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Exercise :

Show that :

$$\int_{-\infty}^{\infty} \frac{x-1}{x^5-1} dx = \frac{4\pi}{5}\sin\bigg(\frac{2\pi}{5}\bigg)$$

Attempt :

$$\int_{-\infty}^{\infty} \frac{x-1}{x^5-1} dx = \int_{-\infty}^{\infty} \frac{1}{x^4 + x^3 + x^2 + x + 1}dx $$

$$x^4 + x^3 + x^2 + x + 1=0 \Leftrightarrow x = \{-(-1)^{1/5},(-1)^{2/5},-(-1)^{3/5},(-1)^{4/5} \}$$

So, the function $f(z) = \frac{1}{x^4 + x^3 + x^2 + x + 1}$ has poles at the points :

$$\{-(-1)^{1/5},(-1)^{2/5},-(-1)^{3/5},(-1)^{4/5} \}$$

Now, I know you have to integrate through a closed curve $C$ and on a line $γ_R$ and then continue on with residues for the poles that reside in this curve, but I am stuck on how to apply it here and I also miss it a bit on how to split the integral for the curve and the line. Most examples I've saw get simpler due to the even function trick, but that cannot be applied here.

I would really appreciate a thorough solution and explanation, since I've just started working on generalized integrals and I have to clear my mind on them.

Thanks for your time !

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    $\begingroup$ $$\int_{-\infty}^{+\infty}\frac{dz}{z^4+z^3+z^2+z+1}$$ equals $2\pi i$ times the sum of residues at the poles in the upper half-plane. $\endgroup$ – Jack D'Aurizio Jun 19 '17 at 8:18
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    $\begingroup$ Your radius should be greater $1$. Otherwise, the singularities are on your curve. But then your right. You just need to compute your residues. $\endgroup$ – Mundron Schmidt Jun 19 '17 at 8:40
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    $\begingroup$ @Lelouch.D.Light It's already solved, but thanks ! No point as well to complicate it this much, since it's not a general form. $\endgroup$ – Rebellos Jun 19 '17 at 9:05
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    $\begingroup$ yeah agree jack's method using residues is lot easier $\endgroup$ – Darthsid1995 Jun 19 '17 at 9:07
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    $\begingroup$ Possible duplicate of An integral of a rational function on the real line $\endgroup$ – Simply Beautiful Art Sep 29 '17 at 20:06
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Using the upper part of the complex plane we get:

$\displaystyle \int_{-\infty}^{+\infty}\frac{x-1}{x^5-1}dx=i2\pi Res(\frac{x-1}{x^5-1},e^{i2\pi/5})+ i2\pi Res(\frac{x-1}{x^5-1},e^{i4\pi/5})$

$\displaystyle = i2\pi\frac{x-1}{x^5-1}(x-e^{i2\pi/5})|_{x\to e^{i2\pi/5}}+ i2\pi\frac{x-1}{x^5-1}(x-e^{i4\pi/5})|_{x\to e^{i4\pi/5}} $

$\displaystyle =\frac{\pi}{4}\frac{1}{\sin(\frac{\pi}{5})\sin(\frac{3\pi}{5})\sin(\frac{4\pi}{5})}=\frac{4\pi}{5}\sin\frac{2\pi}{5}$

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Brute force also helps: $$\int\limits_{-\infty}^{+\infty}\frac{x-1}{x^5-1}dx=\int\limits_{-\infty}^{+\infty}\frac{1}{x^4+x^3+x^2+x+1}dx=$$ $$=\int\limits_{-\infty}^{+\infty}\frac{1}{x^2\left(x^2+\frac{1}{x^2}+x+\frac{1}{x}+1\right)}dx=\int\limits_{-\infty}^{+\infty}\frac{1}{x^2\left(\left(x+\frac{1}{x}\right)^2+x+\frac{1}{x}-1\right)}dx=$$ $$=\int\limits_{-\infty}^{+\infty}\frac{1}{x^2\left(x+\frac{1}{x}-\frac{-1+\sqrt5}{2}\right)\left(x+\frac{1}{x}-\frac{-1-\sqrt5}{2}\right)}dx=$$ $$=\int\limits_{-\infty}^{+\infty}\frac{1}{\left(x^2-\frac{\sqrt5-1}{2}x+1\right)\left(x^2+\frac{\sqrt5+1}{2}x+1\right)}dx=$$ $$=\frac{1}{\sqrt5}\int\limits_{-\infty}^{+\infty}\left(\frac{x+\frac{\sqrt5+1}{2}}{x^2+\frac{\sqrt5+1}{2}x+1}-\frac{x-\frac{\sqrt5-1}{2}}{x^2-\frac{\sqrt5-1}{2}x+1}\right)dx=$$ $$=\frac{1}{\sqrt5}\int\limits_{-\infty}^{+\infty}\left(\frac{\frac{\sqrt5+1}{4}}{x^2+\frac{\sqrt5+1}{2}x+1}-\frac{\frac{\sqrt5-1}{4}}{x^2-\frac{\sqrt5-1}{2}x+1}\right)dx=$$ $$=\frac{\sqrt5+1}{4\sqrt5}\int\limits_{-\infty}^{+\infty}\frac{1}{\left(x+\frac{\sqrt5+1}{4}\right)^2+1-\frac{6+2\sqrt5}{16}}dx-\frac{\sqrt5-1}{4\sqrt5}\int\limits_{-\infty}^{+\infty}\frac{1}{\left(x-\frac{\sqrt5-1}{4}\right)^2+1-\frac{6-2\sqrt5}{16}}dx=$$ $$=\pi\left(\frac{\sqrt5+1}{4\sqrt5\sqrt{\frac{10-2\sqrt5}{16}}}-\frac{\sqrt5-1}{4\sqrt5\sqrt{\frac{10+2\sqrt5}{16}}}\right)=\frac{\pi}{\sqrt5}\left(\frac{\cos36^{\circ}}{\sin36^{\circ}}-\frac{\sin18^{\circ}}{\cos18^{\circ}}\right)=$$ $$=\frac{\pi\cos54^{\circ}}{\sqrt5\sin36^{\circ}\cos18^{\circ}}=\frac{\pi}{\sqrt5\cos18^{\circ}}.$$

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