How can I find the principal value of the following? PV∫−∞∞eixx(x2+x+1)dx

Question

How can I find the principal value of the following?  PV∫−∞∞eixx(x2+x+1)dx

Vasquez · Accepted Answer

With  p=−1+3i2:P.V.∫−∞∞eixx(x2+x+1)dx=∫−∞∞eixx2+x+1[1x+i0++iπδ(x)]dx=∫−∞∞eix(x+i0+)(x−p)(x−p―)dx+iπ=iπ+2πieipp(p−p―)=iπ+2πie−i/2−3/2[(−1+3i)/2](i3)=iπ+2π33e−i/2−3/2−1−3i2=(i−3+3i3e−i/2−3/2)π≈−1.3030+2.3477i

nick1337 · Accepted Answer

You can use the residue theorem. Consider the complex contour integral ∮Cdzeizz(1+z+z2) where C is a semicircle of radius R in the upper half plane with a small semicircular detour about the origin into the upper half plane of radius ϵ. Then the contour integral is equal to PV∫−RRdxeixx(1+x+x2)+iϵ∫π0dϕeiϕeiϵeoϕϵeiθ(1+ϵeiϕ+ϵ2ei2ϕ As R→∞, the third integral vanishes as 1/R3. As ϵ→0, the second integral becomes −iπ. By the residue theorem, the contour integral is equal to i2π times the residue of any poles inside C. The poles of the integrand are at z2+z+1=0⇒z=1±i32 Only z+ is inside C, with residue e−i/2e−3/2ei2π/3(i3) Thus, PV∫−∞∞dxeixx(1+x+x2)=2π3e−3/2e−i(2π/3+1/2)+iπ

karton · Accepted Answer

PV∫−∞+∞eixx(1+x+x2)dx=PV∫0+∞(2i(1+x2)1+x2+x4⋅sin⁡xx−21+x2+x4cos⁡(x)) and both integrals giving the real and imaginary part can be computed by using the Laplace transform. We have: ∫0+∞2cos⁡x1+x2+x4dx=π3e−32(3sin⁡12+3cos⁡12) and: ∫0+∞1+x21+x2+x4⋅sin⁡xxdx=π6(3+23e−32sin⁡12)

How can I find the principal value of the following? PV\int_{-\infty}^\infty\frac{e^{ix}}{x(x^2+x+1)}dx

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New Questions in Calculus 1