let-x-gt-0-find-F-x-arctan-xt-2-1-t-2-dt-

Question Number 38121 by maxmathsup by imad last updated on 22/Jun/18

l e t x > 0 f i n d F (x) = \int_{- \infty}^{+ \infty} \frac{a r c t a n ({x t}^{2})}{1 + t^{2}} d t

Commented by prof Abdo imad last updated on 23/Jun/18

w e h a v e f^{^{'}} (x) = \int_{- \infty}^{+ \infty} \frac{t^{2}}{(1 + t^{2}) (1 + x^{2} t^{4})} d t

l e t c o n s i d e r t h e c o m p l e x f u n c t i o n

φ (z) = \frac{z^{2}}{(z^{2} + 1) (x^{2} z^{4} + 1)} . p o l e s o f φ ?

φ (z) = \frac{z^{2}}{x^{2} (z^{2} + 1) (z^{4} + \frac{1}{x^{2}})}

= \frac{z^{2}}{x^{2} (z - i) (z + i) (z^{2} - \frac{i}{x}) (z^{2} + \frac{i}{x})}

= \frac{z^{2}}{x^{2} (z - i) (z + i) (z - \frac{1}{\sqrt{x}} e^{\frac{i π}{4}}) (z + \frac{1}{\sqrt{x}} e^{\frac{i π}{4}}) (z - \frac{1}{\sqrt{x}} e^{- \frac{i π}{4}}) (z + \frac{1}{\sqrt{x}} e^{- \frac{i π}{4}})}

s o t h e p o l e s o f φ a r e \bar{+} i, \bar{+} \frac{1}{\sqrt{x}} e^{\frac{i π}{4}}, \bar{+} \frac{1}{\sqrt{x}} e^{- \frac{i π}{4}}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, i) + R e s (φ, \frac{1}{\sqrt{x}} e^{\frac{i π}{4}}) + R e s (φ, - \frac{1}{\sqrt{x}} e^{- \frac{i π}{4}})}

R e s (φ, i) = \frac{- 1}{2 {i x}^{2} (1 + \frac{1}{x^{2}})} = \frac{- 1}{2 i (x^{2} + 1)}

Commented by prof Abdo imad last updated on 23/Jun/18

R e s (φ, \frac{1}{\sqrt{x}} e^{i \frac{π}{4}}) = \frac{i}{x^{3} (\frac{i}{x} + 1) (\frac{2}{\sqrt{x}} e^{\frac{i π}{4}}) (\frac{2 i}{x})}

= \frac{\sqrt{x}}{x^{2} (\frac{i}{x} + 1) 4 e^{\frac{i π}{4}}} = \frac{\sqrt{x} e^{- \frac{i π}{4}}}{4 (x i + x^{2})}

R e s (φ, - \frac{1}{\sqrt{x}} e^{- \frac{i π}{4}}) = \frac{- i}{x^{3} (- \frac{i}{x} + 1) (\frac{- 2 i}{x}) (\frac{- 2}{\sqrt{x}} e^{- \frac{i π}{4}})}

= \frac{- \sqrt{x}}{x^{2} (- \frac{i}{x} + 1) 4 e^{- \frac{i π}{4}}} = \frac{- \sqrt{x} e^{\frac{i π}{4}}}{4 (x^{2} - i x)}

\int_{- \infty}^{+ \infty} f (z) d z = 2 i π {\frac{- 1}{2 i (x^{2} + 1)} + \frac{\sqrt{x} e^{- \frac{i π}{4}}}{4 x (x + i)} - \frac{\sqrt{x} e^{\frac{i π}{4}}}{4 x (x - i)}}

= \frac{- π}{x^{2} + 1} + \frac{i π \sqrt{x}}{2 x} {\frac{e^{- \frac{i π}{4}}}{x + i} - \frac{e^{\frac{i π}{4}}}{x - i}}

= \frac{- π}{x^{2} + 1} + \frac{i π \sqrt{x}}{2 x} {\frac{(x - i) e^{- \frac{i π}{4}} - (x + i) e^{\frac{i π}{4}}}{x^{2} + 1}}

= - \frac{π}{x^{2} + 1} - \frac{i π \sqrt{x}}{2 x (x^{2} + 1)} 2 i I m ((x + i) e^{\frac{i π}{4}})

= - \frac{π}{x^{2} + 1} + \frac{π \sqrt{x}}{x (x^{2} + 1)} I m {(x + i) e^{\frac{i π}{4}}} b u t

(x + i) e^{\frac{i π}{4}} = (x + i) (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i)

