Question Number 40984 by prof Abdo imad last updated on 30/Jul/18
$${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:{te}^{−{t}^{\mathrm{2}} } \:{arctan}\left({xt}\right){dt} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{simple}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:{te}^{−{t}^{\mathrm{2}} } {arctantdt}\:{and} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:{t}\:{e}^{−{t}^{\mathrm{2}} } \:{arctan}\left(\mathrm{2}{t}\right){dt} \\ $$$$\left.\mathrm{3}\right){let}\:{u}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:{t}\:{e}^{−{t}^{\mathrm{2}} } {arctan}\left({nt}\right){dt} \\ $$$${find}\:{lim}_{{n}\rightarrow+\infty} {u}_{{n}} \\ $$$${study}\:{the}\:{convergence}\:{of}\:\Sigma\:{u}_{{n}} \\ $$
Commented by math khazana by abdo last updated on 02/Aug/18
![1) by parts f(x)=[−(1/2)e^(−t^2 ) arctan(xt)]_(t=0) ^∞ −∫_0 ^∞ −(1/2) e^(−t^2 ) (x/(1+x^2 t^2 )) dt =(x/2) ∫_0 ^∞ (e^(−t^2 ) /(x^2 t^2 +1)) dt =_(xt =u) (x/2) ∫_0 ^∞ (e^(−(u^2 /x^2 )) /(1+u^2 )) (du/x) =(1/2) ∫_0 ^∞ (e^(−(u^2 /x^2 )) /(1+u^2 )) du =(1/4) ∫_(−∞) ^(+∞) (e^(−(u^2 /x^2 )) /(1+u^2 )) du let ϕ(z) = (e^(−(u^2 /x^2 )) /(1+u^2 )) the poles of ϕ are i and −i residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) =2iπ (e^(1/x^2 ) /(2i)) =π e^(1/x^2 ) ⇒ f(x)=(π/4) e^(1/(x^2 )) with x≠0 2) ∫_0 ^∞ t e^(−t^2 ) arctan(t)dt =f(1)= ((πe)/4) 3) we have u_n =f(n) = (π/4) e^(1/n^2 ) ⇒ lim_(n→+∞) u_n =(π/4) Σ_(n=1) ^∞ u_n =(π/4) Σ_(n=1) ^∞ e^(1/n^2 ) but e^(1/n^2 ) ∼ 1+(1/n^2 ) ⇒ (π/4) Σ_(n=1) ^∞ e^(1/n^2 ) ∼(π/4)Σ_(n=1) ^∞ (1+(1/n^2 )) =+∞⇒ Σ u_n diverges .](https://www.tinkutara.com/question/Q41129.png)
$$\left.\mathrm{1}\right)\:{by}\:{parts}\:{f}\left({x}\right)=\left[−\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{t}^{\mathrm{2}} } {arctan}\left({xt}\right)\right]_{{t}=\mathrm{0}} ^{\infty} \\ $$$$−\int_{\mathrm{0}} ^{\infty} \:−\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{−{t}^{\mathrm{2}} } \:\:\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{2}} }\:{dt} \\ $$$$=\frac{{x}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{t}^{\mathrm{2}} } }{{x}^{\mathrm{2}} {t}^{\mathrm{2}} \:+\mathrm{1}}\:{dt}\:=_{{xt}\:={u}} \:\:\:\:\frac{{x}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−\frac{{u}^{\mathrm{2}} }{{x}^{\mathrm{2}} }} }{\mathrm{1}+{u}^{\mathrm{2}} }\:\frac{{du}}{{x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−\frac{{u}^{\mathrm{2}} }{{x}^{\mathrm{2}} }} }{\mathrm{1}+{u}^{\mathrm{2}} }\:{du}\:\:=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{e}^{−\frac{{u}^{\mathrm{2}} }{{x}^{\mathrm{2}} }} }{\mathrm{1}+{u}^{\mathrm{2}} }\:{du}\:{let} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{e}^{−\frac{{u}^{\mathrm{2}} }{{x}^{\mathrm{2}} }} }{\mathrm{1}+{u}^{\mathrm{2}} }\:\:{the}\:{poles}\:{of}\:\varphi\:{are}\:{i}\:{and}\:−{i}\: \\ $$$${residus}\:{theorem}\:{give}\: \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right)\:=\mathrm{2}{i}\pi\:\frac{{e}^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} }{\mathrm{2}{i}}\:=\pi\:{e}^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{\pi}{\mathrm{4}}\:{e}^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:\:}} \:\:\:\:\:\:{with}\:{x}\neq\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\infty} \:{t}\:{e}^{−{t}^{\mathrm{2}} } \:{arctan}\left({t}\right){dt}\:={f}\left(\mathrm{1}\right)=\:\frac{\pi{e}}{\mathrm{4}} \\ $$$$\left.\mathrm{3}\right)\:{we}\:{have}\:{u}_{{n}} ={f}\left({n}\right)\:=\:\frac{\pi}{\mathrm{4}}\:{e}^{\frac{\mathrm{1}}{{n}^{\mathrm{2}} }} \:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:\:{u}_{{n}} =\frac{\pi}{\mathrm{4}} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:{u}_{{n}} \:=\frac{\pi}{\mathrm{4}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:{e}^{\frac{\mathrm{1}}{{n}^{\mathrm{2}} }} \:\:{but}\:{e}^{\frac{\mathrm{1}}{{n}^{\mathrm{2}} }} \:\sim\:\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\frac{\pi}{\mathrm{4}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:{e}^{\frac{\mathrm{1}}{{n}^{\mathrm{2}} }} \:\sim\frac{\pi}{\mathrm{4}}\sum_{{n}=\mathrm{1}} ^{\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)\:=+\infty\Rightarrow \\ $$$$\Sigma\:{u}_{{n}} \:{diverges}\:. \\ $$$$ \\ $$