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