Compute the cool integral/sum

Calculus Level 2

ln ( α = 1 0 x α 1 e x 2020 2020 ( ( α 1 ) ! ) 2 d x ) = ? \large \ln \left(\sum_{\alpha =1}^\infty \int_0^\infty \frac {x^{\alpha-1}e^{-\frac x{2020}}}{2020((\alpha-1)!)^2} dx \right) = \ ?

Not possible 2019 2020 2018

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1 solution

Chew-Seong Cheong
Aug 11, 2020

X = ln ( α = 1 0 x α 1 e x 2020 2020 ( ( α 1 ) ! ) 2 d x ) Let n = α 1 = ln ( n = 0 0 x n e x 2020 2020 ( n ! ) d x ) Let t = x 2020 d t = d x 2020 = ln ( n = 0 202 0 n ( n ! ) 2 0 t n e t d t ) = ln ( n = 0 202 0 n ( n ! ) 2 Γ ( n + 1 ) ) where Γ ( ) is the gamma function. = ln ( n = 0 202 0 n ( n ! ) 2 n ! ) = ln ( n = 0 202 0 n n ! ) = ln ( e 2020 ) = 2020 \begin{aligned} X & = \ln \left(\sum_{\alpha =1}^\infty \int_0^\infty \frac {x^{\alpha-1}e^{-\frac x{2020}}}{2020((\alpha-1)!)^2} dx \right) & \small \blue{\text{Let }n = \alpha -1} \\ & = \ln \left(\sum_{n=0}^\infty \int_0^\infty \frac {x^n e^{-\frac x{2020}}}{2020(n!)} dx \right) & \small \blue{\text{Let }t = \frac x{2020} \implies dt = \frac {dx}{2020}} \\ & = \ln \left(\sum_{n=0}^\infty \frac {2020^n}{(n!)^2} \blue{\int_0^\infty t^n e^{-t}\ dt}\right) \\ & = \ln \left(\sum_{n=0}^\infty \frac {2020^n}{(n!)^2} \cdot \blue{\Gamma(n+1)}\right) & \small \blue{\text{where }\Gamma(\cdot) \text{ is the gamma function.}} \\ & = \ln \left(\sum_{n=0}^\infty \frac {2020^n}{(n!)^2} \cdot \blue{n!}\right) \\ & = \ln \left(\sum_{n=0}^\infty \frac {2020^n}{n!} \right) \\ & = \ln \left(e^{2020} \right) \\ & = \boxed{2020} \end{aligned}


Reference: Gamma function

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