Determine the closed form for $\int_{0}^{\infty}\sinh(xe^{-x})dx$ Find the closed form for

$$I=\int_{0}^{\infty}\sinh(xe^{-x})dx\tag1$$

Change to
$$I={1\over 2}\int_{0}^{\infty}(e^{xe^{-x}}-e^{-xe^{-x}}dx)\tag2$$
Any hints? I can't go further.

By expanding the hyperbolic sine as a Taylor series we have:

$$ I = \sum_{n\geq 0}\frac{1}{(2n+1)!}\int_{0}^{+\infty}x^{2n+1} e^{-(2n+1)x}\,dx = \color{red}{\sum_{n\geq 0}\frac{1}{(2n+1)^{2n+2}}} \tag{1}$$

with a series recalling sophomore's dream. I won't bet on simple closed forms but in terms of integrals involving the Lambert $W$ function.
