calculate ∫_0 ^∞ ((arctan(x^2 −1))/(x^2 +4))dx


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Question Number 68038 by mathmax by abdo last updated on 03/Sep/19

$c a l c u l a t e \int_{0}^{\infty} \frac{a r c t a n (x^{2} - 1)}{x^{2} + 4} d x$
Commented by mathmax by abdo last updated on 07/Sep/19
$let f(t) =∫_0 ^∞ ((arctan(x^2 −t))/(x^2 +4)) dx⇒f(1) =∫_0 ^∞ ((arctan(x^2 −1))/(x^2 +4))dx f^′ (t) =∫_0 ^∞ (1/((x^2 +4)(1+(x^2 −t)^2 )))dx =∫_0 ^∞ (dx/((x^2 +4)(x^4 −2tx^2 +t^2 +1))) ⇒2f^′ (t) =∫_(−∞) ^(+∞ ) (dx/((x^2 +4)(x^4 −2tx^(2 ) +t^2 +1))) let W(z) =(1/((z^2 +4)(z^4 −2tz^2 +t^(2 ) +1))) poles of W? z^2 +4 =0 ⇒z =+^− 2i z^4 −2tz^2 +t^2 +1 =0 ⇒u^2 −2tu +t^(2 ) +1 =0 with u=z^2 Δ^′ =t^2 −(t^2 +1) =−1 ⇒u_1 =t+i and u_2 =t−i ⇒ u_1 =(√(1+t^2 )) e^(iarctan((1/t))) and u_2 =(√(1+t^2 )) e^(−iarctan((1/t))) z^2 =u_1 ⇒z =+^− (1+t^2 )^(1/4) e^((i/2)arctan((1/t))) z^2 =u_2 ⇒z =+^− (1+t^2 )^(−(i/2)arctan((1/(t )))) ⇒ W(z) =(1/((z−2i)(z+2i)(z−(1+t^2 )^(1/4) e^((i/2)arctan((1/t))) (z+(1+t^2 )^(1/4) e^((i/2)arctan((1/t))) )(z−(1+t^2 )^(1/4) e^(−(i/2)arctan((1/t))) )(z+(1+t^2 )e^(−(i/2)arctan((1/t))) ))) residustheorem ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ{ Res(W,2i) +Res(W,(1+t^2 )^(1/4) e^((i/2)arctan((1/t))) +Res(W,−(1+t^2 )^(1/4) e^(−(i/2)afctan((1/t))) }...be continued...$
$l e t f (t) = \int_{0}^{\infty} \frac{a r c t a n (x^{2} - t)}{x^{2} + 4} d x \Rightarrow f (1) = \int_{0}^{\infty} \frac{a r c t a n (x^{2} - 1)}{x^{2} + 4} d x$ $f^{^{'}} (t) = \int_{0}^{\infty} \frac{1}{(x^{2} + 4) (1 + {(x^{2} - t)}^{2})} d x$ $= \int_{0}^{\infty} \frac{d x}{(x^{2} + 4) (x^{4} - 2 {t x}^{2} + t^{2} + 1)} \Rightarrow 2 f^{^{'}} (t) = \int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} + 4) (x^{4} - 2 {t x}^{2} + t^{2} + 1)}$ $l e t W (z) = \frac{1}{(z^{2} + 4) (z^{4} - 2 {t z}^{2} + t^{2} + 1)} p o l e s o f W ?$ $z^{2} + 4 = 0 \Rightarrow z = \bar{+} 2 i$ $z^{4} - 2 {t z}^{2} + t^{2} + 1 = 0 \Rightarrow u^{2} - 2 t u + t^{2} + 1 = 0 w i t h u = z^{2}$ $Δ^{^{'}} = t^{2} - (t^{2} + 1) = - 1 \Rightarrow u_{1} = t + i a n d u_{2} = t - i \Rightarrow$ $u_{1} = \sqrt{1 + t^{2}} e^{i a r c t a n (\frac{1}{t})} a n d u_{2} = \sqrt{1 + t^{2}} e^{- i a r c t a n (\frac{1}{t})}$ $z^{2} = u_{1} \Rightarrow z = \bar{+} {(1 + t^{2})}^{\frac{1}{4}} e^{\frac{i}{2} \arctan (\frac{1}{t})}$ $z^{2} = u_{2} \Rightarrow z = \bar{+} {(1 + t^{2})}^{- \frac{i}{2} a r c t a n (\frac{1}{t})} \Rightarrow$ $W (z) = \frac{1}{(z - 2 i) (z + 2 i) (z - {(1 + t^{2})}^{\frac{1}{4}} e^{\frac{i}{2} a r c t a n (\frac{1}{t})} (z + {(1 + t^{2})}^{\frac{1}{4}} e^{\frac{i}{2} a r c t a n (\frac{1}{t})}) (z - {(1 + t^{2})}^{\frac{1}{4}} e^{- \frac{i}{2} a r c t a n (\frac{1}{t})}) (z + (1 + t^{2}) e^{- \frac{i}{2} a r c t a n (\frac{1}{t})})}$ $r e s i d u s t h e o r e m \Rightarrow$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π {R e s (W, 2 i) + R e s (W, {(1 + t^{2})}^{\frac{1}{4}} e^{\frac{i}{2} \arctan (\frac{1}{t})}$ $+ R e s (W, - {(1 + t^{2})}^{\frac{1}{4}} e^{- \frac{i}{2} a f c t a n (\frac{1}{t})}} . . . b e c o n t i n u e d . . .$
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