Intimidating at First

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

Let $u = \frac{x}{2018}$ :

$\large \displaystyle S = \ln \left [ \sum_{i=1}^{\infty} \frac{2018^{i-1}}{[(i-1)!]^2} \color{#20A900} \left ( \int_0^{\infty} u^{i-1} e^{-u} du \right ) \color{#333333} \right ]$

$\large \displaystyle S = \ln \left [ \sum_{i=1}^{\infty} \frac{2018^{i-1}}{[(i-1)!]^2} \color{#20A900} \Gamma(i) \color{#333333} \right ]$

$\large \displaystyle S = \ln \left [ \sum_{i=1}^{\infty} \frac{2018^{i-1}}{[(i-1)!]^2} \color{#20A900} (i-1)! \color{#333333} \right ]$

$\large \displaystyle S = \ln \left [ \sum_{i=1}^{\infty} \frac{2018^{i-1}}{(i-1)!} \right ]$

Let $k = i - 1$ :

$\large \displaystyle S = \ln \left [ \sum_{k=0}^{\infty} \frac{2018^{k}}{k!} \right ]$

$\large \displaystyle S = \ln \left [ e^{2018} \right ]$

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