The correct problem is the following one, probably I did some error using Latex.
The equation [math]\frac{2}{\pi r} \int_0^{\frac{\pi }{2}} \exp \left(-\frac{1}{2} \left(\frac{1}{2} \pi a^2 b^2 \left(\cos ^2(\theta )+\frac{1}{\cos ^2(\theta )}-2\right)+\frac{x^2 \cos ^2(\theta )}{\frac{1}{2} \pi a^2 b^2}+2 x \left(1-\cos ^2(\theta )\right)\right)\right) \, d\theta=1[/math]
defines, at least theoretically, [imath]x[/imath] as a function of [imath]a[/imath] and [imath]b[/imath], [imath]x=f(a,b)[/imath]. All values are supposed to be real. Is it possible to find the asymptotic form of [imath]x=f(a,b)[/imath] in the limit of
1) [imath]a\to 0[/imath] and [imath]b\to 0[/imath];
2) [imath]a\to 1[/imath] and [imath]b\to 0[/imath].
I tried the followings for the first case ([imath]a\to 0[/imath] and [imath]b\to 0[/imath]).
Substitute the integrand function with its Taylor series about [imath]a= 0[/imath] and [imath]b=0[/imath],
[math]\exp \left(-\frac{1}{2} \left(\frac{1}{2} \pi a^2 b^2 \left(\cos ^2(\theta )+\frac{1}{\cos ^2(\theta )}-2\right)+\frac{x^2 \cos ^2(\theta )}{\frac{1}{2} \pi a^2 b^2}+2 x \left(1-\cos ^2(\theta )\right)\right)\right) = \exp \left(\frac{-\frac{x^2 \cos ^2(\theta )}{\pi b^2}+O\left(b^2\right)}{a^2}+x \left(\cos ^2(\theta )-1\right)+O\left(a^2\right)\right)[/math]
thus the equation becomes:[math]\frac{2}{\pi a} \int_0^{\frac{\pi }{2}} \exp \left(x \left(\cos ^2(\theta )-1\right)-\frac{x^2 \cos ^2(\theta )}{\pi a^2 b^2}\right) \, d\theta=1[/math]
Solve the integral using the modified Bessel function [imath]I_0[/imath]
[math]\frac{2}{\pi a} \frac{1}{2} \pi e^{-\frac{1}{2} x \left(\frac{x}{\pi a^2 b^2}+1\right)} I_0\left(\frac{1}{2} x \left(\frac{x}{a^2 b^2 \pi }-1\right)\right)=1[/math]
In the limit of [imath]a\to 0[/imath] and [imath]b\to 0[/imath]
[math]\frac{x}{\pi a^2 b^2}+1=\frac{x}{\pi a^2 b^2}-1\to \frac{x}{\pi a^2 b^2}[/math] then last equation becomes
[math]\frac{1}{a} e^{-\frac{x^2}{2 \left(\pi a^2 b^2\right)}} I_0\left(\frac{x^2}{2 \left(a^2 b^2 \pi \right)}\right)=1[/math]
which becomes using the asymptotic form for for [imath]I_0[/imath] for large real [imath]z[/imath]
[math]I_0(z)=\frac{\exp (z)}{\sqrt{2 \pi z}}[/math]
[math]\frac{e^{-\frac{x^2}{2 \left(\pi a^2 b^2\right)}} e^{\frac{x^2}{2 \left(\pi a^2 b^2\right)}}}{a \sqrt{\frac{2 \pi x^2}{2 \left(\pi a^2 b^2\right)}}}=1[/math]
and so on, like in the previous post...
The integrand in your formula can be represented as
[math]f(a,b) = e^{-\left(pa^2b^2 + \frac{q}{a^2b^2}+r \right)}[/math]where [imath]p,q,r[/imath] depend on [imath]\theta[/imath] and [imath]x[/imath]. But I can't recognize the Taylor expansion of [imath]f[/imath] in your subsequent formulae. The integrand in your post #3 looks slightly simpler, i.e., [imath]f(a,b) = e^{-\left(\frac{q}{a^2b^2}+r \right)}[/imath], but I still don't see a correct Taylor expansion either.
BTW, the expressions in your original post and post #3 don't look identical to me -- which one do you need?
Just curious: where do these formulae come from? Your post does not look like a typical homework assighment.
Thanks for your interest. The correct formula is the first one, in the original post, see below. It comes from a heat transfer problem where a gaussian heat source is applied to the surface of a semi-infinite medium. The solution should represent the intercept of a generic isotherm with the x axis. I will try to check the Taylor expansion.
