find the value or ∫_0 ^∞ ((arctan(x^2 ))/(1+x^4 ))dx .


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Question Number 52482 by maxmathsup by imad last updated on 08/Jan/19

$f i n d t h e v a l u e o r \int_{0}^{\infty} \frac{a r c t a n (x^{2})}{1 + x^{4}} d x .$
Commented by Abdo msup. last updated on 09/Jan/19
$let f(t)=∫_0 ^∞ ((arctan(tx^2 ))/(1+x^4 ))dx with t>0 we have f^′ (t)= ∫_0 ^∞ (x^2 /((1+t^2 x^4 )(1+x^4 )))dx ⇒ 2f^′ (x)=∫_(−∞) ^(+∞) (x^2 /((1+t^2 x^4 )(1+x^4 )))dx let consider the complex function ϕ(z)=(z^2 /((t^2 z^4 +1)(z^4 +1))) poles of ϕ? we have ϕ(z)=(z^2 /((tz^2 −i)(tz^2 +i)(z^2 −i)(z^2 +i))) =(z^2 /(((√t)z−e^((iπ)/4) )((√t)z+e^((iπ)/4) )((√t)z−e^(−((iπ)/4)) )((√t)z +e^(−((iπ)/4)) )(z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) =(z^2 /(t^2 (z−(1/(√t))e^((iπ)/4) )(z+(1/(√t))e^((iπ)/4) )(z−(1/(√t))e^(−((iπ)/4)) )(z+(1/(√t))e^(−((iπ)/4)) )(z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) so the poles of ϕ are +^− (1/(√t)) e^((iπ)/4) ,+^− (1/(√t)) e^(−((iπ)/4)) ,+^− e^((iπ)/4) ,+^− e^(−((iπ)/4)) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Σ_(im(z_k )>0) Res(ϕ,z_k ) =2iπ{ Res(ϕ,(1/(√t))e^((iπ)/4) ) +Res(ϕ,−(1/(√t)) e^(−((iπ)/4)) )+Res(ϕ,e^((iπ)/4) )+Res(ϕ,−e^(−((iπ)/4)) )}$
$l e t f (t) = \int_{0}^{\infty} \frac{a r c t a n ({t x}^{2})}{1 + x^{4}} d x w i t h t > 0 w e h a v e$ $f^{^{'}} (t) = \int_{0}^{\infty} \frac{x^{2}}{(1 + t^{2} x^{4}) (1 + x^{4})} d x \Rightarrow$ $2 f^{^{'}} (x) = \int_{- \infty}^{+ \infty} \frac{x^{2}}{(1 + t^{2} x^{4}) (1 + x^{4})} d x l e t c o n s i d e r t h e$ $c o m p l e x f u n c t i o n φ (z) = \frac{z^{2}}{(t^{2} z^{4} + 1) (z^{4} + 1)}$ $p o l e s o f φ ? w e h a v e$ $φ (z) = \frac{z^{2}}{({t z}^{2} - i) ({t z}^{2} + i) (z^{2} - i) (z^{2} + i)}$ $= \frac{z^{2}}{(\sqrt{t} z - e^{\frac{i π}{4}}) (\sqrt{t} z + e^{\frac{i π}{4}}) (\sqrt{t} z - e^{- \frac{i π}{4}}) (\sqrt{t} z + e^{- \frac{i π}{4}}) (z - e^{\frac{i π}{4}}) (z + e^{\frac{i π}{4}}) (z - e^{- \frac{i π}{4}}) (z + e^{- \frac{i π}{4}})}$ $= \frac{z^{2}}{t^{2} (z - \frac{1}{\sqrt{t}} e^{\frac{i π}{4}}) (z + \frac{1}{\sqrt{t}} e^{\frac{i π}{4}}) (z - \frac{1}{\sqrt{t}} e^{- \frac{i π}{4}}) (z + \frac{1}{\sqrt{t}} e^{- \frac{i π}{4}}) (z - e^{\frac{i π}{4}}) (z + e^{\frac{i π}{4}}) (z - e^{- \frac{i π}{4}}) (z + e^{- \frac{i π}{4}})}$ $s o t h e p o l e s o f φ a r e \bar{+} \frac{1}{\sqrt{t}} e^{\frac{i π}{4}}, \bar{+} \frac{1}{\sqrt{t}} e^{- \frac{i π}{4}}, \bar{+} e^{\frac{i π}{4}}, \bar{+} e^{- \frac{i π}{4}}$ $r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \sum_{i m (z_{k}) > 0} R e s (φ, z_{k})$ $= 2 i π {R e s (φ, \frac{1}{\sqrt{t}} e^{\frac{i π}{4}}) + R e s (φ, - \frac{1}{\sqrt{t}} e^{- \frac{i π}{4}}) + R e s (φ, e^{\frac{i π}{4}}) + R e s (φ, - e^{- \frac{i π}{4}})}$
Commented by Abdo msup. last updated on 09/Jan/19

$R e s (φ, \frac{1}{\sqrt{t}} e^{\frac{i π}{4}}) = \frac{1}{t^{3}} \frac{i}{(\frac{2}{\sqrt{t}} e^{\frac{i π}{4}}) (\frac{2 i}{\sqrt{t}} s i n (\frac{π}{4})) (\frac{2}{\sqrt{t}} c o s (\frac{π}{4})) (\frac{i}{t} - i) (\frac{i}{t} + i)}$ $= \frac{1}{8 t^{3} \frac{1}{t \sqrt{t}} \frac{1}{2} (- \frac{1}{t^{2}} + 1)} e^{- \frac{i π}{4}}$ $= \frac{\sqrt{t} e^{- \frac{i π}{4}}}{4 t^{2} (\frac{t^{2} - 1}{t^{2}})} = \frac{\sqrt{t} e^{- \frac{i π}{4}}}{4 (t^{2} - 1)} . . . . b e c o n t i n u e d . . .$
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