I'm not sure how to proceed with the following integral: $$I=\int_{1}^{\infty}\frac{-4}{(2 \cos{x} - 2) x^3}\ \mathrm dx.$$

Mathematica could not find a closed form solution for it and I really have no idea how to go about computing it.

  • 2
    $\begingroup$ It's divergent, $\cos x - 1$ has zeros (of order $2$) at $2k\pi$, $k \in \mathbb{N}$. Since the integrand is non-negative, we could say $I = +\infty$. $\endgroup$ – Daniel Fischer May 31 '16 at 15:16
  • $\begingroup$ @DanielFischer Ah, of course. Thank you. $\endgroup$ – TreFox May 31 '16 at 15:18
  • $\begingroup$ Where did this integral arise? As @DanielFischer points out, as written it is not converging, since it is a multiple of $\int_1^\infty \frac{dx}{(\cos x-1)x^3}$ which diverges (due to $\cos x- 1$). Is there a chance you made a mistake writing it? $\endgroup$ – Clement C. May 31 '16 at 15:18
  • $\begingroup$ @ClementC. It just arose with me playing with some numbers and trying to teach myself limits and infinite series. The integrand is equal to $\frac{1}{n^3 \sin{n}^2}$. I didn't really except a closed form, although I can't believe I didn't realize why the integral was divergent. I'm new to all this :P $\endgroup$ – TreFox May 31 '16 at 16:21

$$I=\int_{1}^{\infty}\frac{-4}{(2 \cos{x} - 2) x^3}\ \mathrm dx$$

The integral here is divergent.

In the denominator, $\cos x -1 = 0$ has zeroes at $x=2n\pi, \,\,\,\, n=0,1,2,..$

So the integrand has singularities at $x=2n\pi, \,\,\,\, n=0,1,2,..$ in the given interval and quite obviously $I \to \infty$.


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