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In a Quantum Mechanics class the professor pointed out a calculation from Zetilli's textbook on Quantum Mechanics (page 630).

There he says that:

$$\int_0^\infty \sin(qx) dx = \lim_{\lambda \to 0} \int_0^\infty e^{-\lambda x} \sin(qx) dx$$

Then he writes the $\sin$ as complex exponentials and reaches the result that the integral is $1/q$. WHAT? I was taught at my calculus course that this integral does not converge. For me this result is plain wrong, and I do not understand it at all. I even don't see why the limit can be imposed like that. Any thoughts on this "trick"?

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    $\begingroup$ I guess you can invert limit and integral sign, since $e^{-\lambda x}\sin{qx}$ is dominated by $e^{-\lambda x}$ which is integrable, so that trick is justified. For the rest, I do not know. $\endgroup$ May 15 '17 at 21:34
  • $\begingroup$ To this mathematician, the analysis does indeed look “plain wrong”. $\endgroup$
    – Lubin
    May 15 '17 at 21:38
  • $\begingroup$ This is perfectly well-defined in the sense of distributions. Multiply the given integral by a smooth test function $\phi(q)$ and integrate over $q$. In that context, the given integral will behave like $1/q$. $\endgroup$ May 15 '17 at 22:33
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    $\begingroup$ The textbook should have had some more explanations here. This integral comes about from considering waves scattering in a Coulomb potential. A problem with the Coulomb potential is that it has infinite range which means that some assumptions in the 'standard' analysis fails to hold. A common trick to deal with this is to assume a finite range, i.e. $U(r) = \frac{e^{-r \lambda}}{r}$ for which the calculation is well defined for all $\lambda > 0$, and then take the range to infinity in end ($\lambda\to 0$). $\endgroup$
    – Winther
    May 15 '17 at 22:46
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Since $$ \newcommand{\Im}{\operatorname{Im}} \int_0^\infty\sin(qx)\,\mathrm{d}x= \Im\left(\int_0^\infty e^{iqx}\,\mathrm{d}x\right)\tag1 $$ in the spirit of analytic continuation, we can look at $$ f(q)=\int_0^\infty e^{iqx}\,\mathrm{d}x\tag2 $$ which is well-defined for $\Im(q)\gt0$, and see how $f$ behaves as $\Im(q)\to0$.

When $\Im(q)\gt0$, we get, via contour integration, $$ \begin{align} f(q) &=\frac iq\int_0^{-iq\infty} e^{-x}\,\mathrm{d}x\tag{3a}\\ &=\frac iq\int_0^\infty e^{-x}\,\mathrm{d}x\tag{3b}\\ &=\frac iq\tag{3c} \end{align} $$ Explanation:
$\text{(3a)}$: substitute $x\mapsto\frac{ix}q$
$\text{(3b)}$: integration over $\gamma=[0,R]\cup[R,-iqR]\cup[-iqR,0]$ is $0$
$\text{(3c)}$: integrate

enter image description here

Then, by analytic continuation, as $\Im(q)\to0$, $\frac iq$ behaves nicely. Therefore, we say for real $q$: $$ \int_0^\infty(\cos(qx)+i\sin(qx))\,\mathrm{d}x=\frac iq\tag4 $$ which says not only that $\int_0^\infty\sin(qx)\,\mathrm{d}x=\frac1q$, but also that $\int_0^\infty\cos(qx)\,\mathrm{d}x=0$ (even though neither of these integrals converges in the classical sense).


Let's verify the integral of $\cos(qx)$: $$ \begin{align} \int_0^\infty\cos(qx)\,\mathrm{d}x &=\int_0^\infty\sin(\pi/2+qx)\,\mathrm{d}x\tag{5a}\\[6pt] &=\int_{\frac\pi{2q}}^\infty\sin(qx)\,\mathrm{d}x\tag{5b}\\ &=\color{#C00}{\int_0^\infty\sin(qx)\,\mathrm{d}x}\color{#090}{-\int_0^{\frac\pi{2q}}\sin(qx)\,\mathrm{d}x}\tag{5c}\\[3pt] &=\color{#C00}{\frac1q}\color{#090}{+\left.\frac1q\cos(qx)\,\right|_0^{\frac\pi{2q}}}\tag{5d}\\[9pt] &=0\tag{5e} \end{align} $$ Explanation:
$\text{(5a)}$: $\cos(x)=\sin(\pi/2+x)$
$\text{(5b)}$: substitute $x\mapsto x-\frac\pi{2q}$
$\text{(5c)}$: difference of integrals
$\text{(5d)}$: integrate
$\text{(5e)}$: evaluate

