Integral representation of $\sum_{k=0}^{n} \frac{x^k}{(k!)^2}$?

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.

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

$$ \boxed{\,f(x)=\color{red}{\frac{1}{\pi}\int_{-\pi/2}^{\pi/2}\cosh\left(2\sqrt{x}\cos\theta\right)\,d\theta}.} \tag{3}$$