I was looking through alternative methods to computing this integral rather than using the classic semi-circular contour avoiding the origin. I came across a method to computing

$$\int_{\infty}^{\infty} \frac{\sin^2(x)}{x^2} \, dx$$

by shifting the poles up by a factor $i \epsilon$ and continuing the calculation as normal.

I want to employ the same idea with the sinc integral, noting that $$\int_{\infty}^{\infty} \frac{\sin(x)}{x} \, dx = \lim_{\epsilon \rightarrow 0}\int_{\infty}^{\infty} \frac{\sin(x)}{x-i\epsilon} \, dx$$

We consider the integral $$\int_{\Gamma} \frac{e^{iz}}{z-i \epsilon}\, dz$$

Where $\Gamma$ is the upper semi-circular arc, $[-R, R] \cup Re^{i[0, \pi]}$. Then we have

\begin{align*} \int_{\Gamma} \frac{e^{iz}}{z-i \epsilon}\, dz &= \int_{-R}^{R} \frac{e^{it}}{t-i \epsilon} \, dt + iR\int_{0}^{\pi} \frac{e^{it}e^{iRe^{it}}}{Re^{it}-i \epsilon}\, dt \\ &= 2 \pi i e^{-\epsilon} \end{align*}

by Residue theorem since there is one pole inside the contour. The integral along the arc vanishes(and is pretty easy to see) as $R \rightarrow \infty$. Letting $\epsilon \rightarrow 0$, we have

$$\lim_{\epsilon \rightarrow 0} \int_{-\infty}^{\infty}\frac{e^{it}}{t-i \epsilon} \, dt = 2 \pi i$$

This is where it gets murky for me. Expanding the integrand, we've

$$\lim_{\epsilon \rightarrow 0}\left(\int_{-\infty}^{\infty} \frac{x \cos (x)}{x^2+\epsilon ^2}-\frac{\epsilon \sin (x)}{x^2+\epsilon ^2} \, dx + i \int_{-\infty}^{\infty} \frac{x \sin (x)}{x^2+\epsilon ^2}+\frac{\epsilon \cos (x)}{x^2+\epsilon ^2} \, dx \right)= 2 \pi i$$

Using the linearity of the limit on the left hand side and equating real and imaginary parts, we have

$$\lim_{\epsilon \rightarrow 0 } \int_{-\infty}^{\infty} \frac{x \sin (x)}{x^2+\epsilon ^2}+\frac{\epsilon \cos (x)}{x^2+\epsilon ^2} \, dx = 2 \pi$$

and yet, at the same time,

$$ \int_{-\infty}^{\infty}\lim_{\epsilon \rightarrow 0 }\left(\frac{x \sin (x)}{x^2+\epsilon ^2}+\frac{\epsilon \cos (x)}{x^2+\epsilon ^2}\right) \, dx = \int_{-\infty}^{\infty}\frac{\sin(x)}{x} \, dx = \pi$$

I'd like to know why the exchange of limit and integral is making the difference here, is there something to do with the convergence of these integrals that is making this technique not work, or is there something I've plainly missed? One thing I can note is that I don't employ that idea I had originally, as the real and imaginary parts of the integrand above don't recover what I want to exploit.

  • $\begingroup$ Why can you perform exchange of limits and integral? $\dfrac{x\sin(x)}{x^2 + \epsilon^2} \sim_{\epsilon \to 0} \dfrac{\sin(x)}{x}$ which is semi-convergent AFAIK. $\endgroup$ – Raito Feb 8 '18 at 9:38
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    $\begingroup$ In math.harvard.edu/~ctm/home/text/others/hardy/sinx/sinx.pdf, G.H. Hardy gives seven different proofs of the value of the integral. $\endgroup$ – awkward Feb 8 '18 at 13:26

I see this as a question about the intuition of the matter (since I think you are already aware of the "classic" actual proof, and you are already aware of that the place where the "fake" proof goes wrong is at the interchange of integral and limit), so allow me to give an answer which aims at an "intuitive" level, even if at the expense of rigorous detail.

Let's focus on the expression that causes the trouble in the end: $\frac{\epsilon \cos(x)}{x^2+\epsilon ^2}$. It causes trouble because of its behavior when $\epsilon$ and $x$ are close to $0$. Well, to be a little more specific, think of $\cos(x)$ as the sum of the main term $1$ and the higher-order part $h(x) = \cos(x)-1 = O(x^2)$; then it's only the main term that causes the trouble. In other words, $\frac{\epsilon}{x^2+\epsilon ^2}$ is the interesting part while $\epsilon \frac{h(x)}{x^2+\epsilon ^2}$ is the uninteresting part (uninteresting because it behaves nicely for $\epsilon$ and $x$ near $0$, and is happy to allow its integral and limit to be interchanged).

The "interesting" expression has the integral $$ \int_{-\infty}^\infty \frac{\epsilon}{x^2+\epsilon ^2} dx = \int_{-\infty}^\infty \frac{1}{\frac{x^2}{\epsilon^2} + 1} \frac{dx}{\epsilon} = \int_{-\infty}^\infty \frac{1}{u^2 + 1} du $$ which doesn't actually depend on $\epsilon$ - it's always $\pi$.

So, what we see is that $\frac{\epsilon}{x^2+\epsilon ^2}$ accounts for the difference between $\pi$ (the actual answer) and $2\pi$ (from the fake proof).

To put it another way, what happens is that $$ \frac{\epsilon}{x^2+\epsilon ^2} = \frac{1}{2i} \left( \frac{1}{x-i\epsilon} - \frac{1}{x+i\epsilon}\right) $$ and $$ \frac{1}{2i} \int_{a}^b \left( \frac{1}{x-i\epsilon} - \frac{1}{x+i\epsilon}\right) dx \approx \frac{1}{2i} \oint \frac1z \, dz, $$ which equals $\pi$ (if $0 \in (a,b)$). In this light, the error of the fake proof is in neglecting to analyze the pole at $0$ (in contrast to the proof involving the "semi-circular contour avoiding the origin" you alluded to, which does tread carefully around the pole).

After a few more drinks, you might say that the true limit of the functions is given by $$ \frac{\epsilon}{x^2+\epsilon ^2} = \frac{1}{2i} \left( \frac{1}{x-i\epsilon} - \frac{1}{x+i\epsilon}\right) \to \pi \delta(x) $$ where $\delta$ is the Dirac delta "function". (The theories of "distributions" and "hyperfunctions" are the elaborations of this idea.) In a sense, the essence of the fake proof's error is in equating $\delta$ with $0$.

The point is, the interchange of the limit and integral doesn't work, as $\frac{\epsilon \cos(x)}{x^2+\epsilon ^2}$ carries the "missing mass" of $\pi$, in spite of its limit being $0$ for every individual $x$ pointwise (except at $x=0$).


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