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Question Number 184027 by SEKRET last updated on 02/Jan/23

∫_0 ^( (𝛑/2)) e^(βˆ’tg^2 (x)) dx = ???

$$\:\:\: \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\boldsymbol{\mathrm{e}}^{βˆ’\boldsymbol{{tg}}^{\mathrm{2}} \left(\boldsymbol{{x}}\right)} \:\boldsymbol{{d}\mathrm{x}}\:=\:??? \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Answered by witcher3 last updated on 03/Jan/23

=∫_0 ^∞ (e^(βˆ’x^2 ) /((1+x^2 )))dxβ‰€βˆ«_0 ^∞ e^(βˆ’x^2 ) dx=((βˆšΟ€)/2)..cv ∫_0 ^∞ (e^(βˆ’ax^2 ) /((1+x^2 )))dx=F(a) fβ€²(a)βˆ’f(a)=βˆ’βˆ«_0 ^∞ e^(βˆ’ax^2 ) dx =∫_0 ^∞ e^(βˆ’(x(√a))^2 ) .dx=βˆ’((βˆšΟ€)/( 2(√a))) fβ€²βˆ’f=βˆ’((βˆšΟ€)/(2(√a))) f(a)=ke^a kβ€²=βˆ’((βˆšΟ€)/2)(e^(βˆ’a) /( (√a)))β‡’k=βˆ’((βˆšΟ€)/2)∫ (e^(βˆ’a) /( (√a)))da=βˆ’(βˆšΟ€)∫e^(βˆ’t^2 ) dt =(Ο€/2)erfc(t)+c,c∈R f(a)=(Ο€/2)erfc((√a))e^(βˆ’a) +ce^(βˆ’a) f(0)=(Ο€/2)⇔c=0 f(a)=∫_0 ^∞ e^(βˆ’atg^2 (x)) dx=(Ο€/2)erfc((√a))e^(βˆ’a) f(1)=∫_0 ^∞ e^(βˆ’tg^2 (x)) dx=(Ο€/(2e))erfc(1)

$$=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{βˆ’{x}^{\mathrm{2}} } }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}\leqslant\int_{\mathrm{0}} ^{\infty} {e}^{βˆ’{x}^{\mathrm{2}} } {dx}=\frac{\sqrt{\pi}}{\mathrm{2}}..{cv} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{e}^{βˆ’{ax}^{\mathrm{2}} } }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}={F}\left({a}\right) \\ $$$${f}'\left({a}\right)βˆ’{f}\left({a}\right)=βˆ’\int_{\mathrm{0}} ^{\infty} {e}^{βˆ’{ax}^{\mathrm{2}} } {dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} {e}^{βˆ’\left({x}\sqrt{{a}}\right)^{\mathrm{2}} } .{dx}=βˆ’\frac{\sqrt{\pi}}{\:\mathrm{2}\sqrt{{a}}} \\ $$$${f}'βˆ’{f}=βˆ’\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{{a}}} \\ $$$${f}\left({a}\right)={ke}^{{a}} \\ $$$${k}'=βˆ’\frac{\sqrt{\pi}}{\mathrm{2}}\frac{{e}^{βˆ’{a}} }{\:\sqrt{{a}}}\Rightarrow{k}=βˆ’\frac{\sqrt{\pi}}{\mathrm{2}}\int\:\frac{{e}^{βˆ’{a}} }{\:\sqrt{{a}}}{da}=βˆ’\sqrt{\pi}\int{e}^{βˆ’{t}^{\mathrm{2}} } {dt} \\ $$$$=\frac{\pi}{\mathrm{2}}{erfc}\left({t}\right)+{c},{c}\in\mathbb{R} \\ $$$${f}\left({a}\right)=\frac{\pi}{\mathrm{2}}{erfc}\left(\sqrt{{a}}\right){e}^{βˆ’{a}} +{ce}^{βˆ’{a}} \\ $$$${f}\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{2}}\Leftrightarrow{c}=\mathrm{0} \\ $$$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{βˆ’{atg}^{\mathrm{2}} \left({x}\right)} {dx}=\frac{\pi}{\mathrm{2}}{erfc}\left(\sqrt{{a}}\right){e}^{βˆ’{a}} \\ $$$${f}\left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{βˆ’{tg}^{\mathrm{2}} \left({x}\right)} {dx}=\frac{\pi}{\mathrm{2}{e}}{erfc}\left(\mathrm{1}\right) \\ $$

Commented by SEKRET last updated on 05/Jan/23

thanks sir. beatiful solution

$$\boldsymbol{\mathrm{thanks}}\:\:\boldsymbol{\mathrm{sir}}.\:\boldsymbol{\mathrm{beatiful}}\:\boldsymbol{\mathrm{solution}} \\ $$

Commented by witcher3 last updated on 05/Jan/23

withe pleasur

$${withe}\:{pleasur} \\ $$

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