I was inspired by this question: Show that :$\sum_{j=0}^{n}j{n\choose j}^2H_{n-j}={1\over 4}{2n\choose n}(4nH_{n}-2nH_{2n}-1)$ to study the following series:
$$\sum _{n=1}^{\infty } \frac{\binom{2 n}{n} \left(4 n H_n-2 n H_{2 n}-1\right) x^{n}}{n!}$$
It turns out that a very curious identity holds:
$$\sum _{n=1}^{\infty } \frac{\binom{2 n}{n} \left(4 n H_n-2 n H_{2 n}-1\right) x^{n+1}}{n!}= \\ =\int_0^{\infty } u^2 \exp \left(-\frac{u^2}{4 x}\right) I_1(u) \left[K_0(u)+(\log (u)+\gamma -\log (2)) I_0(u)\right] \, du$$
I confirmed it numerically using Mathematica.
Usual questions:
What is the best way to prove it? Or just some interesting ways?
Is this identity known?
Can it lead to some interesting consequences?
Before providing my own "proof" I have to warn that it's a result of looking up a random combination of identities (including the linked one), some of them with the help of Mathematica, and most of them also unproven.
Examples:
1) This one I guessed from summing a few cases in Mathematica. $L_j$ are Laguerre polynomials.
$$\sum_{n=0}^\infty \binom{n}{j}^2 \frac{x^n}{n!}=e^x L_j(-x) \frac{x^j}{j!}$$
2) And this one is from G-R.
$$L_j(x)= \frac{2}{j!} e^x \int_0^\infty e^{-t^2} t^{2j+1} J_0(2t \sqrt{x}) dt$$
3) This integral is given by Mathematica.
$$\int_0^1 \frac{I_0(a)-I_0(a \sqrt{t})}{1-t}dt=K_0(a)+\left(\gamma +\log \left(\frac{a}{2}\right)\right) I_0(a)$$
Going by these identities it's not hard to figure out the proof, however, I would like a more clear/concise proof if possible.
Update
WA quite easily sums the following series:
$$\sum_{n=1}^\infty \binom{2n}{n} \frac{y^n}{(n-1)!}=2 y e^{2y} (I_0(2y)+I_1(2y))$$
$$\sum_{n=1}^\infty \binom{2n}{n} \frac{y^n}{n!}=e^{2y} I_0(2y)-1$$
Which leads to another integral representation for the series, using:
$$H_s=\int_0^1 \frac{1-t^s}{1-t}dt$$
