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Is the following correct?

$$\begin{align} \int_{-\infty}^{+\infty} dx\, e^{+iax} &=\int_{-\infty}^{+\infty} dx\,\cos{(ax)}+i\int_{-\infty}^{+\infty}dx\,\sin{(ax)}\\\\ &\overbrace{=}^{x=-y}\int_{+\infty}^{-\infty} d(-y)\,\cos{(a(-y))}+i\int_{+\infty}^{-\infty}d(-y)\,\sin{(a(-y))} \\\\ &=\underbrace{\int_{-\infty}^{+\infty} d{y}\,\cos{(ay)} -i\int_{-\infty}^{+\infty} d{y}\,\sin{(ay)}}_{-\int_{+\infty}^{-\infty}=\int^{+\infty}_{-\infty}}\\\\ &=\int_{-\infty}^{+\infty} d{y}\, e^{-iay}\\\\ &=\int_{-\infty}^{+\infty} d{x}\, e^{-iax}. \end{align}$$

Mark Viola
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upquark31415
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  • The integral dose not exist in any sense if $a$ is real number – Kavi Rama Murthy May 26 '21 at 23:20
  • Must be evaluated in the sense of distributions. As Kavi remarked, it does not converge in the usual sense. As illustrated in the OP: do not try to use the normal integration rules on divergent integrals. – GEdgar May 26 '21 at 23:56
  • My recommendation is to first make the integral convergent. This can be done by multiplying the integrand by a Gaussian with a large width. All your steps then become legitimate. At the end you may take the limit of the width of the Gaussian to infinity. – M. Wind May 27 '21 at 03:03

1 Answers1

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The object written $\int_{-\infty}^\infty e^{\pm iax}\,dx$ is either $(i)$ a divergent integral or $(ii)$ tempered distribution and not an integral.

It is easy to see from even-odd symmetry that for any $L\in \mathbb{R}$ $\int_{-L}^L e^{+ iax}\,dx=\int_{-L}^L e^{- iax}\,dx$, but the limit as $L\to \infty$ of both of these integrals fails to exist (unless the limit is a distributional limit).

However, if we interpret the object as a tempered distribution, then it is, in fact, the Fourier Transform of the constant $1$. And in this case (i.e., in distribution) that the equality $\displaystyle \int_{-\infty}^\infty e^{+iax}\,dx=\int_{-\infty}^\infty e^{-iax}\,dx$ holds.

EVALUATION OF $\displaystyle \int_{-\infty}^\infty e^{\pm iax}\,dx$

If we interpret the object as a Fourier Transform, we can evaluate it as follows. Let $\phi\in \mathbb{S}$ (i.e., a Schwarz Space Function). Then, invoking the Fourier Inverse Theorem, we have

$$\begin{align} \langle \mathscr{F}\{1\}, \phi\rangle&=\langle 1, \mathscr{F}\{\phi\}\rangle\\\\ &=\int_{-\infty}^\infty \int_{-\infty}^\infty \phi(x)e^{\pm iax}\,dx\,da\\\\ &=2\pi \phi(0)\tag1 \end{align}$$

Hence, in distribution we see from $(1)$ that $\mathscr{F}\{1\}=2\pi \delta$ where $\delta$ is the Dirac Delta distribution.

Note that we have tacitly shown that the equality $\displaystyle \int_{-\infty}^\infty e^{+iax}\,dx=\int_{-\infty}^\infty e^{-iax}\,dx$ holds in distribution.

ALTERNATIVE DEVELOPMENT:

Here, we represent the Fourier Transform of $1$ as the distributional limit $\displaystyle \lim_{L\to\infty}\int_{-L}^L e^{\pm iax}\,dx$. Proceeding, we have for any $\phi\in \mathbb{S}$

$$\begin{align} \langle \mathscr{F}\{1\}, \phi\rangle&=\lim_{L\to \infty}\int_{-\infty}^\infty \phi(x) \int_{-L}^L e^{\pm iax} \,da\,dx\tag2\\\\ &=\lim_{L\to \infty}\int_{-\infty}^\infty \phi(x) \frac{2\sin(Lx)}{x}\,dx\\\\ &\overbrace{=}^{\text{IBP}}-\lim_{L\to \infty}\int_{-\infty}^\infty \phi'(x) \int_{-\infty}^x \frac{2\sin(x'L)}{x'}\,dx'\,dx\\\\ &=-\lim_{L\to \infty}\int_{-\infty}^\infty \phi'(x) \int_{-\infty}^{Lx} \frac{2\sin(x')}{x'}\,dx'\,dx\\\\ &\overbrace{=}^{\text{DCT}}-\int_{-\infty}^\infty \phi'(x) \lim_{L\to \infty}\int_{-\infty}^{Lx} \frac{2\sin(x')}{x'}\,dx'\,dx\\\\ &=-\int_{-\infty}^\infty \phi'(x) 2\pi H(x)\,dx\\\\ &=-2\pi \int_0^\infty \phi'(x)\,dx\\\\ &=2\pi \phi(0) \end{align}$$

from which we deduce that in distribution

$$\lim_{L\to \infty}\int_{-L}^L e^{\pm iax}\,dx=2\pi \delta(a)$$

where the limit is taken as in $(2)$.

Mark Viola
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  • Thanks a lot!! I was actually looking for something that would show that the equality holds. My steps are not valid for a divergent integral. Can you please suggest a reference which I can cite? I am from a physics background, so not sure where exactly I might find a good discussion: fourier analysis, complex analysis? Thanks again! – upquark31415 May 29 '21 at 21:18
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    You're welcome. My pleasure. Gelfund and Shilov is a great book on the theory of generalized functions. – Mark Viola May 29 '21 at 21:38