find f(x)= ∫_0 ^∞ arctan(xt^2 )dt with x>0


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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
$we have f^′ (x)=∫_0 ^∞ (t^2 /(1+x^2 t^4 ))dt 2f^′ (x)=∫_(−∞) ^(+∞) (t^2 /(x^2 t^4 +1))dt let consider the complex function ϕ(z) =(z^2 /(x^2 z^4 +1)) we have ϕ(z) = (z^2 /(x^2 { z^4 +(1/x^2 )})) = (z^2 /(x^2 {z^2 −(i/x)}(z^2 +(i/x)))) =(z^2 /(x^2 (z −(1/(√x))e^(i(π/4)) )(z +(1/(√x))e^(i(π/4)) )(z−(1/(√x))e^(−i(π/4)) )(z +(1/(√x))e^(−((iπ)/4)) ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,(1/(√x))e^((iπ)/4) )+Res(ϕ,−(1/(√x))e^(−((iπ)/4)) )} Res(ϕ, (1/(√x))e^(i(π/4)) ) = (i/(x^3 ((2/(√x))e^((iπ)/4) )(((2i)/x)))) = ((√x)/(4x^2 )) e^(−((iπ)/4)) Res(ϕ,−(1/(√x))e^(−((iπ)/4)) )= ((−i)/(x^3 (((−2)/(√x))e^(−((iπ)/4)) )(((−2i)/x)))) =−((√x)/(4x^2 )) e^((iπ)/4) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ .((√x)/(4x^2 )){ e^(−((iπ)/4)) −e^((iπ)/4) } =((iπ(√x))/(2x^2 )){−2isin((π/4))} =((π(√x))/x^2 ) ((√2)/2) = ((π(√x))/(x^2 (√2)))$
$w e h a v e f^{^{'}} (x) = \int_{0}^{\infty} \frac{t^{2}}{1 + x^{2} t^{4}} d t$ $2 f^{^{'}} (x) = \int_{- \infty}^{+ \infty} \frac{t^{2}}{x^{2} t^{4} + 1} d t l e t c o n s i d e r$ $t h e c o m p l e x f u n c t i o n$ $φ (z) = \frac{z^{2}}{x^{2} z^{4} + 1} w e h a v e$ $φ (z) = \frac{z^{2}}{x^{2} {z^{4} + \frac{1}{x^{2}}}}$ $= \frac{z^{2}}{x^{2} {z^{2} - \frac{i}{x}} (z^{2} + \frac{i}{x})}$ $= \frac{z^{2}}{x^{2} (z - \frac{1}{\sqrt{x}} e^{i \frac{π}{4}}) (z + \frac{1}{\sqrt{x}} e^{i \frac{π}{4}}) (z - \frac{1}{\sqrt{x}} e^{- i \frac{π}{4}}) (z + \frac{1}{\sqrt{x}} e^{- \frac{i π}{4}})}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, \frac{1}{\sqrt{x}} e^{\frac{i π}{4}}) + R e s (φ, - \frac{1}{\sqrt{x}} e^{- \frac{i π}{4}})}$ $R e s (φ, \frac{1}{\sqrt{x}} e^{i \frac{π}{4}}) = \frac{i}{x^{3} (\frac{2}{\sqrt{x}} e^{\frac{i π}{4}}) (\frac{2 i}{x})}$ $= \frac{\sqrt{x}}{4 x^{2}} e^{- \frac{i π}{4}}$ $R e s (φ, - \frac{1}{\sqrt{x}} e^{- \frac{i π}{4}}) = \frac{- i}{x^{3} (\frac{- 2}{\sqrt{x}} e^{- \frac{i π}{4}}) (\frac{- 2 i}{x})}$ $= - \frac{\sqrt{x}}{4 x^{2}} e^{\frac{i π}{4}} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π . \frac{\sqrt{x}}{4 x^{2}} {e^{- \frac{i π}{4}} - e^{\frac{i π}{4}}}$ $= \frac{i π \sqrt{x}}{2 x^{2}} {- 2 i s i n (\frac{π}{4})}$ $= \frac{π \sqrt{x}}{x^{2}} \frac{\sqrt{2}}{2} = \frac{π \sqrt{x}}{x^{2} \sqrt{2}}$
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

$l e t p u t L = {l i m}_{x \to + \infty} f (x) \Rightarrow c = L s o$ $f (x) = L - \frac{π}{\sqrt{2 x}} .$
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