Another Typical integral !!!

$\large \displaystyle I = \ln \left ( \sum_{\alpha = 1}^{\infty} \left ( \int_0^{\infty} \frac{x^{\alpha - 1} e^{- \frac{x}{2018}}}{2018[(\alpha-1)!]^2} dx \right ) \right )$

$\large \displaystyle I = \ln \left ( \sum_{\alpha = 1}^{\infty} \frac{1}{2018[(\alpha-1)!]^2} \left ( \int_0^{\infty} x^{\alpha - 1} e^{- \frac{x}{2018}} dx \right ) \right )$

Make $x = 2018u$ , $dx = 2018 du$ :

$\large \displaystyle I = \ln \left ( \sum_{\alpha = 1}^{\infty} \frac{1}{2018[(\alpha-1)!]^2} \left ( \int_0^{\infty} (2018u)^{\alpha - 1} e^{-u} 2018 du \right ) \right )$

$\large \displaystyle I = \ln \left ( \sum_{\alpha = 1}^{\infty} \frac{2018^{\alpha - 1}}{[(\alpha-1)!]^2} \left ( \int_0^{\infty} u^{\alpha - 1} e^{-u} du \right ) \right )$

The integral is definition of the Gamma Function at $\alpha$ , which is equal to $(\alpha-1)!$

$\large \displaystyle I = \ln \left ( \sum_{\alpha = 1}^{\infty} \frac{2018^{\alpha - 1}}{( \alpha-1)!} \right )$

Make $n = \alpha - 1$ :

$\large \displaystyle I = \ln \left ( \sum_{n = 0}^{\infty} \frac{2018^n}{n!} \right )$

From Maclaurin series of exponential function:

$\large \displaystyle I = \ln \left ( e^{2018} \right )$

$\color{#3D99F6} \boxed { \large \displaystyle I = 2018 }$

After integrating, the given expression becomes the logarithm of summation of (2018)^(α-1)/[(α-1)!] from α=1 to infinity, or the logarithm of exp(2018), which is equal to 2018.