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Question Number 124979 by mathmax by abdo last updated on 07/Dec/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\arctan \left({\mathrm{x}}^{\mathrm{n}}\right)\mathrm{with}\mathrm{n}\mathrm{natural}$$$$1)\mathrm{find}{\mathrm{f}}^{\left(\mathrm{n}\right)}\left(0\right)\mathrm{and}{\mathrm{f}}^{\left(\mathrm{n}\right)}\left(1\right)$$$$2)\mathrm{developp}\mathrm{f}\mathrm{at}\mathrm{integr}\mathrm{serie}$$$$3)\mathrm{calculte}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{f}\left(\mathrm{x}\right)}{{\mathrm{x}}^{\mathrm{n}}}\mathrm{dx}\mathrm{with}\mathrm{n}⩾2$$

Answered by Dwaipayan Shikari last updated on 07/Dec/20

$$f\left(x\right)={tan}^{-1}{x}^n$$$${\int }_0^{\mathrm{\infty }}\frac{{tan}^{-1}{x}^n}{x^n}dx=-{\left[n\frac{{tan}^{-1}{x}^n}{x^{n-1}}\right]}_0^{\mathrm{\infty }}+n^2{\int }_0^{\mathrm{\infty }}{x}^{1-n}\frac{x^{n-1}}{1+x^{2n}}dx$$$$=n^2{\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^{2n}}{x}^{2n}=t\Rightarrow 2n{x}^{2n-1}=\frac{dt}{dx}$$$$=\frac{n}{2}{\int }_0^{\mathrm{\infty }}{x}^{1-2n}\frac{1}{1+t}dt=\frac{n}{2}{\int }_0^{\mathrm{\infty }}{t}^{\frac{1}{2n}-1}\frac{1}{1+t}dt\frac{t}{1+t}=u\Rightarrow \frac{1}{{(1+t)}^2}=\frac{du}{dt}$$$$=\frac{n}{2}{\int }_0^1\frac{u^{\frac{1}{2n}-1}}{{(1-u)}^{\frac{1}{2n}-1}}{(1-u)}^{-1}du\frac{u}{1-u}=t$$$$=\frac{n}{2}{\int }_0^1{u}^{\frac{1}{2n}-1}{(1-u)}^{-\frac{1}{2n}}du=\frac{n}{2}\mathrm{\Gamma }\left(\frac{1}{2n}\right)\mathrm{\Gamma }(1-\frac{1}{2n})=\frac{n}{2}.\frac{\pi }{sin\left(\frac{\pi }{2n}\right)}$$

