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Show that $ \int_{-\infty}^{\infty} \frac{x-1}{x^5-1} dx = \frac{4\pi}{5} \sin \frac{2\pi}{5} $.

How we can show that? Please help me.

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    It will be helpful to have your discussion of this integral in order for Readers to respond in terms that you are familiar with. – 2017-01-06
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    The general result here is $$\int_{-\infty}^\infty \frac{x-1}{x^{2n+1}-1}{\rm d}x = \frac{4\pi}{2n+1}\frac{\cos^2\left(\frac{\pi}{2(2n+1)}\right)}{\sin\left(\frac{2\pi}{2n+1}\right)}$$ for all integers $n\geq 1$. This can be found using exactly the same method as outlined in the answer below. – 2017-01-06
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    @Winther Hm, I shall derive that on my own as well... – 2017-01-06

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If you take the counter-clockwise contour of a closed semi-circle with radius $R$ in the complex plane with a contour along the real line from $-R$ to $+R$, we have

$$I_0=\int_{-\infty}^\infty\frac{1-x}{1-x^5}\ dx=P.V.\int_{-\infty}^\infty\frac{1-x}{1-x^5}\ dx$$

$$\lim_{R\to\infty}\oint_{\gamma_R}\frac{1-x}{1-x^5}\ dx=\lim_{R\to\infty}\int_{\text{arc}}\frac{1-x}{1-x^5}\ dx+P.V.\int_{-\infty}^\infty\frac{1-x}{1-x^5}\ dx$$

$$\lim_{R\to\infty}I_{\gamma_R}=\lim_{R\to\infty}I_{\text{arc}}+I_0$$

By Jordan's Lemma, we know that

$$\lim_{R\to\infty}I_{\text{arc}}=0$$

And by Cauchy's residue formula, we know that

$$\lim_{R\to\infty}\oint_{\gamma_r}\frac{1-x}{1-x^5}\ dx=2\pi i\left(\operatorname{Res}(\frac{1-x}{1-x^5},e^{2\pi i/5})+\operatorname{Res}(\frac{1-x}{1-x^5},e^{4\pi i/5})\right)\\=2\pi i\left(-\frac15i\sqrt{\frac12(5+\sqrt5)}\right)=\frac\pi5\sqrt{2(5+\sqrt5)}$$

And $\frac\pi5\sqrt{2(5+\sqrt5)}=\frac{4\pi}5\sin\frac{2\pi}5$, so

$$I_0=\lim_{R\to\infty}I_{\gamma_R}=\frac{4\pi}5\sin\frac{2\pi}5=\frac\pi5\sqrt{2(5+\sqrt5)}$$

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    @MusaMəmmədov No problem :D Though if you are interested in more real methods for deriving this... I really can't see it happening. :-( – 2017-01-07