calculate-F-x-0-dt-1-1-x-1-t-2-2-

Question Number 39389 by maxmathsup by imad last updated on 05/Jul/18

c a l c u l a t e F (x) = \int_{0}^{\infty} \frac{d t}{1 + {(1 + x (1 + t^{2}))}^{2}}

Commented by prof Abdo imad last updated on 24/Jul/18

c h a n g e m e n t t = t a n θ g i v e

F (x) = \int_{0}^{\frac{π}{2}} \frac{(1 + {t a n}^{2} θ)}{1 + {1 + x (1 + {t a n}^{2} θ)}^{2}} d θ

= \int_{0}^{\frac{π}{2}} \frac{1}{{c o s}^{2} θ {1 + \frac{x}{{c o s}^{2} θ}}^{2}} d θ

= \int_{0}^{\frac{π}{2}} \frac{{c o s}^{2} θ}{{x + {c o s}^{2} θ}^{2}} d θ

= \int_{0}^{\frac{π}{2}} \frac{x + {c o s}^{2} θ - x}{{(x + {c o s}^{2} θ)}^{2}} d θ

= \int_{0}^{\frac{π}{2}} \frac{d θ}{x + {c o s}^{2} θ} - x \int_{0}^{\frac{π}{2}} \frac{d θ}{{(x + {c o s}^{2} θ)}^{2}} b u t

\int_{0}^{\frac{π}{2}} \frac{d θ}{x + {c o s}^{2} θ} = \int_{0}^{\frac{π}{2}} \frac{d θ}{x + \frac{1 + c o s (2 θ)}{2}}

= \int_{0}^{\frac{π}{2}} \frac{2 d θ}{2 x + 1 + c o s (2 θ)} =_{2 θ = t} \int_{0}^{π} \frac{d t}{2 x + 1 + c o s (t)}

=_{u = t a n (\frac{t}{2})} \int_{0}^{\infty} \frac{1}{2 x + 1 + \frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}

= \int_{0}^{\infty} \frac{2 d u}{(2 x + 1) u^{2} + 1 - u^{2}}

= \int_{0}^{\infty} \frac{2 d u}{2 {x u}^{2} + 1}

i f x > 0 \int_{0}^{\infty} \frac{2 d u}{1 + 2 {x u}^{2}} =_{\sqrt{2 x} u = α} \int_{0}^{\infty} \frac{2}{1 + α^{2}} \frac{d α}{\sqrt{2 x}}

= \frac{\sqrt{2}}{\sqrt{x}} . \frac{π}{2}

i f x < 0 \int_{0}^{\infty} \frac{2 d u}{1 + 2 {x u}^{2}} = \int_{0}^{\infty} \frac{2 d u}{1 - {(\sqrt{- 2 x} u)}^{2}}

= \int_{0}^{\infty} \frac{2 d u}{(1 - \sqrt{2 x} u) (1 + \sqrt{2 x} u)}

= \int_{0}^{\infty} {\frac{1}{1 - \sqrt{2 x} u} + \frac{1}{1 + \sqrt{2 x} u}} d u

= {[l n ∣ \frac{1 + \sqrt{2 x} u}{1 - \sqrt{2 x} u} ∣]}_{0}^{+ \infty} = 0 s o

\int_{0}^{\frac{π}{2}} \frac{d θ}{x + {c o s}^{2} θ} = \frac{π \sqrt{2}}{\sqrt{2 x}} i f x > 0

\int_{0}^{\frac{π}{2}} \frac{d θ}{x + {c o s}^{2} θ} = 0 i f x < 0

Commented by prof Abdo imad last updated on 24/Jul/18

l e t w (x) = \int_{0}^{\frac{π}{2}} \frac{d θ}{x + {c o s}^{2} θ} \Rightarrow

w^{^{'}} (x) = - \int_{0}^{\frac{π}{2}} \frac{d θ}{{(x + {c o s}^{2} θ)}^{2}} \Rightarrow

\int_{0}^{\frac{π}{2}} \frac{d θ}{{(x + {c o s}^{2} θ)}^{2}} = - w^{^{'}} (x) s o

i f x > 0 w (x) = \frac{π \sqrt{2}}{2 \sqrt{x}} \Rightarrow w^{^{'}} (x) = \frac{π}{\sqrt{2}} (- \frac{1}{2 x \sqrt{x}})

= - \frac{π}{2 \sqrt{2} x \sqrt{x}} \Rightarrow \int_{0}^{\frac{π}{2}} \frac{d θ}{{(x + {c o s}^{2} θ)}^{2}} = \frac{π}{2 \sqrt{2} x \sqrt{x}}

a n d F (x) = \frac{π \sqrt{2}}{2 \sqrt{x}} - x \frac{π}{2 \sqrt{2} x \sqrt{x}}

= \frac{π \sqrt{2}}{2 \sqrt{x}} - \frac{π}{2 \sqrt{2} \sqrt{x}} = \frac{π}{\sqrt{x}} {\frac{\sqrt{2}}{2} - \frac{1}{2 \sqrt{2}}} = \frac{π}{\sqrt{x}} {\frac{2}{4 \sqrt{2}}}

= \frac{π}{2 \sqrt{2 x}}

i f x < 0 w (x) = 0 \Rightarrow \int_{0}^{\frac{π}{2}} \frac{d θ}{{(x + {c o s}^{2} θ)}^{2}} = 0 \Rightarrow

F (x) = 0 b e c a u s e \int_{0}^{\frac{π}{2}} \frac{d θ}{x + {c o s}^{2} θ} = 0