Cauchy Principal Value of $\int_{-\infty}^\infty \frac{e^{ipx}}{x^4-1}\,\mathrm{d}x$

Question

I have to find the Cauchy Principal Value of $$ \int_{-\infty}^{\infty}\frac{\mathrm{e}^{\mathrm{i}px}}{x^{4} - 1}\,\mathrm{d}x $$ There are 4 simple poles at $x=1,-1,i,-i$ so I'm not sure what the best contour to use is because a $D$ shaped contour doesn't work as all the poles are on the axes. I should add there are 2 cases to consider with $p>0$ and $p<0$

Answer 1

The Cauchy Principal Value of the integral of interest is given by

$$\begin{align} \text{PV}\left(\int_{-\infty}^\infty \frac{e^{ipx}}{x^4-1}\,dx\right)&=\lim_{\varepsilon\to 0^+}\left(\int_{-\infty}^{-1-\varepsilon} \frac{e^{ipx}}{x^4-1}\,dx\int_{-1+\varepsilon}^{1-\varepsilon} \frac{e^{ipx}}{x^4-1}\,dx\int_{1+\varepsilon}^\infty \frac{e^{ipx}}{x^4-1}\,dx\right) \end{align}$$

We shall analyze the case for which $p>0$.

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METHODOLOGY $1$:

Now, take $R>1$. If we evaluate the contour integral $\displaystyle \oint_C \frac{e^{ipz}}{z^4-1}\,dz$ where the contour $C$ is comprised of $(i)$ the real line segments from $-R$ to $-1-\varepsilon$, $(ii)$ the semi-circular arc in the third quadrant centered at $-1$ with radius $\varepsilon$ from $-1-\varepsilon$ to $-1+\varepsilon$, $(iii)$ the straight line segment from $-1+\varepsilon$ to $1-\varepsilon$, $(iv)$ the semi-circular arc in the first quadrant centered at $1$ with radius $\varepsilon$ from $1-\varepsilon$ to $1+\varepsilon$, $(v)$ a straight line segment from $1+\varepsilon$ to $R$, and $(vi)$ a semicircular arc from $R$ to $-R$, then the Residue theorem guarantees that

$$\oint_C \frac{e^{ipz}}{z^4-1}\,dz=2\pi i \text{Res}\left(\frac{e^{ipz}}{z^4-1}\,dz, z=i\right)=-\frac{\pi}{2}e^{-p}$$

As $R\to \infty$ and $\varepsilon\to 0^+$, we see that

$$\lim_{R\to\infty\\\varepsilon\to 0^+}\oint_C \frac{e^{ipz}}{z^4-1}\,dz=\text{PV}\left(\int_{-\infty}^\infty \frac{e^{ipx}}{x^4-1}\,dx\right)+\frac\pi2\sin(p)$$

Putting it together, we find that

$$\text{PV}\left(\int_{-\infty}^\infty \frac{e^{ipx}}{x^4-1}\,dx\right)=-\frac\pi2\left(\sin(p)+e^{-p}\right)$$

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METHODOLOGY $2$:

Using partial fraction expansion, we can write

$$\frac{e^{ipx}}{x^4-1}=\frac{e^{ip}}4 \frac{e^{ip(x-1)}}{x-1}-\frac{e^{-ip}}4 \frac{e^{ip(x+1)}}{x+1}+\frac{ie^{-p}}4 \frac{e^{ip(x-i)}}{x-i}-\frac{ie^{p}}4 \frac{e^{ip(x+i)}}{x+i}$$

Then, we have

$$\begin{align} \text{PV}\left(\int_{-\infty}^\infty \frac{e^{ipx}}{x^4-1}\,dx\right)&=\frac{e^{ip}}4 \text{PV}\left(\int_{-\infty}^\infty \frac{e^{ip(x-1)}}{x-1}\,dx\right)\\\\ &-\frac{e^{-ip}}4\text{PV}\left(\int_{-\infty}^\infty \frac{e^{ip(x+1)}}{x+1}\,dx\right)\\\\ &+\frac{ie^{-p}}4\int_{-\infty}^\infty \frac{e^{ip(x-i)}}{x-i}\,dx\\\\ &-\frac{ie^{p}}4\int_{-\infty}^\infty \frac{e^{ip(x+i)}}{x+i}\,dx\tag1 \end{align}$$

The Cauchy Principal values of the first two integrals on the right-hand side of $(1)$ are identical and equal to the value of the integral $\displaystyle \int_{-\infty}^\infty \frac{\sin(px)}{x}\,dx=i\pi\text{sgn}(p)$. For $p>0$ ($p<0$), the Residue Theorem guarantees that the value of the fourth (third) integral in $(4)$ is $0$, while the value of the third (fourth) integral is $2\pi i$ ($-2\pi i$).

