"I am Back" says Integration Part 6

Calculus

$\large \int_{-\infty}^{\infty} {xe^{2x}e^{-{e}^{2x}} \, dx}$

The above Integral can be expressed as $-\dfrac{\gamma}{a},$

where $\gamma$ denotes the Euler-Mascheroni constant $\displaystyle \gamma = \lim_{n\to\infty} \left( - \ln n + \sum_{k=1}^n \dfrac1k \right) \approx 0.5772 .$

Find $a$ .