Integral representation of $\sum_{k=0}^{n} \frac{x^k}{(k!)^2}$? I would like to know if there is any integral representation of the following sum : $\displaystyle{S_n(x) := \sum_{k=0}^{n} \frac{x^k}{(k!)^2}}$.
I'm willing to have an idea of how fast this sums goes to infinity when $x$ is a sequence $(x_n )$ such that $x_n\to \infty$, that is I'm looking for $\lim_{n\to\infty} S_n(x_n)$.
The only thing I know is that $\lim_{n\to\infty} S_n(x) = I_0(2\sqrt{x})$ for $x$ fixed, and $I_0(x)\sim_{x\to \infty} \frac{e^x}{\sqrt{2\pi x}}$ where $I_0$ is the modified Bessel function of the first kind and of order $0$.

Analogous to the integral representation of the modified Bessel function
$$ I_0(2 \sqrt{x}) = \dfrac{2}{\pi} \int_{0}^{\pi/2} \cosh(2 \sqrt{x} \sin(t))\; dt$$
you have
$$S_n(x) =\dfrac{2}{\pi} \int_{0}^{\pi/2} \sum_{k=0}^n \dfrac{4^k  x^k \sin(t)^{2k}}{(2k)!} \; dt$$
I don't know if that will help you.
