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

find f(t) = ∫_0 ^1 arctan(tx^2 )dx with t≥0 developp f at integr serie

$${find}\:\:{f}\left({t}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{arctan}\left({tx}^{\mathrm{2}} \right){dx}\:{with}\:{t}\geqslant\mathrm{0} \\ $$$${developp}\:\:{f}\:{at}\:{integr}\:{serie} \\ $$

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

we have arctan^′ (x)= (1/(1+x^2 )) =Σ_(n=0) ^∞ (−1)^n x^(2n) ⇒ arctan(x)=Σ_(n=0) ^∞ (((−1)^n )/(2n+1))x^(2n+1) with radius of convergence R =1 so f(t) =∫_0 ^1 (Σ_(n=0) ^∞ (((−1)^n )/(2n+1))(tx^2 )^(2n+1) dx) =Σ_(n=0) ^∞ (((−1)^n )/(2n+1))t^(2n+1) ∫_0 ^1 x^(4n+2) dx =Σ_(n=0) ^∞ (((−1)^n )/((2n+1)(4n+3)))t^(2n+1) .

$${we}\:{have}\:\:{arctan}^{'} \left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}} \:\Rightarrow \\ $$$${arctan}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}{x}^{\mathrm{2}{n}+\mathrm{1}} \:{with}\:{radius}\:{of} \\ $$$${convergence}\:\:{R}\:=\mathrm{1}\:\:{so} \\ $$$${f}\left({t}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\left({tx}^{\mathrm{2}} \right)^{\mathrm{2}{n}+\mathrm{1}} {dx}\right) \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}{t}^{\mathrm{2}{n}+\mathrm{1}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{\mathrm{4}{n}+\mathrm{2}} {dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{4}{n}+\mathrm{3}\right)}{t}^{\mathrm{2}{n}+\mathrm{1}} \:\:. \\ $$

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