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 $0$.
 A: 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
$$ \boxed{\,f(x)=\color{red}{\frac{1}{\pi}\int_{-\pi/2}^{\pi/2}\cosh\left(2\sqrt{x}\cos\theta\right)\,d\theta}.} \tag{3}$$
A: 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.
A: Here is a closed form with many integral representations:
\begin{align*}\sum_{k=0}^{n} \frac{x^k}{(k!)^2}&= \sum_{k=0}^\infty \frac{x^k}{(k!)^2}-\sum_{k=n}^\infty \frac{x^k}{(k!)^2}\mathop=^{t\to t+n} \sum_{k=0}^\infty \frac{x^k}{k!^2}-x^n\sum_{k=0}^\infty \frac{x^k \Gamma(k+1)}{\Gamma^2(k+n+1)k!}\\ &= \sum_{k=0}^\infty \frac{x^k}{k!^2}-\frac{x^n}{n!^2}\sum_{k=0}^\infty \frac{x^k (1)_k}{(n+1)_k ^2k!} .\end{align*}
When the regularized $\,_1\text F_2$ hypergeometric function is defined as:
$$\,_1\mathrm{\tilde F}_2(a;b_1,b_2;x)=\sum_{k=0}^\infty \frac{(a)_kx^k}{\Gamma(k+b_1)\Gamma(k+b_2)k!} $$
with the pochhammer symbol.
Therefore, you already know about the Bessel function:
\begin{align*}\boxed{\sum_{k=0}^{n} \frac{x^k}{(k!)^2} =\text I_0\left(2\sqrt x\right)-x^n(n+2)^2\,_1\mathrm{ \tilde F}_2(1;n+2,n+2,x) \\  \qquad\qquad = \text I_0\left(2\sqrt x\right)-\frac{x^n\,_1\mathrm{  F}_2(1;n+2,n+2,x)}{(n+1)!^2}} \end{align*}
where the regularized hypergeometric term just simplifies into a polynomial based on the upper index limit $n$. Please correct me and give me feedback!
