Generating function of exponential integrals

When studying the integral $$f(x,y)=\int_0^\infty \exp\left[-x\mathrm{e}^{-t}-y\cosh t\right]\mathrm d t,$$ I arrived at the handier result (especially for studying the asymptotics) $$f(x,y)=\sum_{n=1}^\infty (-)^{n-1}\frac{(x+\frac y2)^n}{n!}\mathrm{E}_n(\frac y2).$$ (Where $\mathrm{E}_n(x)=\int_1^\infty t^{-n}\mathrm{e}^{-xt}\mathrm d t\;$ is the $n^{\text{th}}$ order exponential integral.) So now $f$ is just the generating function of the exponential integrals. Is there any chance that $f$ has a closed form ?

EDIT Some of the thoughts I have on how to find a closed form.

With a change of variables and an extra term $\mathrm{E}_0(y)=\frac{\mathrm{e}^{-y}}{y}$, I define the exponential generating function of exponential intagrals as $$g(x,y)=\sum_{n=0}(-)^n\frac{x^n}{n!}\mathrm{E}_n(y)$$ We use the relation $\partial_x\mathrm{E}_n(x)=-\mathrm{E}_{n-1}(x)$ for $n\geq1$ ($\partial_x$ denotes the partial derivative with respect to $x$) and we obtain the partial differential equation $$\partial_x\partial_y g(x,y)=g(x,y).$$ Posing $u=x+y$ and $v=x-y$, we can transform this equation into an equation for $h(u,v)=g(x,y)$ $$\partial^2_uh(u,v)-\partial^2_vh(u,v)=h(u,v).$$

This is known as the homogeneous Klein-Gordon equation, it has a lot of known solutions (for instance see the Handbook of partial differential equations by Andrei Polyanin).

EDIT 2

From what I know, we must have $h(u,0)=g(\frac u2,\frac u2)=\mathrm{K}_0(u)$ for $u>0$. We also can find, as @tired wrote, $g(0,y)=\mathrm{E}_1(y)$. We also have $\partial_v h=\frac12\partial_xg-\frac12\partial_yg$ then $$\partial_vh(u,0)=-\int_0^\infty\sinh t\;\exp\left[-u\cosh t\right]\mathrm dt=-\frac{\mathrm{e}^{-u}}{u}=-\mathrm{E}_0(u).$$ As it appears in the Handbook of partial differential equations, the solution is then given $$\begin{split}h(u,v)&=\frac12\mathrm{K}_0(u+v)+\frac12\mathrm{K}_0(u-v) +\frac v2\int_{u-v}^{u+v}\frac{\mathrm{I}_1\left(\sqrt{v^2-(u-\xi)^2}\right)} {\sqrt{v^2-(u-\xi)^2}}\mathrm{K}_0(\xi)\mathrm d\xi\\& \quad+\frac12\int_{u-v}^{u+v}\mathrm{I}_0\left(\sqrt{v^2-(u-\xi)^2}\right)\mathrm{E}_0(\xi)\mathrm d\xi, \end{split}$$ This is far from being the closed form I hoped for. However, the form given by @tired makes me keep hope.

• $E_n(x)$ is the closed form. What do you expect exactly ?. As far as I know, it is related to the incomplete gamma function. – JJacquelin Dec 18 '16 at 9:26
• @JJacquelin. I am looking for a closed form of $f$ or $g$ or $h$. – Tom-Tom Dec 18 '16 at 10:55
• what exactly is the problem? if you know the solutions for this DE what is the point? – tired Dec 19 '16 at 0:26
• @tired. The problem is to find the right solution among a big lot of possible solutions. – Tom-Tom Dec 19 '16 at 10:52
• so you are looking for the correct boundary conditions? – tired Dec 19 '16 at 13:28