Question

$\displaystyle \bf \red{ \ln \bigg( \sum \limits_{ \beta = 1}^{ \infty } \int_{0}^{ \infty } \frac{ {x}^{ \beta - 1} {e}^{ - \frac{x}{2020} } }{2020 [( \beta - 1)!^{} ]^{2} } \: dx \bigg) }$

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Answer 1

Answer:

Step-by-step explanation:

$\int\limits^\infty_0 {x^{\beta-1}\cdot e^{-\frac{x}{2020}} \, dx$

Let,

$\frac{x}{2020} = u \implies dx = 2020\cdot du\\\implies x = 2020\cdot u$

$\text{When } x ~\text{tends to }\infty~\text{then } u~ \text{also tends to }\infty \\\text{and when}~x=0~\text{then}~u =0$

$\int\limits^\infty_0 {x^{\beta-1}\cdot e^{-\frac{x}{2020}} \, dx$

$=2020\int\limits^\infty_0 {{(2020u)}^{\beta-1}\cdot e^{-u} \, du$

$= (2020)^\beta\int\limits^\infty_0 {{(u)}^{\beta-1}\cdot e^{-u} \, du$                  $(~\because~\Gamma(n) = \int\limits^\infty_0 {x^{n-1} e^{-x}} \, dx ~)$

$= (2020)^\beta \Gamma(\beta)$                                       ( $\because~\text{for integer }\beta,~\Gamma(\beta) = (\beta-1)!$ )

$= (2020)^\beta (\beta-1)!$

Now,

$\displaystyle \bf \red{ \ln \bigg( \sum \limits_{\beta = 1}^{ \infty } \int_{0}^{ \infty }\frac{{x}^{\beta-1} e^{-\frac{x}{2020}}}{2020[(\beta-1)!]^2}\,dx\bigg)$

$\displaystyle \bf \red{ \ln \bigg( \sum \limits_{\beta = 1}^{ \infty } \frac{1}{2020[(\beta-1)!]^2}\int_{0}^{ \infty} {{ x}^{\beta-1} e^{-\frac{x}{2020}}}\,dx\bigg)$

$=\displaystyle \bf \red{ \ln \bigg( \sum \limits_{\beta = 1}^{ \infty } \frac{1}{2020[(\beta-1)!]^2}\cdot {(2020)^\beta \cdot (\beta-1)!}\bigg)$

$=\displaystyle \bf \red{ \ln \bigg( \sum \limits_{\beta = 1}^{ \infty } \frac{(2020)^{\beta-1}}{(\beta-1)!}\bigg)$

$= \ln(1 + \frac{2020}{1!} + \frac{(2020)^2}{2!} + \frac{(2020)^3}{3!}+\ldots)$

$=\ln(e^{2020})$                       $(~\because~e^x = \sum_{\n=0}^{\infty} \frac{x^n}{n!}~)$

$= 2020$

Answer 2

$\displaystyle \sf \red{\int\limits^\infty_0 {x^{\beta-1}\cdot e^{-\frac{x}{2020}} \, dx}}$

Let,

$\begin{gathered}\frac{x}{2020} = u \implies dx = 2020\cdot du\\\implies x = 2020\cdot u\end{gathered}$

$\begin{gathered}\text{When } x ~\text{tends to }\infty~\text{then } u~ \text{also tends to }\infty \\\text{and when}~x=0~\text{then}~u =0\end{gathered}$

$\displaystyle \sf \red{\int\limits^\infty_0 {x^{\beta-1}\cdot e^{-\frac{x}{2020}} \, dx}}$

$\displaystyle \sf \red{=2020\int\limits^\infty_0 {{(2020u)}^{\beta-1}\cdot e^{-u} \, du}}$

$\displaystyle{ \sf \red{ = (2020)^\beta\int\limits^\infty_0 {{(u)}^{\beta-1}\cdot e^{-u} \, du (~\because~\Gamma(n) = \int\limits^\infty_0 {x^{n-1} e^{-x}} \, dx ~)}}}$

$\displaystyle \sf \red{= (2020)^\beta \Gamma(\beta)=(2020) </p><p> \: \: \: ( \because~\text{for integer }\beta,~\Gamma(\beta) = (\beta-1)!}$

Now,

$\displaystyle \bf \red{ \ln \bigg( \sum \limits_{\beta = 1}^{ \infty } \int_{0}^{ \infty }\frac{{x}^{\beta-1} e^{-\frac{x}{2020}}}{2020[(\beta-1)!]^2}\,dx\bigg)}$

$\displaystyle \bf \red{ \ln \bigg( \sum \limits_{\beta = 1}^{ \infty } \frac{1}{2020[(\beta-1)!]^2}\int_{0}^{ \infty} {{ x}^{\beta-1} e^{-\frac{x}{2020}}}\,dx\bigg)}$

$\displaystyle \bf \red{ = \ln \bigg( \sum \limits_{\beta = 1}^{ \infty } \frac{1}{2020[(\beta-1)!]^2}\cdot {(2020)^\beta \cdot (\beta-1)!}\bigg)}$

$\displaystyle \bf \red{ = \ln \bigg( \sum \limits_{\beta = 1}^{ \infty } \frac{(2020)^{\beta-1}}{(\beta-1)!}\bigg)}$

${\red{\ln(1 + \frac{2020}{1!} + \frac{(2020)^2}{2!} + \frac{(2020)^3}{3!}+\ldots)}}$

$\sf \red{=\ln(e^{2020})(~\because~e^x = \sum \limits_{n=0}^{\infty} \frac{x^n}{n!}~)}$

=2020