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Question Number 36335 by prof Abdo imad last updated on 31/May/18

$$findf\left(t\right)={\int }_0^{1}arctan\left({tx}^2\right)dxwitht⩾0$$$$developpfatintegrserie$$

Commented by prof Abdo imad last updated on 01/Jun/18

$$wehave{arctan}^{{}^{\prime }}\left(x\right)=\frac{1}{1+x^2}=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{x}^{2n}\Rightarrow$$$$arctan\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{2n+1}{x}^{2n+1}withradiusof$$$$convergenceR=1so$$$$f\left(t\right)={\int }_0^1\left(\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{2n+1}{\left({tx}^2\right)}^{2n+1}dx\right)$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{2n+1}{t}^{2n+1}{\int }_0^1{x}^{4n+2}dx$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{(2n+1)(4n+3)}{t}^{2n+1}.$$

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