find-f-x-0-arctan-xt-2-dt-with-x-gt-0-

Question Number 36336 by prof Abdo imad last updated on 31/May/18

f i n d f (x) = \int_{0}^{\infty} a r c t a n ({x t}^{2}) d t w i t h x > 0

Commented by abdo.msup.com last updated on 01/Jun/18

w e h a v e f^{^{'}} (x) = \int_{0}^{\infty} \frac{t^{2}}{1 + x^{2} t^{4}} d t

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

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

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

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

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {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, \frac{1}{\sqrt{x}} e^{i \frac{π}{4}}) = \frac{i}{x^{3} (\frac{2}{\sqrt{x}} e^{\frac{i π}{4}}) (\frac{2 i}{x})}

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

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

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

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

= \frac{i π \sqrt{x}}{2 x^{2}} {- 2 i s i n (\frac{π}{4})}

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

Commented by abdo.msup.com last updated on 01/Jun/18

\Rightarrow f^{^{'}} (x) = \frac{π}{2 \sqrt{2}} \frac{\sqrt{x}}{x^{2}} \Rightarrow

f (x) = \frac{π}{2 \sqrt{2}} \int \frac{\sqrt{x} d x}{x^{2}} + c b u t

\int \frac{\sqrt{x} d x}{x^{2}} =_{\sqrt{x} = t} \int \frac{t}{t^{4}} 2 t d t

= 2 \int \frac{d t}{t^{2}} = - \frac{2}{t} + λ = \frac{- 2}{\sqrt{x}} + λ \Rightarrow

f (x) = \frac{π}{2 \sqrt{2}} \frac{- 2}{\sqrt{x}} + c = \frac{- π}{\sqrt{2 x}} + c

Commented by abdo.msup.com last updated on 01/Jun/18

w e h a v e f (1) = - \frac{π}{2} + c \Rightarrow

c = \frac{π}{2} + f (1) = \frac{π}{2} + \int_{0}^{\infty} a r c t a n (x^{2}) d x

f (x) = - \frac{π}{2 \sqrt{x}} + \frac{π}{2} + \int_{0}^{\infty} a r c t a n (x^{2}) d x .

t h e v a l u e o f t h i s i n t e g r a l i s e a s y t o

f i n d . (b y p a r t s)

Commented by prof Abdo imad last updated on 01/Jun/18

l e t p u t L = {l i m}_{x \to + \infty} f (x) \Rightarrow c = L s o

f (x) = L - \frac{π}{\sqrt{2 x}} .