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

Answered by Berbere last updated on 19/Apr/24

$$\frac{x}{n}=yA\left(n\right)={\int }_0^n{\left(\frac{2nx}{x^2+n^2}\right)}^{n}dx=n{\int }_0^1{\left(\frac{2y}{1+y^2}\right)}^{n}dy$$$${\int }_0^1{\left(\frac{2y}{1+y^2}\right)}^n$$$$\frac{2\mathit{y}}{1+{\mathit{y}}^2}=\mathit{t}\Rightarrow {ty}^2+t-2y=0$$$$y=\frac{2-\sqrt{4-4t^2}}{2t}=\frac{1-\sqrt{1-t^2}}{t}\Rightarrow dy=-\frac{1}{t^2}+\frac{1}{t^2\sqrt{1-t^2}}$$$$t^2=x$$$$A\left(n\right)=n{\int }_0^1{t}^n(-\frac{1}{t^2}+\frac{1}{t^2\sqrt{1-t^2}})dt=n{\int }_0^1-t^{n-2}dt+n{\int }_0^1\frac{t^{n-2}}{\sqrt{1-t^2}}dt$$$$=-\frac{n}{n-1}+n{\int }_0^1\frac{t^{n-2}}{\sqrt{1-t^2}}dt=-\frac{n}{n-1}+n{\int }_0^1\frac{u^{\frac{n-2}{2}}}{\sqrt{1-u}}.\frac{1}{2\sqrt{u}}du$$$$=-\frac{n}{n-1}+\frac{n}{2}{\int }_0^1{(1-u)}^{\frac{1}{2}-1}{u}^{\frac{n-1}{2}-1}du=-\frac{n}{n-1}+\frac{n}{2}\beta (\frac{1}{2},\frac{n-1}{2})$$$$=-\frac{n}{n-1}+\frac{n}{2}.\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }\left(\frac{n-1}{2}\right)}{\mathrm{\Gamma }\left(\frac{n}{2}\right)}=A\left(n\right);n>1$$$$\mathrm{\Gamma }\left(z\right)\sim \sqrt{2\pi }{z}^{z-\frac{1}{2}}{e}^{-z}$$$$iwillfinishlaterinchelah$$$$$$$$$$

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