Putting it together, we find that

$$\text{PV}\left(\int_{-\infty}^\infty \frac{e^{ipx}}{x^4-1}\,dx\right)=-\frac\pi2 \left(\sin(|p|)+e^{-|p|}\right)$$

as expected!

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{% \left.\mrm{P.V.}\int_{-\infty}^{\infty}{\expo{\ic px} \over x^{4} - 1}\,\dd x\,\right\vert_{\ p\ \in\ \mathbb{R}}} \\[5mm] \stackrel{\mrm{by\ def.}}{=}\,\,\,& \lim_{\epsilon \to 0^{+}}\bracks{% \int_{-\infty}^{- 1 - \epsilon} {\expo{\ic px} \over x^{4} - 1}\,\dd x + \int_{-1 + \epsilon}^{1 - \epsilon} {\expo{\ic px} \over x^{4} - 1}\,\dd x + \int_{1 + \epsilon}^{\infty} {\expo{\ic px} \over x^{4} - 1}\,\dd x} \\[5mm] = &\ \lim_{\epsilon \to 0^{+}}\bracks{% \int_{1 + \epsilon}^{\infty} {\expo{-\ic px} \over x^{4} - 1}\,\dd x + \int_{0}^{1 - \epsilon} {2\cos\pars{px} \over x^{4} - 1}\,\dd x + \int_{1 + \epsilon}^{\infty} {\expo{\ic px} \over x^{4} - 1}\,\dd x} \\[5mm] = &\ 2\,\Re\lim_{\epsilon \to 0^{+}}\bracks{% \int_{0}^{1 - \epsilon} {\expo{\ic\verts{p}x} \over x^{4} - 1}\,\dd x + \int_{1 + \epsilon}^{\infty} {\expo{\ic px} \over x^{4} - 1}\,\dd x} \\[5mm] = &\ 2\,\Re\lim_{\epsilon \to 0^{+}}\left\{% \int_{0}^{1 - \epsilon} {\expo{\ic\verts{p}x} \over x^{4} - 1}\,\dd x + \left.\int_{\pi}^{0}{\expo{\ic\verts{p}z} \over z^{4} - 1}\,\epsilon\expo{\ic\theta}\ic\,\dd\theta \,\right\vert_{\ z\ =\ 1 + \epsilon\exp\pars{\ic\theta}}\right. \\[2mm] &\ \phantom{2\,\Re\lim_{\epsilon \to 0^{+}}\left\{\right.} \left. +\, \int_{1 + \epsilon}^{\infty} {\expo{\ic\verts{p}x} \over x^{4} - 1}\,\dd x\right\} \label{1}\tag{1} \\[2mm] & + \underbrace{2\,\Re\lim_{\epsilon \to 0^{+}}\int_{0}^{\pi} {\expo{\ic\verts{p}}\epsilon\expo{\ic\theta}\ic \over \pars{1 + \epsilon\expo{\ic\theta}}^{4} - 1}\,\dd\theta} _{\ds{=\ -\,{1 \over 2}\,\pi\sin\pars{\verts{p}}}} \label{2}\tag{2} \end{align} $$ \begin{array}{ll} \ds{\Large\bullet} & \mbox{The (\ref{1})-term will be "}closed\mbox{" along a quarter circle in the first quadrant.} \\ \ds{\Large\bullet} & \mbox{The contribution from the arc}\ \ds{R\expo{\ic\pars{0,\pi/2}}}\ \mbox{-whith}\ \ds{R \to \infty}\mbox{- vanishes out.} \\ \ds{\Large\bullet} & \mbox{The integration along the}\ \ds{y}\mbox{-axis }\ \underline{\mbox{is not a real number}}. \\ \ds{\Large\bullet} & \mbox{However, the only additional contribution comes from} \\ & \mbox{the $\underline{indented}$ pole at}\ \ds{z = \expo{\ic\pi/2} = \ic}. Namely, \\ & \ds{-\lim_{\epsilon \to 0^{+}}\int_{\pi/2}^{-\pi/2} {\expo{\ic\verts{p}\ic} \over \pars{\ic + \epsilon\expo{\ic\theta}}^{4} - 1}\epsilon\expo{\ic\theta}\ic\dd\theta = -\,{\pi \over 4}\,\expo{-\verts{p}}} \end{array} $$ Then ( see (\ref{1}) and (\ref{2}) ), \begin{align} &\mbox{} \\ &\bbox[5px,#ffd]{% \left.\mrm{P.V.}\int_{-\infty}^{\infty}{\expo{\ic px} \over x^{4} - 1}\,\dd x\,\right\vert_{\ p\ \in\ \mathbb{R}}} = \bbx{-\,{\pi \over 2}\bracks{% \sin\pars{\verts{p}} + \expo{-\verts{p}}}} \\ & \end{align}