Answered by mathmax by abdo last updated on 07/Dec/20

$$1)\mathrm{f}\left(\mathrm{x}\right)=\arctan \left({\mathrm{x}}^{\mathrm{n}}\right)\Rightarrow {\mathrm{f}}^{\left(1\right)}\left(\mathrm{x}\right)=\frac{{\mathrm{nx}}^{\mathrm{n}-1}}{1+{\mathrm{x}}^{2\mathrm{n}}}\Rightarrow$$$${\mathrm{f}}^{\left(\mathrm{p}\right)}\left(\mathrm{x}\right)=\mathrm{n}{\left(\frac{{\mathrm{x}}^{\mathrm{n}-1}}{{\mathrm{x}}^{2\mathrm{n}}+1}\right)}^{(\mathrm{p}-1)}\mathrm{with}\mathrm{p}⩾1$$$$\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^{\mathrm{n}-1}}{{\mathrm{x}}^{2\mathrm{n}}+1}$$$${\mathrm{z}}^{2\mathrm{n}}+1=0\Rightarrow {\mathrm{z}}^{2\mathrm{n}}={\mathrm{e}}^{\mathrm{i}(2\mathrm{k}+1)\pi }\Rightarrow {\mathrm{z}}_{\mathrm{k}}={\mathrm{e}}^{\mathrm{i}\frac{(2\mathrm{k}+1)\pi }{2\mathrm{n}}}\mathrm{and}\mathrm{k}\in \left[[0,2\mathrm{n}-1]\right]\Rightarrow$$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^{\mathrm{n}-1}}{\prod _{\mathrm{k}=0}^{2\mathrm{n}-1}(\mathrm{x}-{\mathrm{z}}_{\mathrm{k}})}=\sum _{\mathrm{k}=0}^{2\mathrm{n}-1}\frac{{\mathrm{a}}_{\mathrm{k}}}{\mathrm{x}-{\mathrm{z}}_{\mathrm{k}}}\mathrm{we}\mathrm{have}$$$${\mathrm{a}}_{\mathrm{k}}=\frac{{\mathrm{z}}_{\mathrm{k}}^{\mathrm{n}-1}}{2\mathrm{n}{\mathrm{z}}_{\mathrm{k}}^{2\mathrm{n}-1}}=\frac{1}{2\mathrm{n}}\times \frac{{\mathrm{z}}_{\mathrm{k}}^{\mathrm{n}}}{(-1)}=-\frac{1}{2\mathrm{n}}{\mathrm{z}}_{\mathrm{k}}^{\mathrm{n}}=-\frac{1}{2\mathrm{n}}\times {\mathrm{e}}^{\mathrm{i}\frac{(2\mathrm{k}+1)\pi }{2}}$$$$=-\frac{1}{2\mathrm{n}}{\mathrm{e}}^{\mathrm{ik}\pi }{\mathrm{e}}^{\frac{\mathrm{i}\pi }{2}}=-\frac{1}{2\mathrm{n}}{(-1)}^{\mathrm{k}}\mathrm{i}=\frac{\mathrm{i}{(-1)}^{\mathrm{k}}}{2\mathrm{n}}\Rightarrow \mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{i}}{2\mathrm{n}}\sum _{\mathrm{k}=0}^{2\mathrm{n}-1}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{x}-{\mathrm{z}}_{\mathrm{k}}}\Rightarrow$$$${\mathrm{f}}^{\left(\mathrm{p}\right)}\left(\mathrm{x}\right)=\frac{\mathrm{i}}{2}\sum _{\mathrm{k}=0}^{2\mathrm{n}-1}{(-1)}^{\mathrm{k}}{\left\{\frac{1}{\mathrm{x}-{\mathrm{z}}_{\mathrm{k}}}\right\}}^{(\mathrm{p}-1)}$$$$=\frac{\mathrm{i}}{2}\sum _{\mathrm{k}=0}^{2\mathrm{n}-1}{(-1)}^{\mathrm{k}}\frac{{(-1)}^{\mathrm{p}-1}(\mathrm{p}-1)!}{{(\mathrm{x}-{\mathrm{z}}_{\mathrm{k}})}^{\mathrm{p}}}\Rightarrow$$$${\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\mathrm{x}\right)=\frac{\mathrm{i}}{2}\sum _{\mathrm{k}=0}^{2\mathrm{n}-1}{(-1)}^{\mathrm{k}}\times \frac{{(-1)}^{\mathrm{n}-1}(\mathrm{n}-1)!}{{(\mathrm{x}-{\mathrm{z}}_{\mathrm{k}})}^{\mathrm{n}}}\Rightarrow$$$${\mathrm{f}}^{\left(\mathrm{n}\right)}\left(0\right)=\frac{\mathrm{i}{(-1)}^{\mathrm{n}-1}}{2}(\mathrm{n}-1)!\sum _{\mathrm{k}=0}^{2\mathrm{n}-1}\frac{{(-1)}^{\mathrm{k}}}{{(-{\mathrm{z}}_{\mathrm{k}})}^{\mathrm{n}}}$$$$=-\frac{\mathrm{i}}{2}(\mathrm{n}-1)!\sum _{\mathrm{k}=0}^{2\mathrm{n}-1}\frac{{(-1)}^{\mathrm{k}}}{{\mathrm{z}}_{\mathrm{k}}^{\mathrm{n}}}$$$${\mathrm{f}}^{\left(\mathrm{n}\right)}\left(1\right)=\frac{\mathrm{i}}{2}\sum _{\mathrm{k}=0}^{2\mathrm{n}-1}{(-1)}^{\mathrm{k}}\frac{{(-1)}^{\mathrm{n}-1}(\mathrm{n}-1)!}{{(1-{\mathrm{z}}_{\mathrm{k}})}^{\mathrm{n}}}$$

