Charles Kingsley

2022-01-02

How can I find the principal value of the following?
$PV{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{e}^{ix}}{x\left({x}^{2}+x+1\right)}dx$

### Answer & Explanation

Vasquez

With

$\begin{array}{}p=\frac{-1+\sqrt{3}i}{2}:\\ P.V.{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{e}^{ix}}{x\left({x}^{2}+x+1\right)}dx\\ ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{e}^{ix}}{{x}^{2}+x+1}\left[\frac{1}{x+i{0}^{+}}+i\pi \delta \left(x\right)\right]dx\\ ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{e}^{ix}}{\left(x+i{0}^{+}\right)\left(x-p\right)\left(x-\stackrel{―}{p}\right)}dx+i\pi \\ =i\pi +2\pi i\frac{{e}^{ip}}{p\left(p-\stackrel{―}{p}\right)}=i\pi +2\pi i\frac{{e}^{-i/2-\sqrt{3}/2}}{\left[\left(-1+\sqrt{3}i\right)/2\right]\left(i\sqrt{3}\right)}\\ =i\pi +2\pi \frac{\sqrt{3}}{3}{e}^{-i/2-\sqrt{3}/2}\frac{-1-\sqrt{3}i}{2}\\ =\left(i-\frac{\sqrt{3}+3i}{3}{e}^{-i/2-\sqrt{3}/2}\right)\pi \approx -1.3030+2.3477i\end{array}$

nick1337

You can use the residue theorem. Consider the complex contour integral
${\oint }_{C}dz\frac{{e}^{iz}}{z\left(1+z+{z}^{2}\right)}$
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
$PV{\int }_{-R}^{R}dx\frac{{e}^{ix}}{x\left(1+x+{x}^{2}\right)}+iϵ{\int }_{\pi }^{0}d\varphi {e}^{i\varphi }\frac{{e}^{iϵ{e}^{o\varphi }}}{ϵ{e}^{i\theta }\left(1+ϵ{e}^{i\varphi }+{ϵ}^{2}{e}^{i2\varphi }}$
As $R\to \mathrm{\infty }$, the third integral vanishes as $1/{R}^{3}$. As $ϵ\to 0$, the second integral becomes $-i\pi$. By the residue theorem, the contour integral is equal to $i2\pi$ times the residue of any poles inside C. The poles of the integrand are at
${z}^{2}+z+1=0⇒z=\frac{1±i\sqrt{3}}{2}$
Only ${z}_{+}$ is inside C, with residue
$\frac{{e}^{-i/2}{e}^{-\sqrt{3}/2}}{{e}^{i2\pi /3}\left(i\sqrt{3}\right)}$
Thus,
$PV{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}dx\frac{{e}^{ix}}{x\left(1+x+{x}^{2}\right)}=\frac{2\pi }{\sqrt{3}}{e}^{-\sqrt{3}/2}{e}^{-i\left(2\pi /3+1/2\right)}+i\pi$

karton

$PV{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{e}^{ix}}{x\left(1+x+{x}^{2}\right)}dx=PV{\int }_{0}^{+\mathrm{\infty }}\left(\frac{2i\left(1+{x}^{2}\right)}{1+{x}^{2}+{x}^{4}}\cdot \frac{\mathrm{sin}x}{x}-\frac{2}{1+{x}^{2}+{x}^{4}}\mathrm{cos}\left(x\right)\right)$
and both integrals giving the real and imaginary part can be computed by using the Laplace transform. We have:
${\int }_{0}^{+\mathrm{\infty }}\frac{2\mathrm{cos}x}{1+{x}^{2}+{x}^{4}}dx=\frac{\pi }{3}{e}^{-\frac{\sqrt{3}}{2}}\left(3\mathrm{sin}\frac{1}{2}+\sqrt{3}\mathrm{cos}\frac{1}{2}\right)$
and:
${\int }_{0}^{+\mathrm{\infty }}\frac{1+{x}^{2}}{1+{x}^{2}+{x}^{4}}\cdot \frac{\mathrm{sin}x}{x}dx=\frac{\pi }{6}\left(3+2\sqrt{3}{e}^{-\frac{\sqrt{3}}{2}}\mathrm{sin}\frac{1}{2}\right)$

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