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


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Question Number 68470 by mathmax by abdo last updated on 11/Sep/19

$f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{a r c t a n (2 x^{2})}{x^{2} + 4} d x$
Commented by mathmax by abdo last updated on 11/Sep/19
$let I =∫_0 ^∞ ((arctan(2x^2 ))/(x^2 +4))dx changement x =2t give I =∫_0 ^∞ ((artan(8t^2 ))/(4(1+t^2 ))) (2dt) =(1/2) ∫_0 ^∞ ((arctan(8t^2 ))/(1+t^2 ))dt =(1/4)∫_(−∞) ^(+∞) ((arctan(8t^2 ))/(t^2 +1))dt let f(x) =∫_(−∞) ^(+∞) ((arctan(xt^2 ))/(t^2 +1))dt with x≥0 we have f^′ (x) =∫_(−∞) ^(+∞) (t^2 /((1+x^2 t^4 )(t^2 +1)))dt let W(z)=(z^2 /((z^2 +1)(x^2 z^4 +1))) W(z) =(z^2 /((z^2 +1)x^2 (z^4 +(1/x^2 )))) =(z^2 /(x^2 (z−i)(z+i)(z^2 −(i/x))(z^2 +(i/x)))) =(z^2 /(x^2 (z−i)(z+i)(z−(1/((√x) ))e^((iπ)/4) )(z+(1/(√x))e^((iπ)/4) )(z−(1/(√x))e^(−((iπ)/4)) )(z+(1/(√x))e^(−((iπ)/4)) ))) so the poles of W are +^− i and +^− (1/(√x))e^((iπ)/4) and +^− (1/(√x))e^(−((iπ)/4)) ∫_(−∞) ^(+∞) W(z)dz =2iπ{Res(W,i)+Res(W,(1/(√x))e^((iπ)/4) )+Res(W,−(1/(√x))e^(−((iπ)/4)) )} Res(W,i) =((−1)/(x^2 (2i)(1+(1/x^2 )))) =((−1)/(2i(x^2 +1))) Res(W,(1/(√x))e^((iπ)/4) ) =(i/(x^3 (i+1)((2/(√x))e^((iπ)/4) )(i+(i/x)))) =(1/((1+i)x^3 (1+(1/x))(2/(√x))))e^(−((iπ)/4)) =(((√x)e^(−((iπ)/4)) )/(2(1+i)(x^3 +x^2 ))) Res(W,−(1/(√x))e^(−((iπ)/4)) ) =((−i)/(x^3 (−i+1)(−(2/(√x))e^(−((iπ)/4)) )(−i+(i/x)))) =((√x)/(x^3 (i−1)(1−(1/x))2 e^(−((iπ)/4)) )) =(((√x)e^((iπ)/4) )/(2(i−1)(x^3 −x^2 ))) ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ{ −(1/(2i(x^2 +1))) +(((√x)e^(−((iπ)/4)) )/(2(1+i)(x^3 +x^2 )))−(((√x)e^((iπ)/4) )/(2(1−i)(x^3 −x^2 )))} =2iπ{ −(1/(2i(x^2 +1))) +(((√x)(1−i))/(4(x^3 +x^2 )))e^(−((iπ)/4)) −(((√x)(1+i)e^((iπ)/4) )/(4(x^3 −x^2 )))} =−(π/(x^(2 ) +1)) +((iπ(1−i)(√x))/(2(x^3 +x^2 )))e^(−((iπ)/4)) −((iπ(1+i)(√x)e^((iπ)/4) )/(2(x^3 −x^2 ))) ....be continued...$
$l e t I = \int_{0}^{\infty} \frac{a r c t a n (2 x^{2})}{x^{2} + 4} d x c h a n g e m e n t x = 2 t g i v e$ $I = \int_{0}^{\infty} \frac{a r t a n (8 t^{2})}{4 (1 + t^{2})} (2 d t) = \frac{1}{2} \int_{0}^{\infty} \frac{a r c t a n (8 t^{2})}{1 + t^{2}} d t$ $= \frac{1}{4} \int_{- \infty}^{+ \infty} \frac{a r c t a n (8 t^{2})}{t^{2} + 1} d t l e t f (x) = \int_{- \infty}^{+ \infty} \frac{a r c t a n ({x t}^{2})}{t^{2} + 1} d t w i t h x ⩾ 0$ $w e h a v e f^{^{'}} (x) = \int_{- \infty}^{+ \infty} \frac{t^{2}}{(1 + x^{2} t^{4}) (t^{2} + 1)} d t l e t W (z) = \frac{z^{2}}{(z^{2} + 1) (x^{2} z^{4} + 1)}$ $W (z) = \frac{z^{2}}{(z^{2} + 1) x^{2} (z^{4} + \frac{1}{x^{2}})} = \frac{z^{2}}{x^{2} (z - i) (z + i) (z^{2} - \frac{i}{x}) (z^{2} + \frac{i}{x})}$ $= \frac{z^{2}}{x^{2} (z - i) (z + i) (z - \frac{1}{\sqrt{x}} e^{\frac{i π}{4}}) (z + \frac{1}{\sqrt{x}} e^{\frac{i π}{4}}) (z - \frac{1}{\sqrt{x}} e^{- \frac{i π}{4}}) (z + \frac{1}{\sqrt{x}} e^{- \frac{i π}{4}})}$ $s o t h e p o l e s o f W a r e \bar{+} i a n d \bar{+} \frac{1}{\sqrt{x}} e^{\frac{i π}{4}} a n d \bar{+} \frac{1}{\sqrt{x}} e^{- \frac{i π}{4}}$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π {R e s (W, i) + R e s (W, \frac{1}{\sqrt{x}} e^{\frac{i π}{4}}) + R e s (W, - \frac{1}{\sqrt{x}} e^{- \frac{i π}{4}})}$ $R e s (W, i) = \frac{- 1}{x^{2} (2 i) (1 + \frac{1}{x^{2}})} = \frac{- 1}{2 i (x^{2} + 1)}$ $R e s (W, \frac{1}{\sqrt{x}} e^{\frac{i π}{4}}) = \frac{i}{x^{3} (i + 1) (\frac{2}{\sqrt{x}} e^{\frac{i π}{4}}) (i + \frac{i}{x})} = \frac{1}{(1 + i) x^{3} (1 + \frac{1}{x}) \frac{2}{\sqrt{x}}} e^{- \frac{i π}{4}}$ $= \frac{\sqrt{x} e^{- \frac{i π}{4}}}{2 (1 + i) (x^{3} + x^{2})}$ $R e s (W, - \frac{1}{\sqrt{x}} e^{- \frac{i π}{4}}) = \frac{- i}{x^{3} (- i + 1) (- \frac{2}{\sqrt{x}} e^{- \frac{i π}{4}}) (- i + \frac{i}{x})}$ $= \frac{\sqrt{x}}{x^{3} (i - 1) (1 - \frac{1}{x}) 2 e^{- \frac{i π}{4}}} = \frac{\sqrt{x} e^{\frac{i π}{4}}}{2 (i - 1) (x^{3} - x^{2})} \Rightarrow$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π {- \frac{1}{2 i (x^{2} + 1)} + \frac{\sqrt{x} e^{- \frac{i π}{4}}}{2 (1 + i) (x^{3} + x^{2})} - \frac{\sqrt{x} e^{\frac{i π}{4}}}{2 (1 - i) (x^{3} - x^{2})}}$ $= 2 i π {- \frac{1}{2 i (x^{2} + 1)} + \frac{\sqrt{x} (1 - i)}{4 (x^{3} + x^{2})} e^{- \frac{i π}{4}} - \frac{\sqrt{x} (1 + i) e^{\frac{i π}{4}}}{4 (x^{3} - x^{2})}}$ $= - \frac{π}{x^{2} + 1} + \frac{i π (1 - i) \sqrt{x}}{2 (x^{3} + x^{2})} e^{- \frac{i π}{4}} - \frac{i π (1 + i) \sqrt{x} e^{\frac{i π}{4}}}{2 (x^{3} - x^{2})}$ $. . . . b e c o n t i n u e d . . .$
Commented by mathmax by abdo last updated on 11/Sep/19

$a n o t h e r e a s y m e t h o d b u t n e e d a p r o o f$ $l e t I = \int_{0}^{\infty} \frac{a r c t a n (2 x^{2})}{x^{2} + 4} d x \Rightarrow 2 I = \int_{- \infty}^{+ \infty} \frac{a r c t a n (2 x^{2})}{x^{2} + 4} d x l e t$ $φ (z) = \frac{a r c t a n (2 z^{2})}{z^{2} + 4} \Rightarrow φ (z) = \frac{a r c t a n (z^{2})}{(z - 2 i) (z + 2 i)}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, 2 i) a n d R e s (φ, 2 i) = \frac{∣ a r c t a n (- 4) ∣}{4 i}$ $= \frac{a r c t a n (4)}{4 i} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \times \frac{a r c t a n (4)}{4 i} = \frac{π}{2} a r c t a n (4) .$
Commented by mathmax by abdo last updated on 12/Sep/19

$\Rightarrow I = \frac{π}{4} a r c t a n (4) .$
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