calculate ∫_0 ^∞ ((sin(2cos(x^2 +1)))/(1+x^2 ))dx


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Question Number 48172 by Abdo msup. last updated on 20/Nov/18

$c a l c u l a t e \int_{0}^{\infty} \frac{s i n (2 c o s (x^{2} + 1))}{1 + x^{2}} d x$
Commented by Abdo msup. last updated on 25/Nov/18
$let I=∫_0 ^∞ ((sin(2cos(x^2 +1))dx)/(1+x^2 )) we have 2I = ∫_(−∞) ^(+∞) ((sin(2cos(x^2 +1)))/(1+x^2 ))dx =Im( ∫_(−∞) ^(+∞) (e^(2icos(x^(2 ) +1)) /(1+x^2 )))let ϕ(z) =(e^(2i cos(x^2 +1)) /(1+z^2 )) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) =2iπ (e^(2icos(i^2 +1)) /(2i)) =π e^(2i) =π {cos(2)+isin(2)} ⇒2I =πsin(2) ⇒ I =(π/2)sin(2) .$
$l e t I = \int_{0}^{\infty} \frac{s i n (2 c o s (x^{2} + 1)) d x}{1 + x^{2}} w e h a v e$ $2 I = \int_{- \infty}^{+ \infty} \frac{s i n (2 c o s (x^{2} + 1))}{1 + x^{2}} d x$ $= I m (\int_{- \infty}^{+ \infty} \frac{e^{2 i c o s (x^{2} + 1)}}{1 + x^{2}}) l e t φ (z) = \frac{e^{2 i c o s (x^{2} + 1)}}{1 + z^{2}} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i) = 2 i π \frac{e^{2 i c o s (i^{2} + 1)}}{2 i}$ $= π e^{2 i} = π {c o s (2) + i s i n (2)} \Rightarrow 2 I = π s i n (2) \Rightarrow$ $I = \frac{π}{2} s i n (2) .$
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