I came up with the following integrations
$$\int_0^\infty \frac{1}{\sqrt{1+a^2\cosh ^2x}}dx,\quad a>0$$
I want to know the precise asymptotic behavior as $a\to 0^+$. I think it is related to the elliptic integral with complex variables. Can anyone give me help? Thx
Maple says the exact value is $$(1+a^2)^{-1/2}\,\mathrm{EllipticK}((1+a^2)^{-1/2})$$ $$= \ln(4/a) + \bigl(1/4+\ln(a^{1/4}/2^{1/2})\bigr)a^2 + O(a^4\ln a).$$
This checks out numerically. EllipticK is the complete elliptic integral of the first kind.
A human proof of the formula from Brendan McKay's answer. In your integral, denote $\tanh x=t$, then $1-t^2=1/\cosh^2 x$ and $dx=dt/(1-t^2)$, thus your integral equals $$\int_0^1\frac{dt}{(1-t^2)\sqrt{1+\frac{a^2}{1-t^2}}}=\int_0^1\frac{dt}{\sqrt{(1-t^2)(1+a^2-t^2)}}\\= (1+a^2)^{-1/2}\int_0^1\frac{dt}{\sqrt{(1-t^2)(1-\frac{t^2}{1+a^2})}}=(1+a^2)^{-1/2}K((1+a^2)^{-1/2})$$
