Exponential polynomials, Stirling and Bernoulli numbers The polynomials are defined as:
$P_0(x)=1, \ P_{n+1}(x)=x(P_n'(x)+P_n(x))$.
The coeficents of this polynomials are the Stirling numbers of the second kind, thus:
\begin{equation}
P_n(x)=\sum_{k=0}^n\begin{Bmatrix} n\\  k \end{Bmatrix}x^k.
\end{equation}
These polynomials have an interesting property, and it's that I want to prove:
\begin{equation}
\int_{-\infty}^{0}P_n(x)P_m(x)\frac{e^{2x}}{x}dx=(-1)^{n-1}\cfrac{2^{n+m}-1}{n+m}B_{n+m}.
\end{equation}
Something that I've been using and maybe it's useful is the that $P_n(x)e^x=\sum_{k=0}^{\infty}\frac{k^nx^k}{k!}$, so with this we can do the Cauchy product and maybe getting something but, but I haven't gotten anything. For other side, we can change the problem to prove:
\begin{equation}
\sum_{k=0}^{n}\sum_{j=0}^{m}(-1)^{k+j}\begin{Bmatrix} n\\  k \end{Bmatrix}\begin{Bmatrix} m\\  j \end{Bmatrix}\cfrac{(k+j-1)!}{2^{k+j}}=(-1)^{n-1}\cfrac{2^{n+m}-1}{n+m}B_{n+m}.
\end{equation}
 A: Using the exponential generating function for these polynomials:
$$\sum_{n=0}^{\infty}P_n(x)\frac{t^n}{n!}=\exp\big(x(e^t-1)\big),$$
we obtain (for $|y|$, $|z|$ small enough)
$$\sum_{n,m=1}^{\infty}\frac{y^n}{n!}\frac{z^m}{m!}\int_{-\infty}^0 P_n(x)P_m(x)\frac{e^{2x}}{x}\,dx=\int_{-\infty}^0\frac{(e^{xe^y}-e^x)(e^{xe^z}-e^x)}{x}\,dx\\=\ln\frac{(e^y+1)(e^z+1)}{2(e^y+e^z)}=\ln\frac{\cosh(y/2)\cosh(z/2)}{\cosh\big((z-y)/2\big)}$$
(this is a Frullani-type integral). The power series for $\color{DarkBlue}{\ln\cosh}$ is known from the one of $\color{DarkBlue}{\tanh}$:
$$\ln\cosh(w/2)=\sum_{k=2}^{\infty}L_k\frac{w^k}{k!},\qquad L_k=(2^k-1)B_k/k$$
(we keep in mind that $L_k=0$ for odd $k$). Hence,
$$\ln\frac{\cosh(y/2)\cosh(z/2)}{\cosh\big((z-y)/2\big)}=-\sum_{k=2}^{\infty}\frac{L_k}{k!}\sum_{n=1}^{k-1}(-1)^n\binom{k}{n}y^n z^{k-n}\\=\sum_{k=2}^{\infty}\sum_{n=1}^{k-1}(-1)^{n-1}L_k\frac{y^n}{n!}\frac{z^{k-n}}{(k-n)!}=\sum_{n,m=1}^{\infty}(-1)^{n-1}L_{n+m}\frac{y^n}{n!}\frac{z^m}{m!}.$$
The result follows by comparison.
