# Is there any handwavy argument that shows that $\int_{-\infty}^{\infty} e^{-ikx} dk = 2\pi \delta(x)$?

It should not be a good argument but rather a short one and one that convinces a physicist ( so no need for mathematical rigor ) that shows that $\int_{-\infty}^{\infty} e^{-ikx} dk = 2\pi \delta(x)$ holds? It should only refer to basic calculus (especially no fourier transform ) since I am supposed to give a proof of a related relationship about fourier transforms on a physics homework sheet.

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Consider

$$$$\begin{split} f(x) & = \int\limits_{-\infty}^\infty e^{-ikx} \mathrm{d}k\\ \int\limits_{-\epsilon}^\epsilon f(x) \mathrm{d}x = \int\limits_{-\epsilon}^{\epsilon}\int\limits_{-\infty}^\infty e^{-ikx} \mathrm{d}k \mathrm{d}x & = \int\limits_{-\infty}^{\infty}\int\limits_{-\epsilon}^\epsilon e^{-ikx} \mathrm{d}x \mathrm{d}k \\ & = \int\limits_{-\infty}^{\infty}\int\limits_{-\epsilon}^\epsilon \cos(kx) \, \mathrm{d}x \mathrm{d}k \\ & = \int\limits_{-\infty}^{\infty} \frac{2\sin(k\epsilon)}{k} \, \mathrm{d}k \\ & = 2\pi \end{split}$$$$

Hope that was short enough.

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I'd like to make this a tad more rigorous by pointing out that the $x$-integral ought to be from $a+\epsilon$ to $a-\epsilon$ so that when you take the limit $\epsilon \rightarrow 0$, the final integral is $2\pi$ exactly when $a=0$ and is $0$ everywhere else (which is the definition of the delta function). Also, we didn't need to leave off the imaginary part integral since it goes to zero everywhere. – Geoffrey Nov 9 '13 at 16:37
I agree to your first comment. And I left the imaginary part because it is an odd function. – Priyatham Nov 9 '13 at 16:52

$$\int_{-a}^{a} e^{-ikx} dk = \frac{e^{-ikx}}{-ix} |^{+a}_{-a}= i\frac{e^{-iax}-e^{iax}}{x}$$

$$e^{-iax}-e^{iax}=-2i \sin {(ax)}$$

$$\int_{-a}^{a} e^{-ikx} dk = i\frac{e^{-iax}-e^{iax}}{x} = i\frac{-2i \sin {ax}}{x} =2\frac{\sin {ax}}{x} = f_a(x)$$

You need to find the limit. Lets define $\lim\limits_{ a\to \infty } f_a(x) = 2\pi \delta(x)$

$$\lim\limits_{ a\to \infty } \int_{-a}^{+a} e^{-ikx} dk = \lim\limits_{ a\to \infty } 2\frac{\sin {ax}}{x}= 2\pi \delta(x)$$

we can check some graphs for some $a$ values

$a=100$

$a=10^{100}$

Finally If $a\to \infty$ then we get the function $2\pi \delta(x)$ below.

The function for $x\neq0$

$2\pi \delta(x)=0$

but for $x=0$

$\lim\limits_{ x\to 0 } 2\pi \delta(x)=+\infty$

but the function is very special as Priyatham shew.

For $\epsilon>0$ $$\int\limits_{-\epsilon}^\epsilon 2\pi \delta(x) \mathrm{d}x = 2\pi$$

Thus Finally we can get the important result for $\delta(x)$ function that

If $\epsilon>0$ then

$$\int\limits_{-\epsilon}^\epsilon \delta(x) \mathrm{d}x = 1$$

Note: The graphs were taken from www.wolframalpha.com

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