# Evaluate the Cauchy Principal Value of $\int_{-\infty}^{\infty} \frac{\sin x}{x(x^2-2x+2)}dx$

Evaluate the Cauchy Principal Value of $\int_{-\infty}^\infty \frac{\sin x}{x(x^2-2x+2)}dx$

so far, i have deduced that there are poles at $z=0$ and $z=1+i$ if using the upper half plane. I am considering the contour integral $\int_C \frac{e^{iz}}{z(z^2-2z+2)}dz$ I dont know how to input graphs here but it would be intended at the origin with a bigger R, semi-circle surrounding that. So, I have four contour segments.

$\int_{C_R}+\int_{-R}^{-r}+\int_{-C_r}+\int_r^R=2\pi i\operatorname{Res}[f(z)e^{iz}, 1+i]+\pi iRes[f(z)e^{iz},o]$ I think. Ok, so here is where I get stuck. Im not sure how to calculate the residue here, its not a higher pole, so not using second derivatives, not Laurent series. Which method do I use here?

• $0$ is a removable singularity. – hjhjhj57 Dec 10 '14 at 21:15
• No PV necessary, poles are away from real axis. – Ron Gordon Dec 10 '14 at 21:18
• so then, just find the residues and mulitply by $2\pi i$ and @hjhjhj57, since 0 is a removable singularity, I do not need to calculate the residue of that one? just $1+i$? – cele Dec 10 '14 at 22:08
• @cele But when you replace $\sin x$ by $e^{ix}$, you introduce a pole at $0$, and then you need to consider the principal value. Apart from the residue at $1+i$, you also need to look at what happens on the small semicirlce around $0$ when its radius tends to $0$. – Daniel Fischer Dec 10 '14 at 22:28
• @cele That funny looking $\mathfrak{I}$ means "imaginary part of". – user_of_math Dec 11 '14 at 2:49

Write $\sin{x} = (e^{i x}-e^{-i x})/(2 i)$. Then consider the integral

$$PV \oint_{C_{\pm}} dx \frac{e^{\pm i z}}{z (z^2-2 z+2)}$$

where $C_{\pm}$ is a semicircular contour of radius $R$ in the upper/lower half plane with a semicircular detour into the upper/lower half plane of radius $\epsilon$. For $C_{+}$, we have

$$PV \oint_{C_{+}} dz \frac{e^{i z}}{z (z^2-2 z+2)} = \int_{-R}^{-\epsilon} dx \frac{e^{i x}}{x (x^2-2 x+2)}+ i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{e^{i \epsilon e^{i \phi}}}{\epsilon e^{i \phi} (\epsilon^2 e^{i 2 \phi} - 2 \epsilon e^{i \phi}+2)} \\+ \int_{\epsilon}^R dx \frac{e^{i x}}{x (x^2-2 x+2)}+ i R \int_0^{\pi} d\theta \, e^{i \theta} \frac{e^{i R e^{i \theta}}}{R e^{i \theta} (R^2 e^{i 2 \theta} - 2 R e^{i \theta}+2)}$$

For $C_-$, we have

$$PV \oint_{C_{-}} dz \frac{e^{-i z}}{z (z^2-2 z+2)} = \int_{-R}^{-\epsilon} dx \frac{e^{-i x}}{x (x^2-2 x+2)}+ i \epsilon \int_{-\pi}^0 d\phi \, e^{i \phi} \frac{e^{-i \epsilon e^{i \phi}}}{\epsilon e^{i \phi} (\epsilon^2 e^{i 2 \phi} - 2 \epsilon e^{i \phi}+2)} \\+ \int_{\epsilon}^R dx \frac{e^{-i x}}{x (x^2-2 x+2)}- i R \int_0^{\pi} d\theta \, e^{-i \theta} \frac{e^{-i R e^{-i \theta}}}{R e^{-i \theta} (R^2 e^{-i 2 \theta} - 2 R e^{-i \theta}+2)}$$

