find the value of I = ∫_0 ^∞ ((cos(πx^2 ) −sin(πx^2 ))/((1+x^2 )^2 ))dx


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Question Number 43057 by maxmathsup by imad last updated on 06/Sep/18

$f i n d t h e v a l u e o f I = \int_{0}^{\infty} \frac{c o s (π x^{2}) - s i n (π x^{2})}{{(1 + x^{2})}^{2}} d x$
Commented by maxmathsup by imad last updated on 07/Sep/18
$we have I = ∫_0 ^∞ ((cos(πx^2 ))/((1+x^2 )^2 ))dx −∫_0 ^∞ ((sin(πx^2 ))/((1+x^2 )^2 ))dx =H −K we have 2H = ∫_(−∞) ^(+∞) ((cos(πx^2 ))/((1+x^2 )^2 ))dx =Re( ∫_(−∞) ^(+∞) (e^(iπx^2 ) /((1+x^2 )^2 ))dx) let ϕ(z)= (e^(iπz^2 ) /((z^2 +1)^2 )) we have ϕ(z) = (e^(iπz^2 ) /((z−i)^2 (z+i)^2 )) the poles of ϕ are +^− i(doubles) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) but Res(ϕ,i) =lim_(z→i) (1/((2−1)!)){(z−i)^2 ϕ(z)}^((1)) =lim_(z→i) { (e^(iπz^2 ) /((z+i)^2 ))}^((1)) = lim_(z→i) ((2iπz e^(iπz^2 ) (z+i)^2 −2(z+i)e^(iπz^2 ) )/((z+i)^4 )) =lim_(z→i) ((2iπ e^(iπz^2 ) (z+i) −2 e^(iπz^2 ) )/((z+i)^3 )) = ((2iπ e^(−iπ) (2i) −2 e^(−iπ) )/(−8i)) = (((−4π−2)e^(−iπ) )/(−8i)) =(((2π +1)e^(−iπ) )/(4i)) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =2iπ(((2π+1)e^(−iπ) )/(4i)) =(((2π +1)e^(−iπ) )/2) =((2π +1)/2)(−1) ⇒H =−(1/4)(2π+1) also ∫_0 ^∞ ((sin(πx^2 ))/((1+x^2 )^2 ))dx = Im ((1/2) ∫_(−∞) ^(+∞) (e^(iπx^2 ) /((1+x^2 )^2 ))) =0 ⇒ I =−(π/2) −(1/4) .$
$w e h a v e I = \int_{0}^{\infty} \frac{c o s (π x^{2})}{{(1 + x^{2})}^{2}} d x - \int_{0}^{\infty} \frac{s i n (π x^{2})}{{(1 + x^{2})}^{2}} d x = H - K w e h a v e$ $2 H = \int_{- \infty}^{+ \infty} \frac{c o s (π x^{2})}{{(1 + x^{2})}^{2}} d x = R e (\int_{- \infty}^{+ \infty} \frac{e^{i π x^{2}}}{{(1 + x^{2})}^{2}} d x) l e t φ (z) = \frac{e^{i π z^{2}}}{{(z^{2} + 1)}^{2}}$ $w e h a v e φ (z) = \frac{e^{i π z^{2}}}{{(z - i)}^{2} {(z + i)}^{2}} t h e p o l e s o f φ a r e \bar{+} i (d o u b l e s)$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i) b u t R e s (φ, i) = {l i m}_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to i} {\frac{e^{i π z^{2}}}{{(z + i)}^{2}}}^{(1)} = {l i m}_{z \to i} \frac{2 i π z e^{i π z^{2}} {(z + i)}^{2} - 2 (z + i) e^{i π z^{2}}}{{(z + i)}^{4}}$ $= {l i m}_{z \to i} \frac{2 i π e^{i π z^{2}} (z + i) - 2 e^{i π z^{2}}}{{(z + i)}^{3}} = \frac{2 i π e^{- i π} (2 i) - 2 e^{- i π}}{- 8 i}$ $= \frac{(- 4 π - 2) e^{- i π}}{- 8 i} = \frac{(2 π + 1) e^{- i π}}{4 i} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{(2 π + 1) e^{- i π}}{4 i}$ $= \frac{(2 π + 1) e^{- i π}}{2} = \frac{2 π + 1}{2} (- 1) \Rightarrow H = - \frac{1}{4} (2 π + 1) a l s o$ $\int_{0}^{\infty} \frac{s i n (π x^{2})}{{(1 + x^{2})}^{2}} d x = I m (\frac{1}{2} \int_{- \infty}^{+ \infty} \frac{e^{i π x^{2}}}{{(1 + x^{2})}^{2}}) = 0 \Rightarrow I = - \frac{π}{2} - \frac{1}{4} .$
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