Question

Question :
$\color{springgreen}\displaystyle{\rm \int_{0} ^{ \infty } \frac{ {x}^{2020} }{ {2020}^{x} } \: dx}$

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Answer 1

Refer the given attachment

Answer 2

$\green{ \boxed{ \boxed{ \begin{matrix}\displaystyle \rm \red{ \int_{0}^{ \infty } \frac{ {x}^{2020} }{ {2020}^{x} } \: dx } \\ \\ \displaystyle \rm \red{ \int_{0}^{ \infty } \frac{ {x}^{m} }{ {m}^{x} } \: dx } \\ \\ \displaystyle \rm \red{ \int_{0}^{ \infty } { {x}^{m} }{ {m}^{ - x} } \: dx } \\ \\ \displaystyle \rm \red{ \int_{0}^{ \infty } { {x}^{m} }{ {e}^{ - x \ln(m) } } \: dx } \\ \\ \displaystyle \rm \red{ \int_{0}^{ \infty } \bigg( \frac{ {x}^{} }{ { ln(m) }^{} } \bigg)^{m} {e}^{ - t} \frac{1}{ ln(m) } dt } \\ \\ \displaystyle \rm \red{ \int_{0}^{ \infty } \frac{ {x}^{m} }{ { (ln (m) )}^{m + 1} } \: {e}^{ - t} \: dt } \\ \\ \displaystyle \rm \red{ \frac{ 1^{} }{ { (ln (m) )}^{m + 1} } \int_{0}^{ \infty } {t}^{m} {e}^{ - t} \: dt } \\ \\ \displaystyle \rm \red{ \frac{ 1^{} }{ { (ln (m) )}^{m + 1} } m !} \\ \\ \displaystyle \rm \red{ \frac{ {2020!} }{( ln(2020)) {}^{2021} } } \end{matrix}}}}$