In both cases, we take the limits as $R \to \infty$ and $\epsilon \to 0$. Note that, in both cases, the respective fourth integrals have a magnitude bounded by

$$\frac{2}{R^2} \int_0^{\pi/2} d\theta \, e^{-R \sin{\theta}} \le \frac{2}{R^2} \int_0^{\pi/2} d\theta \, e^{-2 R \theta/\pi}\le \frac{\pi}{R^3}$$

The respective second integrals of $C_{\pm}$, on the other hand, become equal to $\mp i \frac{\pi}{2}$. Thus,

$$PV \oint_{C_{\pm}} dz \frac{e^{\pm i z}}{z (z^2-2 z+2)} = PV \int_{-\infty}^{\infty} dx \frac{e^{\pm i x}}{x (x^2-2 x+2)} \mp i \frac{\pi}{2}$$

On the other hand, the respective contour integrals are each equal to $\pm i 2 \pi$ times the sum of the residues of the poles inside their contours. (For $C_-$, there is a negative sign because the contour was traversed in a clockwise direction.) The poles of the denominator are at $z_{\pm}=1 \pm i$. Thus,

$$PV \int_{-\infty}^{\infty} dx \frac{e^{\pm i x}}{x (x^2-2 x+2)} \mp i \frac{\pi}{2} = \pm i 2 \pi \frac{e^{\pm i (1 \pm i)}}{(1 \pm i) (2) (\pm i)}$$

Taking the difference between the two results and dividing by $2 i$, we get that

$$\int_{-\infty}^{\infty} dx \frac{\sin{x}}{x (x^2-2 x+2)} = \frac{\pi}{2} \left (1+\frac{\sin{1}-\cos{1}}{e} \right )$$

Note that we may drop the $PV$ because the difference between the integrals removes the pole at the origin.


Zeros of $\ds{x^{2} - 2x + 2}$ are given by $\ds{r_{\pm} = 1 \pm \ic=\root{2}\exp\pars{\pm\ic\,{\pi \over 4}}}$.

\begin{align}&\color{#66f}{\large% \int_{-\infty}^{\infty}{\sin\pars{x} \over x\pars{x^{2} - 2x + 2}}\,\dd x} =\int_{-\infty}^{\infty}{1 \over x^{2} - 2x + 2}\,\ \overbrace{\half\int_{-1}^{1}\expo{\ic kx}\,\dd k} ^{\ds{=\ \dsc{\sin\pars{x} \over x}}}\ \,\dd x \\[5mm]&=\half\,\Re\int_{-1}^{1}\int_{-\infty}^{\infty} {\expo{\ic\verts{k}x} \over x^{2} - 2x + 2}\,\dd x\,\dd k =\half\,\Re\int_{-1}^{1}2\pi\ic\lim_{x\ \to\ r_{+}}\bracks{\pars{x - r_{+}} {\expo{\ic\verts{k}x} \over x^{2} - 2x + 2}}\,\dd k \\[5mm]&=\pi\,\Re\int_{-1}^{1} {\expo{\ic\verts{k}r_{+}} \over 2r_{+} - 2}\,\dd k ={\pi \over 2}\,\Re\bracks{% \ic\int_{-1}^{1}{\expo{\pars{-1 + \ic}\verts{k}} \over \ic}\,\dd k} =\pi\,\Re\bracks{% \int_{0}^{1}\expo{\pars{-1 + \ic}k}\,\dd k} \\[5mm]&=\pi\,\Re\bracks{\expo{\pars{-1 + \ic}} - 1\over -1 + \ic} =\pi\,\Re{\bracks{\expo{-1}\cos\pars{1} - 1 + \expo{-1}\sin\pars{1}\ic} \pars{-1 - \ic}\over 2} \\[5mm]&={\pi \over 2}\bracks{-\expo{-1}\cos\pars{1} + 1 + \expo{-1}\sin\pars{1}} =\color{#66f}{\large% {\pi \over 2}\bracks{1 + {\sin\pars{1} - \cos\pars{1} \over \expo{}}}} \end{align}