How come, when we integrate a complex exponential from $ -\infty $ to $ \infty $, we get a scaled delta function? $$ \begin{align} \int_{-\infty}^{\infty} e^{i k x} \; dk & = 2 \pi \delta \left ( x \right ) \end{align} $$ Specifically, why do we say that the integral converges for $ x \neq 0 $ to $ 0 $? Doesn't it just continue to oscillate?

  • $\begingroup$ What is the value of the integral if $x=0$? $\endgroup$ – Alex S Jun 29 '15 at 22:00
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    $\begingroup$ If $ x = 0 $, then the exponential would just be $ e^{i k 0} = 1 $, and $ \int_{-\infty}^{\infty} 1 dx = \infty $. $\endgroup$ – QuantumFool Jun 29 '15 at 22:01
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    $\begingroup$ The integral is not convergent (either being seen as a generalised Riemann integral or as a Lebesque integral). Actually this equality is an equality for distributions, it says that the Fourier trasnform of the constant unit function (abusively noted as an integral) is a Delta distribution. How much do you know about distributions ? $\endgroup$ – Sylvain L. Jun 29 '15 at 22:05
  • $\begingroup$ @SylvainL. I would like to know the answer to this, but I don't know much about distributions. The result above was derived in my lecture notes by saying that $$I=\lim_{a \rightarrow \infty}\int_{-a}^{a} e^{ik x} dx = \lim_{a \rightarrow \infty}\left(\frac{2\pi\sin(ax)}{\pi x}\right) = 2\pi\delta (x)$$ which actually satisfied me at the time, because intuitively, $\frac{\sin(ax)}{\pi x}$ becomes the delta function as $a\rightarrow \infty$, until I noticed that $I$ oscillates when $x \neq 0$ $\endgroup$ – texasflood Aug 7 '15 at 7:18

Following Sylvain's comment look up the formulas for Fourier transform $F(w)=\int f(t) e^{-iwt}{\rm {d}}t$ and the inverse transform $ f(t)=1/{2\pi}\int F(w) e^{iwt} { \rm {d}}w$ and combine them to write \begin{equation} \begin{split} F(\hat w)=\int_{-\infty}^\infty f(t) e^{-i\hat{w}t}\rm {d}t & = \int_{-\infty}^\infty \frac{1}{2\pi}\int_{-\infty}^\infty F(w) e^{iwt} {\rm {d}} w\, e^{-i\hat{w}t}\,\rm {d}t \\ & = \frac{1}{2\pi}\int_{-\infty}^\infty F(w)\,{\rm {d}} w\int_{-\infty}^\infty e^{iwt} \, e^{-i\hat{w}t}\,\rm {d}t\\ & = \frac{1}{2\pi}\int_{-\infty}^\infty F(w)\,{\rm {d}} w\int_{-\infty}^\infty e^{i(w-\hat{w})t}\,\rm {d}t \end{split} \end{equation} Knowing the definition of the delta-function $f(y)=\int f(x) \delta(x-y) {\rm d} x$ one can see that in this case $$ F(\hat w) = \int_{-\infty}^{\infty} F(w) \delta (w-\hat w) {\rm d}w $$ and thus identifying the right integral with $$\frac{1}{2\pi}\int_{-\infty}^\infty e^{i(w-\hat{w})t}\,\rm {d}t= \delta (w- \hat w) $$


The integral is not meant to be taken in the space of functions; it is meant to be taken over the space of distributions.

If I write $[f(x)]$ when I mean to view $f(x)$ as a distribution rather than a function, then this integral is saying

$$ \int_{-\infty}^{\infty} [e^{ikx}] \, \mathrm{d}k = 2 \pi \delta(x) $$

(note that $\delta(x)$ is a distribution, not a function, so I don't need to put brackets around it)

Two distributions are equal if and only if they have the same value when convolved with test functions. The above equation is asserting, for every test function $f(x)$, you have

$$ \int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} [e^{ikx}] \, \mathrm{d}k \right) f(x) \, \mathrm{d}x = \int_{-\infty}^{\infty} 2 \pi \delta(x)f(x) \, \mathrm{d} x$$

What the test functions are precisely can vary with context; in this case they are probably meant to be the rapidly decreasing functions (aka Schwartz functions).

IIRC, distributions coming from functions of two variables satisfy

$$ \int_{-\infty}^{\infty}\left( \int_{-\infty}^{\infty} [g(x,y)] \, \mathrm{d}y \right)f(x) \mathrm{d}x = \int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} g(x,y) f(x)\, \mathrm{d}x \right) \mathrm{d}y $$

(note the change of order of the integration variables)

Consequently, The above equation is asserting

$$ \int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} e^{ikx} f(x) \, \mathrm{d}x \right) \mathrm{d}k =2 \pi f(0)$$

Note, in particular, that the inner integral is that of ordinary functions, and the integrand is integrable.


The answer of @Statics attacks with the argument that "if you think Fourier Transformation is correct, then you should accept this definition of Dirac Delta Function." But why the Fourier Transformation works at the first place, is because we have this Dirac Delta definition. So the argument using FT isn't sound to me.

The distribution explanation is correct. And the delta function makes sense under an integral sign. We can arrive this formula by constructing a sequence of distribution $\delta_n(x)$, such that: $$\lim_{n\rightarrow\infty}\int^\infty_{-\infty}\delta_n(x-a)f(x)dx = f(a)$$ Then the limit of the sequence is $\delta(x)$, i.e. $$\lim_{n\rightarrow\infty}\delta_n(x) = \delta(x)$$

There are many ways to construct $\delta_n(x-a)$. One common example is: $$\delta_n(x-a)=\sqrt{\frac{n}{\pi}}e^{-n(x-a)^2}$$ $$=\frac{1}{2\pi}\int^\infty_{-\infty}e^{-\frac{k^2}{4n}}e^{i(x-a)k}dk$$

As you can see, the integral converge to $$\frac{1}{2\pi}\int^\infty_{-\infty}e^{i(x-a)k}dk$$ as $n\rightarrow\infty$. And integrate $f(x)$ with a Gaussian distribution will give the value $f(a)$ (For a rigorous prove, check textbooks on functional analysis). Therefore, we can safely use $$\frac{1}{2\pi}\int^\infty_{-\infty}e^{i(x-a)k}dk=\delta(x-a)$$

This is a sketch of proof, and definitely not a rigorous one. But hopefully, you can get some flavor out of it.


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