0-pi-2-e-sec-2-d-

Question Number 100026 by Ar Brandon last updated on 24/Jun/20

\int_{0}^{\frac{π}{2}} e^{- \sec^{2} θ} d θ

Answered by mathmax by abdo last updated on 24/Jun/20

I = \int_{0}^{\frac{π}{2}} e^{- \frac{1}{\cos^{2} θ}} d θ we have 1 + \tan^{2} θ = \frac{1}{\cos^{2} θ} \Rightarrow I = \int_{0}^{\frac{π}{2}} e^{- (1 + \tan^{2} θ)} d θ

changement \tan θ = x give I = \int_{0}^{\infty} e^{- (1 + x^{2})} \frac{dx}{1 + x^{2}}

= \int_{0}^{\infty} \frac{e^{- (1 + x^{2})}}{1 + x^{2}} dx let f (a) = \int_{0}^{\infty} \frac{e^{- a (1 + x^{2})}}{1 + x^{2}} dx (a > 0) \Rightarrow

f^{^{'}} (a) = - \int_{0}^{\infty} e^{- a (1 + x^{2})} dx = - e^{- a} \int_{0}^{\infty} e^{- {ax}^{2}} dx =_{x \sqrt{a} = u} - e^{- a} \int_{0}^{\infty} e^{- u^{2}} \frac{du}{\sqrt{a}}

= - \frac{e^{- a}}{\sqrt{a}} \int_{0}^{\infty} e^{- u^{2}} du = - \frac{\sqrt{π}}{2} \times \frac{e^{- a}}{\sqrt{a}} \Rightarrow f (a) = - \frac{\sqrt{π}}{2} \int_{0}^{a} \frac{e^{- t}}{\sqrt{t}} dt + c

=_{\sqrt{t} = α} - \frac{\sqrt{π}}{2} \int_{0}^{\sqrt{a}} \frac{e^{- α^{2}}}{α} (2 α) d α + c = - \sqrt{π} \int_{0}^{\sqrt{a}} e^{- α^{2}} d α + c

c = \lim_{a \to 0} f (a) = \frac{π}{2} \Rightarrow f (a) = \frac{π}{2} - \sqrt{π} \int_{0}^{\sqrt{a}} e^{- α^{2}} d α \Rightarrow

I = f (1) = \frac{π}{2} - \sqrt{π} \int_{0}^{1} e^{- α^{2}} d α but e^{- α^{2}} = \sum_{n = 0}^{\infty} \frac{{(- α^{2})}^{n}}{n!} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n!} α^{2 n} \Rightarrow

I = \frac{π}{2} - \sqrt{π} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n!} \int_{0}^{1} α^{2 n} d α

I = \frac{π}{2} - \sqrt{π} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1) (n!)}

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