Question-197365

Question Number 197365 by mokys last updated on 15/Sep/23

Answered by witcher3 last updated on 15/Sep/23

x=((1−t)/(1+t)) ∫_0 ^1 (((((1−t)/(1+t)))^(1−p) (((2t)/(1+t)))^p )/(((2/(1+t)))^3 )).(2/((1+t)^2 ))dt =∫_0 ^1 (1−t)^(1−p) t^p .2^(p−2) =2^(p−2) ∫_0 ^1 t^(p+1−1) (1−t)^(2−p−1) dt =2^(p−2) β(p+1,2−p) =2^(p−2) .((Γ(p+1)Γ(2−p))/(Γ(3)))=(2^(p−2) /2).pΓ(p).(1−p)Γ(1−p) =((2^p p(1−p).Γ(p)Γ(1−p))/8)=((2^p p(1−p))/8).(π/(sin(pπ)))

$$\mathrm{x}=\frac{\mathrm{1}−\mathrm{t}}{\mathrm{1}+\mathrm{t}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\frac{\mathrm{1}−\mathrm{t}}{\mathrm{1}+\mathrm{t}}\right)^{\mathrm{1}−\mathrm{p}} \left(\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}}\right)^{\mathrm{p}} }{\left(\frac{\mathrm{2}}{\mathrm{1}+\mathrm{t}}\right)^{\mathrm{3}} }.\frac{\mathrm{2}}{\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{2}} }\mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\mathrm{t}\right)^{\mathrm{1}−\mathrm{p}} \mathrm{t}^{\mathrm{p}} .\mathrm{2}^{\mathrm{p}−\mathrm{2}} \\ $$$$=\mathrm{2}^{\mathrm{p}−\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{t}^{\mathrm{p}+\mathrm{1}−\mathrm{1}} \left(\mathrm{1}−\mathrm{t}\overset{\mathrm{2}−\mathrm{p}−\mathrm{1}} {\right)}\mathrm{dt} \\ $$$$=\mathrm{2}^{\mathrm{p}−\mathrm{2}} \beta\left(\mathrm{p}+\mathrm{1},\mathrm{2}−\mathrm{p}\right) \\ $$$$=\mathrm{2}^{\mathrm{p}−\mathrm{2}} .\frac{\Gamma\left(\mathrm{p}+\mathrm{1}\right)\Gamma\left(\mathrm{2}−\mathrm{p}\right)}{\Gamma\left(\mathrm{3}\right)}=\frac{\mathrm{2}^{\mathrm{p}−\mathrm{2}} }{\mathrm{2}}.\mathrm{p}\Gamma\left(\mathrm{p}\right).\left(\mathrm{1}−\mathrm{p}\right)\Gamma\left(\mathrm{1}−\mathrm{p}\right) \\ $$$$=\frac{\mathrm{2}^{\mathrm{p}} \mathrm{p}\left(\mathrm{1}−\mathrm{p}\right).\Gamma\left(\mathrm{p}\right)\Gamma\left(\mathrm{1}−\mathrm{p}\right)}{\mathrm{8}}=\frac{\mathrm{2}^{\mathrm{p}} \mathrm{p}\left(\mathrm{1}−\mathrm{p}\right)}{\mathrm{8}}.\frac{\pi}{\mathrm{sin}\left(\mathrm{p}\pi\right)} \\ $$$$ \\ $$

Commented by mokys last updated on 15/Sep/23

$${can}\:{you}\:{solve}\:{by}\:{resideo} \\ $$

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