let-f-x-0-arctan-xt-2-dt-find-a-explicite-form-of-f-x-

Question Number 41279 by math khazana by abdo last updated on 04/Aug/18

l e t f (x) = \int_{0}^{\infty} a r c t a n ({x t}^{2}) d t .

f i n d a e x p l i c i t e f o r m o f f^{^{'}} (x)

Commented by maxmathsup by imad last updated on 05/Aug/18

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

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

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

= \frac{z^{2}}{x^{2} (t - \frac{e^{i \frac{π}{4}}}{\sqrt{x}}) (t + \frac{e^{\frac{i π}{4}}}{\sqrt{x}}) (t - \frac{e^{- \frac{i π}{4}}}{\sqrt{x}}) (t + \frac{e^{- \frac{i π}{4}}}{\sqrt{x}})} s o t h e p o l e s o f φ a r e

\bar{+} \frac{e^{\frac{i π}{4}}}{\sqrt{x}} a n d \bar{+} \frac{e^{- \frac{i π}{4}}}{\sqrt{x}} r e s i d u s t h e o r e m g i v e

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

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

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

= \frac{\sqrt{x}}{x^{2}} \frac{\sqrt{2}}{2} a n d f^{^{'}} (x) = \frac{1}{2} \int_{- \infty}^{+ \infty} φ (z) d z = \frac{\sqrt{2}}{4 x \sqrt{x}} w i t h x > 0

i f x < w e p u t x = - u \Rightarrow f (x) = f (- u) = \int_{0}^{\infty} a r c t a n (- {u t}^{2}) d t

= - f (u) \Rightarrow f^{^{'}} (x) = - f^{^{'}} (u) = - \frac{\sqrt{2}}{4 u \sqrt{u}} = - \frac{\sqrt{2}}{- 4 x \sqrt{- x}} = \frac{\sqrt{2}}{4 x \sqrt{- x}} .