I'm asked to show that as $x \to \infty$ the modified Bessel function $$ I_n(x) :=\frac{1}{\pi}\int_o^\pi e^{x \cos \theta} \cos n \theta \ d\theta \sim \frac{e^x}{\sqrt{2\pi x}}. $$
I am not quite sure how to go about this. I have tried integration by parts without much success as terms all seem to go to zero. Any suggestions as to a method of approach would be appreciated.
Here's an attempt to derive the asymptotics from the integral itself.
First, we rewrite the exponent, using double angle formulas:
$$\cos \theta=1-2 \sin^2 \frac{\theta}{2}$$
And making a substitution: $t= \frac{\theta}{2}$. This gives us the following integral:
$$I_n=\frac{2 e^x}{\pi} \int_0^{\pi/2} e^{-2 x \sin^2t} \cos (2nt) dt$$
Now naturally, we make a substituion:
$$u=\sin t, \qquad t= \arcsin u$$
I won't specify the exact form of $ \cos (2nt)$ in terms of $u$, I will just denote:
$$f(u)=\cos (2n \arcsin u), \qquad |f(u)| \leq 1$$
Then we can write the integral as:
$$I_n=\frac{2 e^x}{\pi} \int_0^1 e^{-2 x u^2} f(u) \frac{du}{\sqrt{1-u^2}}$$
Now the final trick will be the substitution:
$$v= \sqrt{x} ~u$$
We get:
Here's when it gets tricky. I have no idea how to justify the asymptotics, but suppose if for large $x$ we can set the upper limit of the integral to $\infty$, but simultaneously assume $x-v^2 \asymp x$ under the radical sign and also discount $f$ somehow, then we would get:
$$I_n\asymp \frac{2 e^x}{\pi \sqrt{x}} \int_0^\infty e^{-2 v^2} dv=\frac{e^x}{ \sqrt{2\pi x}}$$
Which is correct, but the argument I used is not justified in the slightest. So there's still some work to be done.