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Question Number 100026 by Ar Brandon last updated on 24/Jun/20
∫_0 ^(π/2) e^(−sec^2 θ) dθ
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{e}^{−\mathrm{sec}^{\mathrm{2}} \theta} \mathrm{d}\theta \\ $$
Answered by mathmax by abdo last updated on 24/Jun/20
I =∫_0 ^(π/(2 )) e^(−(1/(cos^2 θ))) dθ we have 1+tan^2 θ =(1/(cos^2 θ)) ⇒ I =∫_0 ^(π/2) e^(−(1+tan^2 θ)) dθ changement tanθ =x give I =∫_0 ^∞ e^(−(1+x^2 )) (dx/(1+x^2 )) = ∫_0 ^∞ (e^(−(1+x^2 )) /(1+x^2 ))dx let f(a) =∫_0 ^∞ (e^(−a(1+x^2 )) /(1+x^2 ))dx (a>0) ⇒ f^′ (a) =−∫_0 ^∞ e^(−a(1+x^2 )) dx =−e^(−a) ∫_0 ^∞ e^(−ax^2 ) dx =_(x(√a)=u) −e^(−a) ∫_0 ^∞ e^(−u^2 ) (du/( (√a))) =−(e^(−a) /( (√a))) ∫_0 ^∞ e^(−u^2 ) du =−((√π)/2) ×(e^(−a) /( (√a))) ⇒f(a) =−((√π)/2) ∫_0 ^a (e^(−t) /( (√t)))dt +c =_((√t)=α) −((√π)/2) ∫_0 ^(√a) (e^(−α^2 ) /α)(2α)dα +c =−(√π)∫_0 ^(√a) e^(−α^2 ) dα +c c=lim_(a→0) f(a) =(π/2) ⇒f(a) =(π/2)−(√π)∫_0 ^(√a) e^(−α^2 ) dα ⇒ I =f(1) =(π/2)−(√π)∫_0 ^1 e^(−α^2 ) dα but e^(−α^2 ) =Σ_(n=0) ^∞ (((−α^2 )^n )/(n!)) =Σ_(n=0) ^∞ (((−1)^n )/(n!)) α^(2n) ⇒ I =(π/2)−(√π)Σ_(n=0) ^∞ (((−1)^n )/(n!)) ∫_0 ^1 α^(2n) dα I=(π/2)−(√π)Σ_(n=0) ^∞ (((−1)^n )/((2n+1)(n!)))
$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}\:}} \:\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \theta}} \:\:\mathrm{d}\theta\:\:\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta\:=\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \theta}\:\Rightarrow\:\mathrm{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{e}^{−\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta\right)} \:\mathrm{d}\theta \\ $$$$\mathrm{changement}\:\mathrm{tan}\theta\:=\mathrm{x}\:\:\mathrm{give}\:\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{e}^{−\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)} \:\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{e}^{−\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:\:\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{e}^{−\mathrm{a}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:\:\:\:\:\left(\mathrm{a}>\mathrm{0}\right)\:\Rightarrow \\ $$$$\mathrm{f}^{'} \left(\mathrm{a}\right)\:=−\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{a}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)} \mathrm{dx}\:=−\mathrm{e}^{−\mathrm{a}} \:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{ax}^{\mathrm{2}} } \mathrm{dx}\:=_{\mathrm{x}\sqrt{\mathrm{a}}=\mathrm{u}} \:\:−\mathrm{e}^{−\mathrm{a}} \:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{u}^{\mathrm{2}} } \frac{\mathrm{du}}{\:\sqrt{\mathrm{a}}} \\ $$$$=−\frac{\mathrm{e}^{−\mathrm{a}} }{\:\sqrt{\mathrm{a}}}\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{u}^{\mathrm{2}} } \:\mathrm{du}\:=−\frac{\sqrt{\pi}}{\mathrm{2}}\:×\frac{\mathrm{e}^{−\mathrm{a}} }{\:\sqrt{\mathrm{a}}}\:\Rightarrow\mathrm{f}\left(\mathrm{a}\right)\:=−\frac{\sqrt{\pi}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{a}} \:\frac{\mathrm{e}^{−\mathrm{t}} }{\:\sqrt{\mathrm{t}}}\mathrm{dt}\:+\mathrm{c} \\ $$$$=_{\sqrt{\mathrm{t}}=\alpha} \:−\frac{\sqrt{\pi}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{a}}} \:\frac{\mathrm{e}^{−\alpha^{\mathrm{2}} } }{\alpha}\left(\mathrm{2}\alpha\right)\mathrm{d}\alpha\:+\mathrm{c}\:=−\sqrt{\pi}\int_{\mathrm{0}} ^{\sqrt{\mathrm{a}}} \:\mathrm{e}^{−\alpha^{\mathrm{2}} } \mathrm{d}\alpha\:+\mathrm{c} \\ $$$$\mathrm{c}=\mathrm{lim}_{\mathrm{a}\rightarrow\mathrm{0}} \:\:\mathrm{f}\left(\mathrm{a}\right)\:=\frac{\pi}{\mathrm{2}}\:\Rightarrow\mathrm{f}\left(\mathrm{a}\right)\:=\frac{\pi}{\mathrm{2}}−\sqrt{\pi}\int_{\mathrm{0}} ^{\sqrt{\mathrm{a}}} \mathrm{e}^{−\alpha^{\mathrm{2}} } \:\mathrm{d}\alpha\:\Rightarrow \\ $$$$\mathrm{I}\:=\mathrm{f}\left(\mathrm{1}\right)\:=\frac{\pi}{\mathrm{2}}−\sqrt{\pi}\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{e}^{−\alpha^{\mathrm{2}} } \mathrm{d}\alpha\:\:\mathrm{but}\:\mathrm{e}^{−\alpha^{\mathrm{2}} } \:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\alpha^{\mathrm{2}} \right)^{\mathrm{n}} }{\mathrm{n}!}\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}!}\:\alpha^{\mathrm{2n}} \:\Rightarrow \\ $$$$\mathrm{I}\:=\frac{\pi}{\mathrm{2}}−\sqrt{\pi}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}!}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\alpha^{\mathrm{2n}} \:\mathrm{d}\alpha \\ $$$$\mathrm{I}=\frac{\pi}{\mathrm{2}}−\sqrt{\pi}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}+\mathrm{1}\right)\left(\mathrm{n}!\right)} \\ $$
Commented by Ar Brandon last updated on 24/Jun/20
wow ! thank you
Commented by mathmax by abdo last updated on 24/Jun/20
you are welcome
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$

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