Prove that $ \sum_{k=0}^n \frac{(-1)^k}{k!}\binom{n}{k}=e\int_0^\infty \frac{t^ne^{-t}}{n!} J_0(2\sqrt{t})\;\mathrm{d}t$ using only real analysis. Here $J_0$ is the Bessel function.
Yesterday I ask this question on MSE. After some time some answers were given, only one (answer 2) being without mistakes. Then I tried to understand it.
This answer use complex analysis to prove the problem from the title:

$\displaystyle \sum_{k=0}^n \frac{(-1)^k}{k!}\binom{n}{k}=e\int_0^\infty \frac{t^ne^{-t}}{n!} J_0(2\sqrt{t})\;\mathrm{d}t$

(this formula is a special case of https://dlmf.nist.gov/18.10.E9, but here is not presented a proof)
and the problem is that I don't know complex analysis. So, I ask for a solution to this problem using only real analysis.
My approach.
Denote LHS by $I_n$ and RHS by $\mathcal{J}_n$. Then by using power series expansion for $J_0$ we obtain
\begin{align*}
\mathcal{J}_n&=\frac{e}{n!}\int_0^\infty e^{-t}t^n \sum_{k=0}^\infty
\frac{(-1)^k}{k!^2}t^k\;\mathrm{d}t\\&=
\frac{e}{n!}\sum_{k=0}^\infty \int_0^\infty
               e^{-t}t^{n+k}\frac{(-1)^k}{k!^2}\;\mathrm{d}t\\
&=\frac{e}{n!}\sum_{k=0}^\infty
\frac{(-1)^k}{k!^2}\int_0^\infty e^{-t}t^{n+k}\;\mathrm{d}t
\\&=\frac{e}{n!}\sum_{k=0}^\infty\frac{(-1)^k}{k!^2}\Gamma(n+k+1)\\
&=\frac{e}{n!}\sum_{k=0}^\infty\frac{(-1)^k}{k!^2}(n+k)!\\&=
e\sum_{k=0}^\infty \frac{(-1)^k}{k!}\binom{n+k}{k}.
\end{align*}
Remains to show that $e\sum\limits_{k=0}^\infty \dfrac{(-1)^k}{k!}\dbinom{n+k}{k}=I_n$.
Here I stopped. We have two problems:

*

*At the second equality we use Fubini's theorem without knowing that the integral converges.

*I cannot prove that $I_n=e\sum\limits_{k=0}^\infty \dfrac{(-1)^k}{k!}\dbinom{n+k}{k}$ (here my ideas are to use $e=\sum\limits_{n\geqslant 0}\frac1{n!}$ and Cauchy product of two series).

EDIT. The answer to the question is in the now edited answer 2 here.
 A: Calculating the generating function for the LHS has already been done following your link. It suffices to calculate the generating function for the RHS i.e.
$$\sum_{n=0}^\infty \mathcal{J}_n x^n = e\sum_{n=0}^\infty x^n \sum_{k=0}^\infty \frac{(-1)^k}{k!} \binom{n+k}{k}  = e \sum_{k=0}^\infty \frac{(-1)^k}{k!} \sum_{n=0}^\infty \binom{n+k}{k} x^n \\ = \frac{e}{1-x} \sum_{k=0}^\infty \frac{(-1)^k}{k!} \frac{1}{(1-x)^k} = \frac{e}{1-x} \, e^{-\frac{1}{1-x}} \, .$$
Here we used the fact that $$\sum_{n=0}^\infty \binom{n+k}{k} x^n = \frac{1}{(1-x)^{k+1}} \,.$$

Interchanging summation and integration is justified, since $$\left|\sum_{k=0}^n \frac{(-t)^k}{k!^2} \right| \leq \sum_{k=0}^n \frac{t^k}{k!^2} \leq \sum_{k=0}^\infty \frac{t^k}{k!^2} \leq \left( \sum_{k=0}^\infty \frac{t^{k/2}}{k!} \right)^2 = e^{2\sqrt{t}}$$
for any integer $n$. You can then make use of the DCT.
A: May be a few identities could help.
$$\sum_{k=0}^n \frac{(-1)^k}{k!}\binom{n}{k}\,x^k=\, _1F_1(-n;1;x)=L_n(x)$$ ($L_n(x)$ being Laguerre polynomial) and
$\lim_{n\to \infty } \, L_n(1) =0$
On the other side, the last expression you wrote
$$\sum\limits_{k=0}^\infty \dfrac{(-1)^k}{k!}\dbinom{n+k}{k}x^k=\, _1F_1(n+1;1;-x)=L_{-n-1}(-x)$$
So, it remains to prove that
$$\, _1F_1(-n;1;1)=e\,\, _1F_1(n+1;1;-1)$$ or that
$$L_n(1)=e\,L_{-n-1}(-1)$$
Both are (numerically) true but, at this pont, I am stuck.
