0
$\begingroup$

At this forum there is an abundance of questions regarding the convergence of integrals and sums of infinite series. The mathematicians who answer these questions emphasize that only under strict conditions an integral or a sum can be considered convergent. If the conditions are not met, the integral or sum must be categorized as undetermined or divergent.

For example: due to symmetry cancellation of terms takes place in integrals (or sums) of odd functions. It is argued by the experts that this property is of no consequence for non-convergent integrals or sums. Despite the fact that cancellation takes place the integral or sum must be considered undetermined.

I wonder how this applies to the following integral and sum:

$$I = \int _{-\infty}^{+\infty}\sin(cx)dx$$

$$S = \sum _{n=-\infty}^{+\infty}\sin(cn)$$

They appear in Fourier analysis and are routinely assigned the value $0$.

$\endgroup$
2
  • 1
    $\begingroup$ Put a big fat $\operatorname{v.p.}$ in front of them, and everything's fine. Without saying you want the principal value, those are meaningless expressions (unless $c = 0$). $\endgroup$
    – Daniel Fischer
    Jun 27, 2015 at 18:12
  • 1
    $\begingroup$ What he said, except he meant pv. (At least in English.) The point: For example, $\lim_{A\to\infty}\int_{-A}^A\sin(x)\,dx=0$. But $\int_{-\infty}^\infty \sin(x)\,dx$ is meaningless. (You say these things routinely appear in Fourier analysis and are assigned the value $0$? Not in Fourier analysis written by mathematicians, unless they've been careful to explain that the pv is intended...) $\endgroup$
    – David C. Ullrich
    Jun 27, 2015 at 18:17

1 Answer 1

Reset to default
2
$\begingroup$

The improper Riemann integral $\int_{-\infty}^{\infty} \sin cx\,dx$ is interpreted as

$$\int_{-\infty}^{\infty}\sin cx\,dx=\lim_{L_{-}\to \infty} \int_{-L_{-}}^0\sin cx\,dx+\lim_{L_{+}\to \infty} \int_0^{L_{+}}\sin cx\,dx \tag 1$$

Neither integral in $(1)$ converges and neither does their sum. However, if one interprets the improper integral in the sense of Cauchy Principal value, then $L_{-}=L_{+}=L$ and we have

$$\int_{-\infty}^{\infty}\sin cx\,dx=\lim_{L\to \infty} \int_{-L}^{L}\sin cx\,dx=0$$

where the odd symmetry immediately gives the result.

$\endgroup$
5
  • $\begingroup$ Of course $$\mathrm{pv} \int_{-\infty}^\infty x^{2015}\,dx = 0$$ also. But perhaps a mathematician would say: so what? $\endgroup$
    – GEdgar
    Jun 27, 2015 at 18:37
  • $\begingroup$ @Dr.MV Thank you. Are you saying that in Fourier analysis one tacitly assumes that $L_+ = L_-$? Otherwise the elementary integral above, and also the Dirac $\delta$-function, would be undefined. $\endgroup$
    – M. Wind
    Jun 27, 2015 at 18:40
  • $\begingroup$ Often in applied fields such as physics, we assume the PV. $\endgroup$
    – Mark Viola
    Jun 27, 2015 at 19:05
  • $\begingroup$ @gedgar I agree. Yet, in applied areas, such as physics and engineering, the PV is assumption is based on the physics of the problem. Those practitioners might say "so what" that without the PV, the integrals are rendered meaningless. $\endgroup$
    – Mark Viola
    Jun 27, 2015 at 19:08
  • $\begingroup$ @m.wind You're welcome. My pleasure. On the topic of distributions (aka, Generalized Functions) such as the Dirac Delta, that is a rich and beautiful theory aside from classical analysis. $\endgroup$
    – Mark Viola
    Jun 27, 2015 at 19:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .