Question

$\huge \displaystyle \sf \int^{1}_{0} \frac{ {x}^{ {e}^{ \pi } - 1 } - {x}^{ {e}^{ \gamma } - 1 } }{ ln( \sqrt[2020]{x} ) } \: dx$

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Answer 1

Answer:

Step-by-step explanation:

$\huge \displaystyle\sf\int^{1}_{0}\frac{{x}^{{e}^\pi-1} - {x}^{{e}^\gamma-1}}{\ln(\sqrt[2020]{x})}dx$

= $\huge \displaystyle\sf\int^{1}_{0}\frac{{x}^{{e}^\pi-1} - {x}^{{e}^\gamma-1}}{\ln({x}^\frac{1}{2020})}dx$

= $\huge \displaystyle\sf\int^{1}_{0}\frac{{x}^{{e}^\pi-1} - {x}^{{e}^\gamma-1}}{\frac{1}{2020}\ln({x})}dx$

= $2020\huge \displaystyle\sf\int^{1}_{0}\frac{{x}^{{e}^\pi-1} - {x}^{{e}^\gamma-1}}{\ln({x})}dx$

= $2020\huge \displaystyle\sf\int^{1}_{0}\frac{{x}^{{e}^\pi} - {x}^{{e}^\gamma}}{x\cdot \ln({x})}dx$  --------------- (1)

Let,

$\ln(x) = t \implies \frac{dx}{x} = {dt}$

$\implies x = e^t$

$\text{for}~~x = 0, t~tends~to -\infty~\text{and for}~~x=1,t=0$

From (1) we get,

$2020\huge \displaystyle\sf\int^{0}_{-\infty}\frac{{e}^{mt} - {e}^{nt}}{t} {dt}$                      ( let, $e^\pi = m$ and $e^\gamma = n$ )

$= 2020\int_{-\infty}^{0}\int_{n}^{m}e^{rt}dr\,dt\\ = 2020\int_{n}^{m}\int_{-\infty}^{0}e^{rt}dtdr \\\\= 2020\int_{n}^{m}\frac{dr}{r} \\\\= 2020\log(\frac{m}{n})$

$=2020 \log(\frac{e^\pi}{e^\gamma})\\= 2020(\log(e^\pi) - \log(e^\gamma))\\= 2020(\pi - \gamma)$