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


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Question Number 51552 by maxmathsup by imad last updated on 28/Dec/18

$c a l c u l a t e \int_{0}^{+ \infty} \frac{a r c t a n (1 + x^{2})}{x^{2} + 4} d x$
Commented by Abdo msup. last updated on 29/Dec/18
$let I =∫_0 ^∞ ((arctan(1+x^2 ))/(x^2 +4))dx I =_(x=2t) ∫_0 ^∞ ((arctan(1+4t^2 ))/(4(1+t^2 )))2dt =(1/2) ∫_0 ^∞ ((arctan(1+4t^2 ))/(1+t^2 ))dt let consider f(x)=∫_0 ^∞ ((arctan(1+xt^2 ))/(1+t^2 ))dt with x>0 f^′ (x) =∫_0 ^∞ (t^2 /((1+x^2 t^4 )(1+t^2 )))dt ⇒2f(x)=∫_(−∞) ^(+∞) (t^2 /((x^2 t^4 +1)(t^2 +1)))dt let ϕ(z)=(z^2 /((x^2 z^4 +1)(z^2 +1))) ⇒ ϕ(z)= (z^2 /((xz^2 −i)(xz^2 +i)(z−i)(z+i))) =(z^2 /(x^2 (z^2 −(i/x))(z^2 +(i/x))(z−i)(z+i))) = (z^2 /(x^2 (z−(1/(√x))e^((iπ)/4) )(z +(1/((√)x))e^((iπ)/4) )(z−(i/(√x))e^(−((iπ)/4)) )(z+(i/(√x))e^(−((iπ)/4)) )(z−i)(z+i))) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,(1/(√x))e^((iπ)/4) )+Res(ϕ,−(1/(√x))e^(−((iπ)/4)) ) +Res(ϕ,i)} ...be continued....$
$l e t I = \int_{0}^{\infty} \frac{a r c t a n (1 + x^{2})}{x^{2} + 4} d x$ $I =_{x = 2 t} \int_{0}^{\infty} \frac{a r c t a n (1 + 4 t^{2})}{4 (1 + t^{2})} 2 d t$ $= \frac{1}{2} \int_{0}^{\infty} \frac{a r c t a n (1 + 4 t^{2})}{1 + t^{2}} d t l e t c o n s i d e r$ $f (x) = \int_{0}^{\infty} \frac{a r c t a n (1 + {x t}^{2})}{1 + t^{2}} d t w i t h x > 0$ $f^{^{'}} (x) = \int_{0}^{\infty} \frac{t^{2}}{(1 + x^{2} t^{4}) (1 + t^{2})} d t$ $\Rightarrow 2 f (x) = \int_{- \infty}^{+ \infty} \frac{t^{2}}{(x^{2} t^{4} + 1) (t^{2} + 1)} d t l e t$ $φ (z) = \frac{z^{2}}{(x^{2} z^{4} + 1) (z^{2} + 1)} \Rightarrow$ $φ (z) = \frac{z^{2}}{({x z}^{2} - i) ({x z}^{2} + i) (z - i) (z + i)}$ $= \frac{z^{2}}{x^{2} (z^{2} - \frac{i}{x}) (z^{2} + \frac{i}{x}) (z - i) (z + i)}$ $= \frac{z^{2}}{x^{2} (z - \frac{1}{\sqrt{x}} e^{\frac{i π}{4}}) (z + \frac{1}{\sqrt{} x} e^{\frac{i π}{4}}) (z - \frac{i}{\sqrt{x}} e^{- \frac{i π}{4}}) (z + \frac{i}{\sqrt{x}} e^{- \frac{i π}{4}}) (z - i) (z + i)}$ $r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, \frac{1}{\sqrt{x}} e^{\frac{i π}{4}}) + R e s (φ, - \frac{1}{\sqrt{x}} e^{- \frac{i π}{4}}) + R e s (φ, i)}$ $. . . b e c o n t i n u e d . . . .$
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