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Question Number 33885 by math khazana by abdo last updated on 26/Apr/18
developp at integr serie f(x)= ∫_0 ^x sin(t^2 )dt .
$${developp}\:{at}\:{integr}\:{serie}\:{f}\left({x}\right)=\:\int_{\mathrm{0}} ^{{x}} {sin}\left({t}^{\mathrm{2}} \right){dt}\:. \\ $$
Commented by prof Abdo imad last updated on 27/Apr/18
we know that sin(u) = Σ_(n=0) ^∞ (((−1)^n u^(2n+1) )/((2n+1)!)) let find the radius of convergence ∣(u_(n+1) /u_n )∣=(((2n+1)!)/((2n+3)!)) = (((2n+1)!)/((2n+3)(2n+2)(2n+1)!)) = (1/((2n+3)(2n+2))) →0 (n→+∞) so R=+∞ ⇒ sin(t^2 )= Σ_(n=0) ^∞ (((−1)^n t^(4n+2) )/((2n+1)!)) ⇒ f(x)=∫_0 ^x ( Σ_(n=0) ^∞ (((−1)^n )/((2n+1)!))t^(4n+2) )dt =Σ_(n=0) ^∞ (((−1)^n )/((2n+1)!)) ∫_0 ^x t^(4n+2) dt f(x)= Σ_(n=0) ^∞ (((−1)^n )/((4n+3)(2n+1)!))x^(4n+3) .
$${we}\:{know}\:{that}\:{sin}\left({u}\right)\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} {u}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$${let}\:{find}\:{the}\:{radius}\:{of}\:{convergence} \\ $$$$\mid\frac{{u}_{{n}+\mathrm{1}} }{{u}_{{n}} }\mid=\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{\left(\mathrm{2}{n}+\mathrm{3}\right)!}\:=\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{\left(\mathrm{2}{n}+\mathrm{3}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$$=\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{3}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)}\:\rightarrow\mathrm{0}\:\left({n}\rightarrow+\infty\right)\:{so}\:{R}=+\infty \\ $$$$\Rightarrow\:{sin}\left({t}^{\mathrm{2}} \right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} \:{t}^{\mathrm{4}{n}+\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:\Rightarrow \\ $$$${f}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \left(\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{t}^{\mathrm{4}{n}+\mathrm{2}} \right){dt} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:\int_{\mathrm{0}} ^{{x}} \:{t}^{\mathrm{4}{n}+\mathrm{2}} {dt} \\ $$$${f}\left({x}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{4}{n}+\mathrm{3}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)!}{x}^{\mathrm{4}{n}+\mathrm{3}} \:. \\ $$$$ \\ $$

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