# Contour integral of $\int_{-\infty}^{\infty}\frac{xdx}{(x-1)^{4}-1}$

I'm having a little trouble solving this contour integral. I've found the singularities to be at 0, 2, and $1\pm i$. I'm not quite sure how to evaluate the integral when there are two poles on the contour instead of just one. Any help is appreciated.

• Calculate the residues ... Cauchy's Residue theorem etc... Feb 7 '17 at 20:30
• Maybe, you're asking for the $\textit{Principal Value}$. Feb 7 '17 at 20:31
• @JohnPage Just to clarify there is an $x$ in the numerator ? ... and the in denominator, that will cancel so your integral can be rewritten as $\int _{-\infty}^{\infty} \frac{dx}{(x-2)(x^2-2x+2)}$ So there is only one singularity on the contour. Feb 7 '17 at 21:22
• There is a subtle but important difference between $\text{PV}\int_{-\infty}^{+\infty}\frac{dx}{x}=0$ and $\int_{-\infty}^{+\infty}\frac{dx}{x}$, that simply does not exist. The same applies to the question. Feb 7 '17 at 21:22


Assuming the question asks for the Principal Value $\pars{~\mrm{P.V.}~}$:

\begin{align} &\mrm{P.V.}\int_{-\infty}^{\infty}{x\,\dd x \over \pars{x - 1}^{4} - 1} \\[5mm] \stackrel{\mrm{def.}}{=} &\ \lim_{\epsilon \to 0^{+}}\bracks{% \int_{-\infty}^{-\epsilon}{x\,\dd x \over \pars{x - 1}^{4} - 1} + \int_{\epsilon}^{2 - \epsilon}{x\,\dd x \over \pars{x - 1}^{4} - 1} + \int_{2 + \epsilon}^{\infty}{x\,\dd x \over \pars{x - 1}^{4} - 1}} \\[5mm] = &\ \lim_{\epsilon \to 0^{+}}\bracks{% \int_{-\infty}^{-1 - \epsilon}{x + 1 \over x^{4} - 1}\,\dd x + \int_{-1 + \epsilon}^{1 - \epsilon}{x + 1 \over x^{4} - 1}\,\dd x + \int_{1 + \epsilon}^{\infty}{x + 1 \over x^{4} - 1}\,\dd x} \\[5mm] = &\ \lim_{\epsilon \to 0^{+}}\bracks{% \int_{-\infty}^{-1 - \epsilon}{\dd x \over \pars{x - 1}\pars{x^{2} + 1}}\,\dd x + 2\int_{0}^{1 - \epsilon}{\dd x \over x^{4} - 1} - \int_{0}^{1/\pars{1 + \epsilon}}{x \over \pars{x - 1}\pars{x^{2} + 1}}\,\dd x} \\[1cm] = &\ -\int_{1}^{\infty}{\dd x \over \pars{x + 1}\pars{x^{2} + 1}} \\[5mm] + &\ \lim_{\epsilon \to 0^{+}}\braces{% \int_{0}^{1 - \epsilon}\bracks{% {2 \over x^{4} - 1} - {x \over \pars{x - 1}\pars{x^{2} + 1}}}\,\dd x - \int_{1 - \epsilon}^{1/\pars{1 + \epsilon}} {x \over \pars{x - 1}\pars{x^{2} + 1}}\,\dd x} \\[1cm] & = -\int_{1}^{\infty}{\dd x \over \pars{x + 1}\pars{x^{2} + 1}} - \int_{0}^{1}{x + 2 \over \pars{x + 1}\pars{x^{2} + 1}}\,\dd x \\[3mm] - & \ \underbrace{% \lim_{\epsilon \to 0^{+}}\int_{1 - \epsilon}^{1/\pars{1 + \epsilon}} {x \over \pars{x - 1}\pars{x^{2} + 1}}\,\dd x}_{\ds{=\ 0}} = \bbx{-\ds{\pi \over 2}} \end{align}

I left to you the integral evaluations and the limit evaluation.

• @JackD'Aurizio I checked it with $\texttt{Mathematica}$. $$\texttt{Integrate[x/((x - 1)^4 - 1), {x, -Infinity, Infinity}, PrincipalValue -> True]}$$ Feb 7 '17 at 21:32
• Sorry, my bad: I previously simplified $\frac{x+1}{x^2-1}$ as $\frac{1}{x+1}$ instead of $\frac{1}{x-1}$, now fixed. Feb 7 '17 at 21:35
• @JackD'Aurizio It still works: $-\pi/2$. Thanks. Feb 7 '17 at 21:39
• Sure, sure, you are right. It was me not double-checking my computations. (+1) to you. Feb 7 '17 at 21:40

$$\text{PV}\int_{-\infty}^{+\infty}\frac{x}{(x-1)^4-1}\,dx = \text{PV}\int_{-\infty}^{+\infty}\frac{x+1}{x^4-1}\,dx = \text{PV}\int_{-\infty}^{+\infty}\frac{dx}{(x-1)(x^2+1)}$$ equals, by partial fraction decomposition, $$\frac{1}{2}\,\text{PV}\int_{-\infty}^{+\infty}\frac{dx}{x-1}-\frac{1}{2}\,\text{PV}\int_{-\infty}^{+\infty}\frac{x+1}{x^2+1}\,dx = \color{red}{-\frac{\pi}{2}}.$$

• Ah, good use shifting the integral and PFD. Feb 7 '17 at 21:37