Answered by mathmax by abdo last updated on 07/Dec/20

$$2)\mathrm{f}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{f}}^{\left(\mathrm{n}\right)}\left(0\right)}{\mathrm{n}!}{\mathrm{x}}^{\mathrm{n}}\mathrm{and}{\mathrm{f}}^{)\mathrm{n})}\left(0\right)\mathrm{is}\mathrm{known}$$

Answered by mathmax by abdo last updated on 07/Dec/20

$$3){\mathrm{A}}_{\mathrm{n}}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{f}\left(\mathrm{x}\right)}{{\mathrm{x}}^{\mathrm{n}}}\mathrm{dx}\Rightarrow {\mathrm{A}}_{\mathrm{n}}={\int }_0^{\mathrm{\infty }}\frac{\arctan \left({\mathrm{x}}^{\mathrm{n}}\right)}{{\mathrm{x}}^{\mathrm{n}}}\mathrm{dx}={\int }_0^{\mathrm{\infty }}{\mathrm{x}}^{-\mathrm{n}}\arctan \left({\mathrm{x}}^{\mathrm{n}}\right)\mathrm{dx}$$$$\mathrm{by}\mathrm{psrts}{\mathrm{A}}_{\mathrm{n}}={\left[\frac{1}{1-\mathrm{n}}{\mathrm{x}}^{1-\mathrm{n}}\arctan \left({\mathrm{x}}^{\mathrm{n}}\right)\right]}_0^{\mathrm{\infty }}-{\int }_0^{\mathrm{\infty }}\frac{1}{1-\mathrm{n}}{\mathrm{x}}^{1-\mathrm{n}}\times \frac{{\mathrm{nx}}^{\mathrm{n}-1}}{1+{\mathrm{x}}^{2\mathrm{n}}}\mathrm{dx}$$$$=\frac{\mathrm{n}}{\mathrm{n}-1}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{1+{\mathrm{x}}^{2\mathrm{n}}}\mathrm{but}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{1+{\mathrm{x}}^{2\mathrm{n}}}{=}_{{\mathrm{x}}^{2\mathrm{n}}=\mathrm{t}\to \mathrm{x}={\mathrm{t}}^{\frac{1}{2\mathrm{n}}}}{\int }_0^{\mathrm{\infty }}\frac{1}{2\mathrm{n}}{\mathrm{t}}^{\frac{1}{2\mathrm{n}}-1}\frac{\mathrm{dt}}{1+\mathrm{t}}$$$$=\frac{1}{2\mathrm{n}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1}{2\mathrm{n}}-1}}{1+\mathrm{t}}\mathrm{dt}=\frac{1}{2\mathrm{n}}\times \frac{\pi }{\sin \left(\frac{\pi }{2\mathrm{n}}\right)}\Rightarrow {\mathrm{A}}_{\mathrm{n}}=\frac{\mathrm{n}}{\mathrm{n}-1}\times \frac{\pi }{2\mathrm{nsin}\left(\frac{\pi }{2\mathrm{n}}\right)}$$$$\Rightarrow {\mathrm{A}}_{\mathrm{n}}=\frac{\pi }{2(\mathrm{n}-1)\sin \left(\frac{\pi }{2\mathrm{n}}\right)}(\mathrm{n}⩾2)$$

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