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 $0$.

Since by De Moivre's formula
$$\binom{2k}{k}= \frac{4^k}{\pi}\int_{-\pi/2}^{\pi/2}\cos(\theta)^{2k}\,d\theta \tag{1}$$
we have
$$f(x)=\sum_{k\geq 0}\frac{x^k}{k!^2}=\int_{-\pi/2}^{\pi/2}\sum_{k\geq 0}\frac{(4x)^k\cos(\theta)^{2k}}{\pi(2k)!}\,d\theta \tag{2}$$
hence
$$ \boxed{\,f(x)=\color{red}{\frac{1}{\pi}\int_{-\pi/2}^{\pi/2}\cosh\left(2\sqrt{x}\cos\theta\right)\,d\theta}.} \tag{3}$$
