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$$ P V \int\limits _{-\infty} ^{\infty} \frac {1}{x(x^4-1)} dx .$$

I would be very grateful if you could help me solving this integral.
I tried to let $ \displaystyle f(z) = \frac{1}{z(z^{4}-1)} $ and integrate around a large closed circle in the upper-half complex plane indented at the origin, with 3 poles on the real axis $ \displaystyle +1,-1,0 $ and a singularity point in $ +i $ My result was :

$$ P V \int\limits _{-\infty} ^{\infty} \frac {1}{x(x^4-1)} dx = 0 . $$

Is that correct?

My attempt:
$P V \int\limits _{-\infty} ^{\infty} \frac {1}{x(x^4-1)} dx = \oint_{\gamma} \frac{1}{z(z^4-1)} dz = 2\pi i Res(f(z),+i) + \pi i Res(f(z),+1) + \pi i Res(f(z),-1) + \pi i Res(f(z),0) = \frac{\pi i}{2} + \frac{\pi i}{4} +\frac{\pi i}{4} - \pi i = 0. $

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  • $\begingroup$ TMM has changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. $\endgroup$ – Lord_Farin Jun 10 '13 at 13:54
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    $\begingroup$ Did you make sure that for your residue, the points on the contour are piiRes and then in the contour the result is 2pi i Res $\endgroup$ – yankeefan11 Jun 10 '13 at 14:19
  • $\begingroup$ Yes,i considered the residue as you suggested. $\endgroup$ – Matt Jun 10 '13 at 14:28
  • $\begingroup$ I did the integral out myself and got zero as well. $\endgroup$ – yankeefan11 Jun 10 '13 at 14:36
  • $\begingroup$ But the integrand is not even. Is your method still working? I am also new in complex variables. $\endgroup$ – Brain Zhang Jun 10 '13 at 14:39
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I get zero, but here you will see that the only residue computation is at $z=i$. Indent about each pole on the real axis into a semicircular contour above the real axis. Each indentation is a semicircle above the real axis of radius $\epsilon$. You then get

$$PV \int_{-\infty}^{\infty} \frac{dx}{x (x^4-1)} + i \epsilon \int_{\pi}^0 d\phi \frac{e^{\phi}}{(-1+\epsilon e^{i \phi}) (-4 \epsilon e^{i \phi})} + i \epsilon \int_{\pi}^0 d\phi \frac{e^{\phi}}{(\epsilon e^{i \phi})(-1+\epsilon^4 e^{i 4 \phi})} + \\ i \epsilon \int_{\pi}^0 d\phi \frac{e^{\phi}}{(1-\epsilon e^{i \phi}) (4 \epsilon e^{i \phi})} = \frac{i 2 \pi}{i (-4 i)}$$

The RHS is $i 2 \pi$ times the residue of the pole at $z=i$. This simplifies to

$$PV \int_{-\infty}^{\infty} \frac{dx}{x (x^4-1)} - i \frac{\pi}{4} + i \pi - i \frac{\pi}{4} = i \frac{\pi}{2}$$

From this, the PV of the integral may be deduced to be zero.

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  • $\begingroup$ Really helpful explanation, thanks for your help! $\endgroup$ – Matt Jun 10 '13 at 17:59
  • $\begingroup$ @Matt: you're welcome. Please remember to accept a solution if you found it useful by clicking the checkmark to the left. $\endgroup$ – Ron Gordon Jun 10 '13 at 18:00

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