Prove that $\int_{-\infty}^{\infty} \frac{\sin x}{(1-x+x^2)^2} dx = \frac{2\pi(\sqrt 3+2)}{3\sqrt 3}e^{\frac{-\sqrt 3}{2}} \sin \frac{1}{2}$

by anirudh b   Last Updated January 14, 2018 13:20 PM

Prove that $\int_{-\infty}^{\infty} \frac{\sin x}{(1-x+x^2)^2} dx = \frac{2\pi(\sqrt 3+2)}{3\sqrt 3}e^{\frac{-\sqrt 3}{2}} \sin \frac{1}{2}$

My attempt: consider $C:$ semi-circle with center as origin with Radius $R\to \infty$ in upper XY-plane

Now given integral = Imaginary part of $\int_{C:|z| <R} \frac{e^{iz}}{(1-z+z^2)^2} dz$

THis has 2 poles of order 2 :$\alpha = e^{i\frac{\pi}{3}}=\frac{1+i \sqrt 3}{2}, \beta = e^{i\frac{5\pi}{3}}=\frac{-1+i \sqrt 3}{2}$

Note that we have $\alpha - \beta = 1$

calculating residue at $[z=\alpha]$ = $Res_{[z = \alpha]} f(z) = \frac{d}{dz}\frac{(z-\alpha)^2 e^{iz}}{(1-z+z^2)^2} = \frac{d}{dz}\frac{ e^{iz}}{(z-\beta)^2} = \frac{e^{iz}i}{(z-\beta)^2}- \frac{2e^{iz}}{(z-\beta)^3}=e^{i \alpha}(i-2) = e^{-\frac{\sqrt 3}{2}} e^{\frac{i}{2}}(i-2)$ ---------------->(1)

Similarly $Res_{[z = \beta]} f(z) = \frac{d}{dz}\frac{(z-\beta)^2 e^{iz}}{(1-z+z^2)^2} = \frac{d}{dz}\frac{ e^{iz}}{(z-\alpha)^2} = \frac{e^{iz}i}{(z-\alpha)^2}- \frac{2e^{iz}}{(z-\alpha)^3}=e^{i \alpha}(i-2) = e^{-\frac{\sqrt 3}{2}} e^{\frac{-i}{2}}(i+2)$

Integral along C = 0;

value of given integral = $2 \pi i$*(Sum of residues) = $-4\pi e^{-\frac{\sqrt 3}{2}}(\cos \frac{1}{2}-2\sin\frac{1}{2})$

But the answer i got is different from what is given.

When i saw solution,

he calculated residue using laurent expansion. let $z-\alpha = t, z = t+\alpha$

$f(\alpha +t)=\frac{e^{i(\alpha+t)}}{(t^2+(2\alpha -1)t+\alpha^2-\alpha +1)^2} = \frac{e^{i(\alpha+t)}}{(t^2+(2\alpha -1)t)^2} = \frac{e^{i(\alpha+t)}}{t^2(2\alpha-1)^2t^2} (1+\frac{t}{2\alpha-1})^{-2}$

In above, coefficient of 1/t = residue at $\alpha = \frac{e^{-\frac{\sqrt 3}{2}} e^{i/2}(\sqrt 3+2)}{i3\sqrt 3}$ ---------------->(2)

can you pls point me to why such difference i am getting? I have cross checked my solution but could not find any error. Pls enlighten

Also pls tell me is there any alternative way to solve this problem.

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$$\int_{-\infty}^{+\infty}\frac{e^{ix}}{(1-x+x^2)^2}\,dx = 2\pi i\operatorname*{Res}_{z=\frac{1+\sqrt{-3}}{2}}\frac{e^{iz}}{(1-z+z^2)^2}=\frac{2\pi}{9} \left(3+2 \sqrt{3}\right) e^{-\frac{\sqrt{3}}{2}}e^{\frac{i}{2}}$$ and by considering the imaginary part of the RHS the claim is proved. When you compute the LHS through the residue theorem, only the pole in the upper half plane matters, since the ML lemma is not fulfilled in the lower half plane, so you have to consider a semi-circular contour.

Jack D'Aurizio
January 14, 2018 13:15 PM

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