2}{dg^2}I(g)=-2\ln(g-1)+O(1);\,\,g\to1^+$$
Addendum
As $\ln(g-1)$ is a slowly growing function, it might make sense to evaluate next asymptotic term $\sim O(1)$. $$I_2=-\int_0^{2\pi}\frac{(g+\cos k)^2}{\big(1+g^2 + 2g \cos(k)\big)^\frac32}\,dk\sim-2^{-3/2}\int_0^{2\pi}\sqrt{1+\cos k}\,dk=-2\tag{1}$$ The constant term in $I_1$ $$\frac4{1+g}\ln(1+g)+\frac4{1+g}\int_0^\infty\frac{\ln\big(t+\sqrt{\big(\frac{1-g}{1+g}\big)^2+t^2}\big)}{(1+t^2)^\frac32}\,t\,dt$$ $$\sim2\ln2+2\int_0^\infty\frac{\ln(2t)}{(1+t^2)^\frac32}\,t\,dt=6\ln2\tag{2}$$ Using (1) and (2) $$\boxed{\,\,J(g)=\frac{d^2}{dg^2}I(g)=-2\ln(g-1)+6\ln2-2+o(g-1);\,\,g\to1^+\,\,}$$