= \frac{1}{\sqrt{2}} (x + i) (1 + i) = \frac{1}{\sqrt{2}} (x + x i + i - 1)

= \frac{1}{\sqrt{2}} {x - 1 + i (x + 1)} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = - \frac{π}{x^{2} + 1} + \frac{π \sqrt{x}}{\sqrt{2} x (x^{2} + 1)} (x + 1) = f^{^{'}} (x)

\Rightarrow f (x) = - π a r c t a n (x) + \frac{π}{\sqrt{2}} \int_{.}^{x} \frac{(t + 1) \sqrt{t}}{t (t^{2} + 1)} d t + c

Commented by prof Abdo imad last updated on 23/Jun/18

c h a n g e m e n t \sqrt{t} = u g i v e

\int_{.}^{x} \frac{(t + 1) \sqrt{t}}{t (t^{2} + 1)} d t = \int_{.}^{\sqrt{x}} \frac{(u^{2} + 1) u}{u^{2} (u^{4} + 1)} 2 u d u

= \int_{.}^{\sqrt{x}} \frac{2 (1 + u^{2})}{1 + u^{4}} d u \dots b e c p n t i n u e d \dots

Commented by math khazana by abdo last updated on 26/Jun/18

l e t d r c o m p o s e F (u) = \frac{2 u^{2} + 2}{u^{4} + 1}

F (u) = \frac{2 u^{2} + 2}{{(u^{2} + 1)}^{2} - 2 u^{2}} = \frac{2 u^{2} + 2}{(u^{2} + \sqrt{2} u + 1) (u^{2} - \sqrt{2} u + 1)}

= \frac{a u + b}{u^{2} + \sqrt{2} u + 1} + \frac{c u + d}{u^{2} - \sqrt{2} u + 1}

F (- u) = F (u) \Rightarrow \frac{- a u + b}{u^{2} - \sqrt{2} u + 1} + \frac{- c u + d}{u^{2} + \sqrt{2} u + 1} = F (u)

\Rightarrow c = - a a n d b = d \Rightarrow

F (u) = \frac{a u + b}{u^{2} + \sqrt{2} u + 1} + \frac{- a u + b}{u^{2} - \sqrt{2} u + 1}

F (0) = 2 = 2 b \Rightarrow b = 1 \Rightarrow

F (u) = \frac{a u + 1}{u^{2} + \sqrt{2} u + 1} + \frac{- a u + 1}{u^{2} - \sqrt{2} u + 1}

F (1) = 2 = \frac{a + 1}{2 + \sqrt{2}} + \frac{- a + 1}{2 - \sqrt{2}}

= \frac{(2 - \sqrt{2}) a + 2 - \sqrt{2} - (2 + \sqrt{2}) a + 2 + \sqrt{2}}{2}

= \frac{- 2 \sqrt{2} a + 4}{2} = - \sqrt{2} a + 2 = 2 \Rightarrow a = 0 \Rightarrow

F (u) = \frac{1}{u^{2} + \sqrt{2} u + 1} + \frac{1}{u^{2} - \sqrt{2} u + 1}

\int F (u) d u = \int \frac{d u}{u^{2} + 2 \frac{\sqrt{2}}{2} u + \frac{1}{2} + \frac{1}{2}}

+ \int \frac{d u}{u^{2} - 2 \frac{\sqrt{2}}{2} u + \frac{1}{2} + \frac{1}{2}}

= \int \frac{d u}{{(u + \frac{1}{\sqrt{2}})}^{2} + \frac{1}{2}} + \int \frac{d u}{{(u - \frac{1}{\sqrt{2}})}^{2} + \frac{1}{2}}

= I + J

I =_{u + \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} t} \int \frac{1}{\frac{1}{2} (t^{2} + 1)} \frac{d t}{\sqrt{2}} = \sqrt{2} a r c t a n (u \sqrt{2} + 1)

J =_{u - \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} t} \sqrt{2} a r c t a n (u \sqrt{2} - 1) \Rightarrow

\int_{.}^{\sqrt{x}} F (u) d u = \sqrt{2} a r c t a n (\sqrt{2 x} + 1) + \sqrt{2} a r c t a n (\sqrt{2 x} - 1)

Commented by math khazana by abdo last updated on 26/Jun/18

\Rightarrow F (x) = - π a r c t a n (x) + π a r c t a n (\sqrt{2 x} + 1)

+ π a r c t a n (\sqrt{2 x} - 1) + λ

λ = {l i m}_{x \to 0} F (x) = 0 \Rightarrow

F (x) = π {a r c t a n (\sqrt{2 x} + 1) + a r c t a n (\sqrt{2 x} - 1) - a r c t a n x}