I am evaluating the improper integral $\int_{-\infty}^\infty{\frac{\sin^3 x}{x^3}dx}$; I am also told to show that this is equal to its principal value, and use this fact to evaluate the integral.

I have already evaluated the integral in the past by seeing this as the imaginary part of the integral $\int_{-\infty}^\infty{\frac{-e^{3ix}+3e^{ix}+2}{4x^3}dx}$ and solving this by using a dimpled semicircle of radius $R$ centered at the origin (with the dimple a semicircle of radius $\epsilon<R$ also centered at the origin). Using this methodology provides the correct result of $3\pi/4$.

Knowing about the Cauchy Principal value is making me re-evaluate the use of contour integration to evaluate real integrals. It seems to me that the above use of the dimpled semi-circle implicitly assumed that the integral was equal to its principal value since both sides of the integral on the real line were treated with the same values $R$ and $\epsilon$.

My questions seem to be concerned with the better/more acceptable approach to dealing with this issue, then: In justifying that the principal value of the integral is equal to the integral itself, should I

  1. just realize that the integrand is analytic everywhere within the contour or everywhere but finitely-many points, at which point I can deform the contour in any way I see fit, meaning that I wouldn't need to have semicircles of different sizes on either side of the real line?
  2. or should I first use the contour integration method to evaluate just a single side of the integral, and then the second side will follow from the fact that the integrand is even?

Personally, I feel the first method makes better use of the properties of the integrand, while the second is not as generalizable since for non-even integrands this would boil down to just carrying out the contour integration on both sides separately, which takes a little more work and doesn't really seem illuminating at all.

  • $\begingroup$ Perhaps this might help: math.stackexchange.com/questions/406939/… $\endgroup$ – Ron Gordon Mar 1 '14 at 16:07
  • $\begingroup$ I'm not sure, as I haven't tried it, but couldn't this be solved with u-substitution where $U = x^3$? $\endgroup$ – recursive recursion Mar 1 '14 at 16:50
  • $\begingroup$ I'm not interested in solving it; I have already done that. I just wanted to know which of the two explanations of the fact that the principal value of the integral is equal to the integral itself are considered better. $\endgroup$ – Hayden Mar 1 '14 at 19:02

The explanation I'd use is the following: we know that $\frac{\sin^3(x)}{x^3}$ is an integrable function through standard analysis (observe that the $\sin^3(x)$ term decays like $x^3$ near 0 whereas the $x^3$ in the denominator guarantees the convergence of the integral away from 0).

Now, for fixed $\epsilon > 0$, define $f_{\epsilon}(x) := \frac{\sin^3(x)}{x^3}$ for $|x| > \epsilon$ and $0$ otherwise, and define $f(x) := \frac{\sin^3(x)}{x^3}$. It should not be hard to show that $f_{\epsilon} \to f$ uniformly as $\epsilon \to 0^+$. So, given $R > 0$,

$\displaystyle \int_{\epsilon < |x| < R} \frac{\sin^3(x)}{x^3} dx \to \int_{|x| < R} \frac{\sin^3(x)}{x^3} dx$ as $\epsilon \to 0^+$.


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