“Hideous” Integral

$\large I = \int_{0}^{\infty} e^{x+e^x+e^{e^x}-e^{e^{e^{x}}-e}}dx =e^e \int_{0}^{\infty} \orange{\frac{e^{x+e^x+e^{e^x}}}{e^e}}\cdot e^{-\blue{e^{e^{e^{x}}-e}}}dx$ Let $\displaystyle u = \blue{e^{e^{e^{x}}-e}} \implies du = \orange{\frac{e^{x+e^x+e^{e^x}}}{e^e}}$ : $\implies I = e^e\int_{u=1}^{u=\infty}e^{-u}du = e^e\cdot -e^{-u}\bigg|_{1}^{\infty} = \lim_{u \to \infty}\left(\frac{-e^e }{e^{u}}\right) + e^e\cdot e^{-1}$ The limit at the front tends to $0$ because $e^u$ on the denominator tends to $\infty$ as $u$ tends to $\infty$ , leaving us with $I = 0 + e^{e-1} = e^{\green{1} \cdot e + \blue{1} \cdot e^{i\pi}} \; \; \; \; \; \blue{(e^{i \pi} = -1)}$ $\implies \green{\alpha} + \blue{\beta} = \green{1} + \blue{1} = \boxed{2}$

This one is actually not difficult...