Expressing the solution of integral using Special functions Gamma or/and Beta functions?

Question

$$\int_{0}^{\infty}(1+x)^{2020}*\frac{1+x+x^2+...+x^{2020}}{2020^{{x}^{2020}}}dx$$

How can this integral be solved?

Answer 1

$$=\int_{0}^{\infty }{\frac{{{(1+x)}^{2020}}}{{{2020}^{{{x}^{2020}}}}}}\left( \sum\nolimits_{n=0}^{2020}{{{x}^{n}}} \right)dx$$

$$=\sum\nolimits_{n=0}^{2020}{\int_{0}^{\infty }{\frac{{{(1+x)}^{2020}}}{{{2020}^{{{x}^{2020}}}}}{{x}^{n}}dx}}$$

$$=\sum\nolimits_{n=0}^{2020}{\sum\nolimits_{k=0}^{2020}{\left( \begin{align} & 2020 \\ & \ \ k \\ \end{align} \right)\int_{0}^{\infty }{\frac{{{x}^{k+n}}}{{{2020}^{{{x}^{2020}}}}}dx}}}$$

$$=\frac{1}{2020}\sum\nolimits_{n=0}^{2020}{\sum\nolimits_{k=0}^{2020}{\left( \begin{align} & 2020 \\ & \ \ k \\ \end{align} \right)}}\frac{\Gamma \left( \frac{n+k+1}{2020} \right)}{{{\left( \ln 2020 \right)}^{\frac{n+k+1}{2020}}}}$$ for the integral in the third line: $$=\int_{0}^{\infty }{{{x}^{k+n}}{{2020}^{-{{x}^{2020}}}}dx}=\int_{0}^{\infty }{{{x}^{k+n}}{{e}^{-{{x}^{2020}}\ln 2020}}dx}$$ use $$u=\ln 2020\ {{x}^{2020}}$$ or $$x={{\left( \frac{u}{\ln 2020} \right)}^{\frac{1}{2020}}}$$