Question-206568

Question Number 206568 by universe last updated on 18/Apr/24

Answered by Berbere last updated on 19/Apr/24

(x/n)=y A(n)=∫_0 ^n (((2nx)/(x^2 +n^2 )))^n dx=n∫_0 ^1 (((2y)/(1+y^2 )))^n dy ∫_0 ^1 (((2y)/(1+y^2 )))^n ((2y)/(1+y^2 ))=t⇒ty^2 +t−2y=0 y=((2−(√(4−4t^2 )))/(2t))=((1−(√(1−t^2 )))/t)⇒dy=−(1/t^2 )+(1/(t^2 (√(1−t^2 )))) t^2 =x A(n)=n∫_0 ^1 t^n (−(1/t^2 )+(1/(t^2 (√(1−t^2 )))))dt=n∫_0 ^1 −t^(n−2) dt+n∫_0 ^1 (t^(n−2) /( (√(1−t^2 ))))dt =−(n/(n−1))+n∫_0 ^1 (t^(n−2) /( (√(1−t^2 ))))dt=−(n/(n−1))+n∫_0 ^1 (u^((n−2)/2) /( (√(1−u)))).(1/(2(√u)))du =−(n/(n−1))+(n/2)∫_0 ^1 (1−u)^((1/2)−1) u^(((n−1)/2)−1) du=−(n/(n−1))+(n/2)β((1/2),((n−1)/2)) =−(n/(n−1))+(n/2).((Γ((1/2))Γ(((n−1)/2)))/(Γ((n/2))))=A(n);n>1 Γ(z)∼(√(2π))z^(z−(1/2)) e^(−z) i willfinish later inchelah

$$\frac{{x}}{{n}}={y}\:\:{A}\left({n}\right)=\int_{\mathrm{0}} ^{{n}} \left(\frac{\mathrm{2}{nx}}{{x}^{\mathrm{2}} +{n}^{\mathrm{2}} }\right)^{{n}} {dx}={n}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{2}{y}}{\mathrm{1}+{y}^{\mathrm{2}} }\right)^{{n}} {dy} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{2}{y}}{\mathrm{1}+{y}^{\mathrm{2}} }\right)^{{n}} \\ $$$$\frac{\mathrm{2}\boldsymbol{{y}}}{\mathrm{1}+\boldsymbol{{y}}^{\mathrm{2}} }=\boldsymbol{{t}}\Rightarrow{ty}^{\mathrm{2}} +{t}−\mathrm{2}{y}=\mathrm{0} \\ $$$${y}=\frac{\mathrm{2}−\sqrt{\mathrm{4}−\mathrm{4}{t}^{\mathrm{2}} }}{\mathrm{2}{t}}=\frac{\mathrm{1}−\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{{t}}\Rightarrow{dy}=−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+\frac{\mathrm{1}}{{t}^{\mathrm{2}} \sqrt{\mathrm{1}−{t}^{\mathrm{2}} }} \\ $$$${t}^{\mathrm{2}} ={x} \\ $$$${A}\left({n}\right)={n}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{n}} \left(−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+\frac{\mathrm{1}}{{t}^{\mathrm{2}} \sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\right){dt}={n}\int_{\mathrm{0}} ^{\mathrm{1}} −{t}^{{n}−\mathrm{2}} {dt}+{n}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{n}−\mathrm{2}} }{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt} \\ $$$$=−\frac{{n}}{{n}−\mathrm{1}}+{n}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{n}−\mathrm{2}} }{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}=−\frac{{n}}{{n}−\mathrm{1}}+{n}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{\frac{{n}−\mathrm{2}}{\mathrm{2}}} }{\:\sqrt{\mathrm{1}−{u}}}.\frac{\mathrm{1}}{\mathrm{2}\sqrt{{u}}}{du} \\ $$$$=−\frac{{n}}{{n}−\mathrm{1}}+\frac{{n}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {u}^{\frac{{n}−\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {du}=−\frac{{n}}{{n}−\mathrm{1}}+\frac{{n}}{\mathrm{2}}\beta\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{{n}−\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=−\frac{{n}}{{n}−\mathrm{1}}+\frac{{n}}{\mathrm{2}}.\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{{n}−\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{{n}}{\mathrm{2}}\right)}={A}\left({n}\right);{n}>\mathrm{1} \\ $$$$\Gamma\left({z}\right)\sim\sqrt{\mathrm{2}\pi}{z}^{{z}−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−{z}} \\ $$$${i}\:{willfinish}\:{later}\:{inchelah} \\ $$$$ \\ $$$$ \\ $$

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