1

Let $k$ and $l$ be two real numbers. We know that $\int^{\infty}_{-\infty} e^{ikx} d x = 2 \pi \delta(k)$. Here $i$ is the imaginary unit and $\delta(\cdot)$ is the Dirac delta function. What is $\int^{\infty}_{-\infty} e^{(ik+l)x} d x$?

Stanley
  • 195
  • Why is there a $k$ on the left side but not the right side? – cpiegore Apr 06 '16 at 23:13
  • Sorry for the typo. RHS should be $2 \pi \delta (k)$. – Stanley Apr 06 '16 at 23:15
  • This is just speculation but $ik+l = i(k-li)$ so $\int^{\infty}_{-\infty} e^{(ik+l)x} dx = 2\pi\delta(k-li)$ – cpiegore Apr 06 '16 at 23:22
  • This may work if the delta function is defined in the complex plane. However, usually we think that the delta function is defined in the real number line, which is zero everywhere except at zero. – Stanley Apr 06 '16 at 23:30
  • how do you know that $\int_{-\infty}^\infty e^{ikx} dx = 2 \pi \delta(x)$ ? (hint : regularization) – reuns Apr 06 '16 at 23:36
  • Refer to this post http://math.stackexchange.com/questions/177091/int-infty-infty-eikxdx-equals-what?rq=1 or http://dfcd.net/articles/delta.pdf. – Stanley Apr 06 '16 at 23:41
  • the Fourier transform of $1$ is the Dirac because when $\epsilon \to 0$ : $\int_{-\infty}^\infty e^{ikx} e^{-\epsilon^2 x^2} dx$ converges to $2 \pi \delta(x)$ in the sense of distributions. fortunately, computing $\int_{-\infty}^\infty e^{ikx} e^{l x} e^{-\epsilon^2 x^2} dx$ isn't so complicated by completing the square and using the Fourier transform of the gaussian. hence you'll see if the Fourier transform of $e^{l x}$ exists in the sense of distributions – reuns Apr 06 '16 at 23:48
  • The problem in this question is that both of these integrals are an exemple of abuse of notation. How do you define $\delta$? as a measure? as a distribution? as a generalised function? What is the meaning of the second integral? It's clearly not an usual integral, as these are not convergent. So, what is the definition of this object? – Tryss Apr 06 '16 at 23:52
  • @Tryss : you are abusing too :-) you should refer to my definition as Stanley probably thinks to the "magic definition" such that the Fourier transform is invertible (hence he is thinking to the distributions, and in this particular case it reduces to what I wrote) – reuns Apr 06 '16 at 23:55
  • The Dirac delta function is well known. See this https://en.wikipedia.org/wiki/Dirac_delta_function – Stanley Apr 07 '16 at 00:03
  • @Stanley : Is your exact question "what's the Fourier transform of $e^{lx}$ (as a tempered distribution)?" – Tryss Apr 07 '16 at 00:15
  • @Tryss: Yes. I should mention that both $k$ and $l$ are real numbers. If we calculate the Fourier transform of $e^{i l x}$, then the answer is simply $2πδ(k+l)$. But what is the Fourier transform of $e^{l x}$? – Stanley Apr 07 '16 at 00:46
  • divergent integral ... $\int^{\infty}_{-\infty} e^{ikx} d x = 2 \pi \delta(k)$ ... Of course in physics you may write this. But in mathematics you should say in what sense it should be interpreted. – GEdgar Apr 07 '16 at 00:48
  • OK. I write $\int^{\infty}{-\infty} e^{i k x} d x = 2 \pi \delta (k)$ in the sense that $\lim\limits{a \rightarrow \infty} \int^a_{-a} e^{i k x} d x = 2 \pi \delta (k)$. – Stanley Apr 07 '16 at 00:51

0 Answers0