Jtdylktuy

Find two approximations for the integral ($x>0$)
begin{equation}
I(x) = frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}}
e^{x cos^2(theta)}dtheta
end{equation}
one for small $x$ (keeping up to linear order in $x$) and one for large values of $x$ (keeping only the leading order term).

For the small x expansion, I tried using a Maclaurin series for the function in the exponential, and it seemed to work...I am unsure what to do for large x or how to check my answer. I know this is likely related to the Gamma/Gaussian function, but am fairly novice at problems of this form. Any help deeply appreciated!

For large $x$, Laplace's method seems like the best option. Or its counterpart, the Watson's lemma.

Let's transform the integral first:

$$ I(x) = frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}}
e^{x cos^2(theta)}dtheta =frac{1}{pi}int_0^{frac{pi}{2}}
e^{x cos^2(theta)}dtheta=frac{e^{x}}{pi} int_0^{frac{pi}{2}}
e^{-x sin^2(theta)}dtheta= \ =frac{e^{x}}{pi} int_0^1
frac{e^{-x s^2}}{sqrt{1-s^2}} d s=frac{e^{x}}{2pi} int_0^1
frac{e^{-x t}}{sqrt{t}sqrt{1-t}} d t$$

Now for the latter integral, the main contribution for $x to infty$ will be given by $t to0$. And the form of it allows us to use the Watson's lemma directly (see the Wikipedia link).

Using the notation from the article, we have:

$$phi(t)=frac{1}{sqrt{t}sqrt{1-t}}=t^{-1/2} g(t)$$

Where $g(t)$ can be expanded into Taylor series around $0$. Additionally, $int_0^1 |phi(t)| dt=pi<infty$. So the lemma conditions are satisfied.

Then we can represent:

$$int_0^1
frac{e^{-x t}}{sqrt{t}sqrt{1-t}} d t asymp sum_{n=0}^infty frac{Gamma(n+1/2) g^{(n)}(0)}{n! ~x^{n+1/2}} $$

How about this:
$$I(x)=frac{1}{2pi}int_{-pi/2}^{pi/2}e^{xcos^2theta}dtheta=frac{1}{2pi}int_{-pi/2}^{pi/2}e^{x(1-sin^2theta)}dtheta$$
$$theta=pi/2-phi$$
$$dtheta=-dphi$$
$$I(x)=frac{1}{2pi}int_0^pi e^{xsin^2phi}dphi$$

EDIT:

taking the other approach, I have:
$$I'(x)=frac{1}{2pi}int_{-pi/2}^{pi/2}cos^2theta e^{xcos^2theta}dtheta$$
then using $u=costheta$ you can obtain:
$$I'(x)=frac1piint_0^1frac{u^2}{sqrt{1-u^2}}e^{xu^2}du$$
now using $v=sqrt{1-u^2}$ we can get:
$$I'(x)=frac1piint_0^1sqrt{1-v^2}e^{x(1-v^2)}dv$$
which can be re-written as:
$$I'(x)=frac1piint_0^1sqrt{1-v^2}e^{-(sqrt{x}v)^2}dv$$
if you continue and differentiate once again you get:
$$I''(x)=frac{1}{2pi x^2}int_0^xbeta e^beta dbeta$$
so:
$$I(x)=frac1{2pi}iintfrac{(x-1)e^x+1}{x^2}dxdx$$