What is integral of $e^{\cos^2x}dx$?

Question

I have tried to integrate $e^{\cos ^2x}dx$ from $0 \to 1$ , Wolfram alpha showed me that is equal 2.11 as shown here but really i'm interested for it's closed form using sompe special standard function , my attempt is to use this basic $$ \int \frac{e^u}{f'(x)}=\int\exp(f(x)) du. $$but it's seems don't works then any way to find it's integral ?

Answer 1

Such function does not have an elementary primitive by Risch algorithm, but on the other hand the Fourier series of $e^{\cos(2x)/2}$ is not a nightmare: $$ e^{\cos(2x)/2} = I_0\left(\tfrac{1}{2}\right)+2\sum_{n\geq 1} I_{n}\left(\tfrac{1}{2}\right)\cos(2nx)\tag{1} $$ especially since the involved values of Bessel functions can be computed through very simple continued fractions. The RHS of $(1)$ can be termwise integrated over $(0,1)$ et voilà: $$ \int_{0}^{1}e^{\cos^2(x)}\,dx = \sqrt{e}\,I_0\left(\tfrac{1}{2}\right)+\sqrt{e}\,\sum_{n\geq 1}I_{n}\left(\tfrac{1}{2}\right)\frac{\sin(2n)}{n}.\tag{2}$$

I leave to you an interesting exercise - to compute the following limit: $$ \lim_{n\to +\infty}\int_{0}^{1}\exp\left[\cos^2(nx)\right]\,dx.$$