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Is the following proposition true?

Proposition. For any $a,k\in \mathbb{R}$, \begin{equation} \int_a^{\infty} dx e^{ikx} = 2\pi \delta(k). \end{equation} (End)

I think it is true based on the following argument. First, let me change the integration variable from $x$ to $x'$ by \begin{equation} x' = x-\lambda, \end{equation} with $\lambda>0$. Then, \begin{equation} \int_a^{\infty} dx e^{ikx} =\int_{a-\lambda}^{\infty} dx' e^{ik(x'+\lambda)} = e^{ik\lambda} \int_{a-\lambda}^{\infty} dx' e^{ikx'}. \end{equation} Now, by letting $\lambda\rightarrow +\infty$, the factors in the last expression approach, as distribution of $k$, \begin{equation} e^{ik\lambda} \rightarrow \begin{cases} 0 & \text{ for } k\not=0\\ 1 & \text{ for } k =0 \end{cases} , \end{equation} and \begin{equation} \int_{a-\lambda}^{\infty} dx' e^{ikx'} \rightarrow \int_{-\infty}^{\infty} dx' e^{ikx'} =2\pi\delta(k). \end{equation} Hence, \begin{equation} e^{ik\lambda}\int_{a-\lambda}^{\infty} dx' e^{ikx'} \rightarrow 2\pi\delta(k). \end{equation} By this, I think the proposition is true.

However, I don't see this representation in the books or web pages. (The lower bound $a$ of the integral is always written as $-\infty$.) I guess it may be wrong, and I would like someone point out mistakes in my argument.

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  • $\begingroup$ No, this can not be correct. One simple way to see this, is to differentiate your initial expression with respect to $a$. Then the left hand side becomes equal to $-e^{ika}$, whereas the right hand is equal to zero. $\endgroup$
    – M. Wind
    Aug 9, 2023 at 3:57
  • $\begingroup$ It's valid for $a=-\infty$. $\endgroup$
    – md2perpe
    Aug 9, 2023 at 5:00

1 Answer 1

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Some of the mistakes:

  1. First, the function $e^{ikx}$ is not integrable, hence one should be cautious with with these kind of formal manipulations. The formula with $a=−\infty$ is true only if you know that the integral in this case is not a true integral but a shortcut for the Fourier transform in the sense that distributions.

  2. The convergence of $e^{ikx}$ as a distribution of $k$ is also false: it converges to $0$. There is no meaning to the "value when $k=0$". The proof of this fact follows easily from the Riemann-Lebesgue lemma which implies in particular that for any $\varphi\in C^\infty_c$, $$ \langle e^{ikx},\varphi\rangle = \int_{\Bbb R} e^{ikx}\,\varphi(x)\,\mathrm d x \underset{k\to\infty}\to 0. $$

  3. The limit of a product is not the product of limits for distributions.

The right solution:

The good way to interpret the formula "$\int_a^\infty e^{-ikx}\,\mathrm d x$" is as the Fourier transform in the sense of distributions of the function $\mathbf{1}_{x>a}$, that is defined as the distribution such that for any test function $\varphi\in C^\infty_c$ $$ \langle \mathcal{F}(\mathbf{1}_{x>a}),\varphi\rangle = \langle \mathbf{1}_{x>a},\widehat{\varphi}\rangle = \int_a^\infty \widehat{\varphi}(k)\,\mathrm d k. $$ Now, $\mathbf{1}_{x>a} = H(x-a)$ where $H$ is the Heaviside function. Using the well-known formula of the Fourier transform of the Heaviside function, $$ \mathcal{F}(\mathbf{1}_{x>a}) = e^{-ika} \mathcal{F}(H) = \pi\,e^{-ika} \left(\delta_0 + \mathrm{vp}(\tfrac{1}{i\pi k})\right), $$ where $\mathrm{vp}$ denotes the Cauchy principal value. In your case, you have $k$ instead of $-k$ in the integral, so you get $$ ``\int_a^\infty e^{ikx}\,\mathrm d x" = \mathcal{F}(\mathbf{1}_{x>a})(-k) = \pi\,e^{ika} \left(\delta_0 - \mathrm{vp}(\tfrac{1}{i\pi k})\right). $$

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  • $\begingroup$ Thank you. Could you please elaborate more about point 1 for the mistakes? Do you mean that the change of integration variable is restricted in some way for integrals defining a distribution? It looks to me that, in $\mathcal{F}(1_{x>a}) = e^{-ika} \mathcal{F}(H)$, you shifted the variable by $a$, whereas I shifted it by $\lambda$ which was later let to $\infty$. $\endgroup$
    – norio
    Aug 9, 2023 at 10:31
  • $\begingroup$ Is $\delta_0$ the same thing as $\delta(k)$, the Dirac delta distribution that picks up the value of a test function at $0$? $\endgroup$
    – norio
    Aug 9, 2023 at 10:32
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    $\begingroup$ I am just saying that one should be careful, some things work as integrals, some other not. Indeed, the shifting property also works for the Fourier transform in the sense of distributions, but it has to be proved from the definition using test functions. An example of danger with non-absolutely convergent integrals is the integral $I = \int_{-\infty}^\infty x\,\mathrm d x$, since for example $$ \int_{-n}^{n} x\,\mathrm d x \underset{n\to\infty}{\to} 0 $$ but $$ \int_{-n}^{n+1} x\,\mathrm d x \underset{n\to\infty}{\to} \infty. $$ Yes, $\delta(k) = \delta_0$ is the Dirac delta at $0$. $\endgroup$
    – LL 3.14
    Aug 9, 2023 at 10:49

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