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 $\int^{\infty}_{-\infty} e^{(ik+l)x} d x$?

  • $\begingroup$ Why is there a $k$ on the left side but not the right side? $\endgroup$ – cpiegore Apr 6 '16 at 23:13
  • $\begingroup$ Sorry for the typo. RHS should be $2 \pi \delta (k)$. $\endgroup$ – Stanley Apr 6 '16 at 23:15
  • $\begingroup$ This is just speculation but $ik+l = i(k-li)$ so $\int^{\infty}_{-\infty} e^{(ik+l)x} dx = 2\pi\delta(k-li)$ $\endgroup$ – cpiegore Apr 6 '16 at 23:22
  • $\begingroup$ 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. $\endgroup$ – Stanley Apr 6 '16 at 23:30
  • $\begingroup$ how do you know that $\int_{-\infty}^\infty e^{ikx} dx = 2 \pi \delta(x)$ ? (hint : regularization) $\endgroup$ – reuns Apr 6 '16 at 23:36

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