Question Number 166756 by HongKing last updated on 27/Feb/22
$$\mathrm{Evaluate}: \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:\frac{\left(\mathrm{x}\:+\:\mathrm{1}\right)^{\mathrm{2020}} }{\left(\mathrm{x}\:+\:\mathrm{2}\right)^{\mathrm{2022}} }\:\mathrm{dx}\:=\:? \\ $$
Answered by Mathspace last updated on 27/Feb/22
![x+2=t ⇒I=∫_2 ^∞ (((t−1)^(2020) )/t^(2022) )dt =∫_2 ^∞ ((Σ_(k=0) ^(2020) C_(2020) ^k t^k (−1)^(2020−k) )/t^(2022) )dt =Σ_(k=0) ^(2020) (−1)^k C_(2020) ^k t^(k−2022) dt =Σ_(k=0) ^(2020) (−1)^k C_(2020) ^k [(1/(k−2021))t^(k−2022) ]_2 ^∞ =−Σ_(k=0) ^(2020) (((−1)^k C_(2020) ^k )/(k−2021)) 2^(k−2022)](https://www.tinkutara.com/question/Q166772.png)
$${x}+\mathrm{2}={t}\:\Rightarrow{I}=\int_{\mathrm{2}} ^{\infty} \frac{\left({t}−\mathrm{1}\right)^{\mathrm{2020}} }{{t}^{\mathrm{2022}} }{dt} \\ $$$$=\int_{\mathrm{2}} ^{\infty} \:\frac{\sum_{{k}=\mathrm{0}} ^{\mathrm{2020}} {C}_{\mathrm{2020}} ^{{k}} {t}^{{k}} \left(−\mathrm{1}\right)^{\mathrm{2020}−{k}} }{{t}^{\mathrm{2022}} }{dt} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{2020}} \left(−\mathrm{1}\right)^{{k}} \:{C}_{\mathrm{2020}} ^{{k}} \:{t}^{{k}−\mathrm{2022}} {dt} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{2020}} \left(−\mathrm{1}\right)^{{k}} \:{C}_{\mathrm{2020}} ^{{k}} \left[\frac{\mathrm{1}}{{k}−\mathrm{2021}}{t}^{{k}−\mathrm{2022}} \right]_{\mathrm{2}} ^{\infty} \\ $$$$=−\sum_{{k}=\mathrm{0}} ^{\mathrm{2020}} \frac{\left(−\mathrm{1}\right)^{{k}} {C}_{\mathrm{2020}} ^{{k}} }{{k}−\mathrm{2021}}\:\mathrm{2}^{{k}−\mathrm{2022}} \\ $$
Answered by mathsmine last updated on 27/Feb/22
$$=\int_{\mathrm{0}} ^{\infty} \left(\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}\right)^{\mathrm{2020}} .\left(\frac{\mathrm{1}}{{x}+\mathrm{2}}\right)^{\mathrm{2}} {dx} \\ $$$${u}=\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}=\mathrm{1}−\frac{\mathrm{1}}{\left({x}+\mathrm{2}\right)}\Rightarrow{du}=\frac{{dx}}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} {u}^{\mathrm{2020}} {du}=\frac{\mathrm{1}}{\mathrm{2021}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2021}} }\right) \\ $$