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  • $\begingroup$ This is indeed very interesting. I like it better than the physics explanation, which is kind of obscure... Especially since in my question I could have added a regularisation term $e^{-\lambda x^2}$ and the result would have been different. $\endgroup$ Dec 29 '20 at 15:56
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    $\begingroup$ @VladimirVargas: When I use $e^{-\lambda x^2}$, I get $$\lim_{\lambda\to0^+}\sqrt{\frac\pi{4\lambda}}e^{-\frac{q^2}{4\lambda}}\operatorname{erfi}\left(\frac{q}{\sqrt{4\lambda}}\right)=\frac1q$$ $\endgroup$
    – robjohn
    Dec 29 '20 at 19:08
  • $\begingroup$ Oh, I must have made a mistake when I did this integral some years ago then. Pretty sure I got a different result lol. Thanks a lot! $\endgroup$ Dec 29 '20 at 21:46
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I think this is called either dimensional reduction, exponential regularization, or laplace (transform) regularization. The integral does not converge in any classical sense, but we are defining it via the above exponential regularization. This is a fairly common technique (trick?) in physics, where integrals tend to diverge without such considerations. The idea is that the integral with the exponential defines an analytic function for $\lambda>0$ and can be "extended" to $\lambda\rightarrow 0$.

This is not entirely unjustified because rarely do physical quantities oscillate as such to infinity. For example in an electromagnetic cavity, even though we assume an infinite sequence of modes exist, their energies would be ridiculous so there has to be some kind of cutoff or regularization. Feynman integrals are notorious for this kind of divergence (ultraviot and infrared divergenve especially), so that's why there's a flurry of regularization techniques to assign sensible values to them.

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  • $\begingroup$ My whole life was just lie... But wait, this would mean that a monochromatic wave can exist and has a physical sense oO $\endgroup$ May 15 '17 at 21:42
  • $\begingroup$ But is this well defined? I mean, the same function can be obtained as the limit of many different sequences of integrable functions. Is the limit integral independent of the sequence? $\endgroup$ May 15 '17 at 21:43
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    $\begingroup$ It is not independent of the sequence at all. Try $e^{-\lambda x^2}$ instead. $\endgroup$
    – Alex R.
    May 15 '17 at 21:44
  • $\begingroup$ But why do all of this? In QM class, the first we are thaught is that monochromatic waves do not exist and it manifests in the fact that it doesn't satisfy the normalization condition since the integral we talk about is divergent and now they try to give it a sense... $\endgroup$ May 15 '17 at 21:46
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    $\begingroup$ @OussamaBoussif $f(x)=\sin(qx)1_{x > 0}$ has a Fourier transform in the sense of distributions. $\hat{f}(\xi)$ is the distribution such that $\hat{f} \ast a e^{-a^2 \pi\xi^2}$ is the Fourier transform of $f(x) e^{-\pi x^2/a^2}$ for every $a > 0$ $\endgroup$
    – reuns
    May 15 '17 at 21:53
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Not an answer, just a long comment

If we consider the sum of a series in a different way (Cesàro summation), then $$\sum_{n=1}^{\infty} \sin n x=\frac{1}{2} \cot \left(\frac{x}{2}\right)$$ Indeed $$S_k=\sum _{i=1}^k \sin (i x)\csc \left(\frac{x}{2}\right) \sin \left(\frac{k x}{2}\right) \sin \left(\frac{1}{2} (k+1) x\right)$$ and $$\frac{1}{n}\sum _{k=1}^n \csc \left(\frac{x}{2}\right) \sin \left(\frac{k x}{2}\right) \sin \left(\frac{1}{2} (k+1) x\right)=\\=\frac{1}{4n}\csc \left(\frac{x}{2}\right) \left(2 n \cos \left(\frac{x}{2}\right)-\csc \left(\frac{x}{2}\right) \sin ((n+1) x)+\sin (x) \csc \left(\frac{x}{2}\right)\right)$$

and finally $$\underset{n\to \infty }{\text{lim}}\frac{1}{4n}\csc \left(\frac{x}{2}\right) \left(2 n \cos \left(\frac{x}{2}\right)-\csc \left(\frac{x}{2}\right) \sin ((n+1) x)+\sin (x) \csc \left(\frac{x}{2}\right)\right)=\frac{1}{2} \cot \left(\frac{x}{2}\right)$$

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The mean value of the integral as the upper bound goes to infinity is $1/q$. The Abel regularization in your source is essentially finding the mean value.

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