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Question Number 52356 by Tawa1 last updated on 06/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Jan/19

Commented by Tawa1 last updated on 07/Jan/19

$Sir please help .$
Commented by MJS last updated on 07/Jan/19

$this {can}^{'} t be solved using elementary calculus$ $not sure if these help :$ $\tan^{- 1} θ = \frac{i}{2} (\ln (1 - i θ) - \ln (1 + i θ)) \Rightarrow$ $\Rightarrow \tan^{- 1} x^{2} = \frac{i}{2} (\ln (1 - i x^{2}) - \ln (1 + i x^{2})) \Rightarrow$ $\Rightarrow \tan^{- 1} x^{2} = \frac{i}{2} \ln \frac{1 - i x^{2}}{1 + i x^{2}} = \frac{i}{2} \ln \frac{1 - x^{4} + 2 i x^{2}}{x^{4} + 1} =$ $[\ln (a + b i) = \frac{1}{2} \ln (a^{2} + b^{2}) + i (sign (b) \frac{π}{2} - \tan^{- 1} \frac{a}{b})]$ $= \frac{π}{4} + \frac{1}{2} \tan^{- 1} \frac{x^{4} - 1}{2 x^{2}}$ $x^{8} + x^{4} + 1 = (x^{4} + x^{2} + 1) (x^{4} - x^{2} + 1) =$ $= (x^{2} + x + 1) (x^{2} - x + 1) (x^{2} + \sqrt{3} x + 1) (x^{2} - \sqrt{3} x + 1$ $= (x + \frac{1 + i \sqrt{3}}{2}) (x - \frac{1 + i \sqrt{3}}{2}) (x + \frac{1 - i \sqrt{3}}{2}) (x - \frac{1 - i \sqrt{3}}{2}) (x + \frac{\sqrt{3} + i}{2}) (x - \frac{\sqrt{3} + i}{2}) (x + \frac{\sqrt{3} - i}{2}) (x - \frac{\sqrt{3} - i}{2})$
Commented by Tawa1 last updated on 07/Jan/19

$God bless you sir$
Commented by Abdo msup. last updated on 07/Jan/19

$l e t I = \int_{0}^{\infty} \frac{a r c t a n (x^{2})}{x^{8} + x^{4} + 1} d x c h a n g e m e n t x^{2} = t a n t$ $g i v e I = \int_{0}^{\frac{π}{2}} \frac{t}{{t a n}^{4} t + {t a n}^{2} t + 1} \frac{1 + {t a n}^{2} t}{2 \sqrt{1 + {t a n}^{2} t}} d t$ $= \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{t \sqrt{1 + {t a n}^{2} t}}{1 + {t a n}^{2} t + {t a n}^{4} t} d t$ $= \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{t}{c o s t (\frac{1}{{c o s}^{2} t} + {(1 + {t a n}^{2} t - 1)}^{2})} d t$ $= \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{t}{\frac{1}{c o s t} + c o s t {(\frac{1}{{c o s}^{2} t} - 1)}^{2}} d t$ $= \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{t}{\frac{1}{c o s t} + c o s s t (\frac{1}{{c o s}^{4} t} - \frac{2}{{c o s}^{2} t} + 1)} d t$ $= \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{t}{\frac{1}{c o s t} + \frac{1}{{c o s}^{3} t} - \frac{2}{c o s t} + c o s t} d t$ $= \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{t {c o s}^{3} t}{{c o s}^{2} t + 1 - 2 {c o s}^{2} t + {c o s}^{4} t} d t$ $= \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{t {c o s}^{3} t}{{c o s}^{4} t - {c o s}^{2} t + 1} d t . . . b e c o n t i n u e d . . .$
Commented by Abdo msup. last updated on 07/Jan/19
$let F(α)=∫_0 ^∞ ((arctan(αx^2 ))/(1+x^4 +x^8 )) dx with α>0 we have F^′ (α) =∫_0 ^∞ (x^2 /((1+α^2 x^4 )(x^8 +x^4 +1)))dx ⇒2F^′ (α) =∫_(−∞) ^(+∞) (x^2 /((x^8 +x^4 +1)(α^2 x^4 +1)))dx let consider the complex function ϕ(z)=(z^2 /((z^8 +z^4 +1)(α^2 z^4 +1))) poles of ϕ? z^8 +z^4 +1=0 ⇔1+z^4 +(z^4 )^2 =0 ⇔((1−z^(12) )/(1−z^4 )) =0 ⇔ z^(12) =1 and z^4 ≠ 1 the roots of z^(12) =1 are z_k =e^(i((kπ)/6)) and k∈[[0,11]] but only we must take 8 roots let search.. z_0 =1 (to eliminate) z_1 =e^((iπ)/6) and z_1 ^4 ≠1 z_2 =e^((iπ)/3) and z_2 ^4 ≠1 z_3 =e^((iπ)/2) and z_3 ^4 =1 (to eliminate) z_4 =e^(i((2π)/3)) and z_4 ^4 ≠1 z_5 =e^((i5π)/6) and z_5 ^4 ≠1 z_6 =e^(iπ) and z_6 ^4 =1 (to eliminate) z_7 =e^((i7.π)/6) and z_7 ^4 ≠1 z_8 = e^((i4π)/3) and z_8 ^4 ≠1 z_9 = e^((i3π)/2) and z_9 ^4 =1 (to eliminate) z_(10) =e^((i5π)/3) and z_(10) ^4 ≠1 z_(11) =e^((i11π)/6) ⇒ z^8 +z^4 +1 =(z−z_1 )(z−z_2 )(z−z_4 )(z−z_5 )(z−z_7 )(z−z_8 )(z−z_(10) )(z−z_(11) ) also we have α^2 z^4 +1 =α^2 (z^4 +(1/α^2 )) =α^2 (z^4 −((i/α))^2 )=α^2 (z^2 −(i/α))(z^2 +(i/α)) =α^2 (z−(1/(√α)) e^((iπ)/4) )(z+(1/(√α)) e^((iπ)/4) )(z−(1/(√α)) e^(−((iπ)/4)) )(z+(1/(√α))e^(−((iπ)/4)) ) so the poles of ϕ are +^− (1/(√α)) e^((iπ)/4) and +^− (1/(√α)) e^(−((iπ)/4)) also tbe complex z_k with k ∈{1,2,4,5,7,8,10,11} after we use Residus theorem ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Σ_(im(z_k )>0) Res(ϕ ,z_k ) ...be continued ...$
$l e t F (α) = \int_{0}^{\infty} \frac{a r c t a n (α x^{2})}{1 + x^{4} + x^{8}} d x w i t h α > 0 w e$ $h a v e F^{^{'}} (α) = \int_{0}^{\infty} \frac{x^{2}}{(1 + α^{2} x^{4}) (x^{8} + x^{4} + 1)} d x$ $\Rightarrow 2 F^{^{'}} (α) = \int_{- \infty}^{+ \infty} \frac{x^{2}}{(x^{8} + x^{4} + 1) (α^{2} x^{4} + 1)} d x$ $l e t c o n s i d e r t h e c o m p l e x f u n c t i o n$ $φ (z) = \frac{z^{2}}{(z^{8} + z^{4} + 1) (α^{2} z^{4} + 1)} p o l e s o f φ ?$ $z^{8} + z^{4} + 1 = 0 \Leftrightarrow 1 + z^{4} + {(z^{4})}^{2} = 0 \Leftrightarrow \frac{1 - z^{12}}{1 - z^{4}} = 0 \Leftrightarrow$ $z^{12} = 1 a n d z^{4} \neq 1 t h e r o o t s o f z^{12} = 1 a r e$ $z_{k} = e^{i \frac{k π}{6}} a n d k \in [[0, 11]] b u t o n l y w e m u s t t a k e$ $8 r o o t s l e t s e a r c h . .$ $z_{0} = 1 (t o e l i m i n a t e)$ $z_{1} = e^{\frac{i π}{6}} a n d z_{1}^{4} \neq 1$ $z_{2} = e^{\frac{i π}{3}} a n d z_{2}^{4} \neq 1$ $z_{3} = e^{\frac{i π}{2}} a n d z_{3}^{4} = 1 (t o e l i m i n a t e)$ $z_{4} = e^{i \frac{2 π}{3}} a n d z_{4}^{4} \neq 1$ $z_{5} = e^{\frac{i 5 π}{6}} a n d z_{5}^{4} \neq 1$ $z_{6} = e^{i π} a n d z_{6}^{4} = 1 (t o e l i m i n a t e)$ $z_{7} = e^{\frac{i 7 . π}{6}} a n d z_{7}^{4} \neq 1$ $z_{8} = e^{\frac{i 4 π}{3}} a n d z_{8}^{4} \neq 1$ $z_{9} = e^{\frac{i 3 π}{2}} a n d z_{9}^{4} = 1 (t o e l i m i n a t e)$ $z_{10} = e^{\frac{i 5 π}{3}} a n d z_{10}^{4} \neq 1$ $z_{11} = e^{\frac{i 11 π}{6}} \Rightarrow$ $z^{8} + z^{4} + 1 = (z - z_{1}) (z - z_{2}) (z - z_{4}) (z - z_{5}) (z - z_{7}) (z - z_{8}) (z - z_{10}) (z - z_{11})$ $a l s o w e h a v e α^{2} z^{4} + 1 = α^{2} (z^{4} + \frac{1}{α^{2}})$ $= α^{2} (z^{4} - {(\frac{i}{α})}^{2}) = α^{2} (z^{2} - \frac{i}{α}) (z^{2} + \frac{i}{α})$ $= α^{2} (z - \frac{1}{\sqrt{α}} e^{\frac{i π}{4}}) (z + \frac{1}{\sqrt{α}} e^{\frac{i π}{4}}) (z - \frac{1}{\sqrt{α}} e^{- \frac{i π}{4}}) (z + \frac{1}{\sqrt{α}} e^{- \frac{i π}{4}})$ $s o t h e p o l e s o f φ a r e$ $\bar{+} \frac{1}{\sqrt{α}} e^{\frac{i π}{4}} a n d \bar{+} \frac{1}{\sqrt{α}} e^{- \frac{i π}{4}} a l s o t b e c o m p l e x$ $z_{k} w i t h k \in {1, 2, 4, 5, 7, 8, 10, 11} a f t e r w e u s e$ $R e s i d u s t h e o r e m$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \sum_{i m (z_{k}) > 0} R e s (φ, z_{k})$ $. . . b e c o n t i n u e d . . .$
Commented by Tawa1 last updated on 07/Jan/19

$God bless you sir . waiting . . .$
Commented by Abdo msup. last updated on 07/Jan/19

$t h a n k s s i r t h i s i s t h e w a y f o r t h i s i n t e g r a l . . . .$
	Answered by MJS last updated on 07/Jan/19

	$\approx . 269949$
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