# Evaluating an integral and showing function behaves like Dirac delta function $\delta(x)$

From my quantum mechanics textbook, I'm given this problem:

Problem: The family of functions $\delta_L(x)$ is defined by $$\delta_L (x) = \frac{1}{2 \pi} \int_{-L}^{+L} e^{ikx} dk.$$ Evaluate the integral and show that $\delta_L (x)$ behaves like the Dirac delta function $\delta(x)$ as $L \rightarrow \infty$.

Attempt: We have \begin{align*} \delta_L(x) &= \frac 1 {2\pi} \bigg( \int_{-L}^L \cos(kx) \, dk + i \int_{-L}^L \sin(kx) \, dk \bigg) \\ &= \frac{1}{2 \pi} \bigg( \frac{1}{x} \big(\sin(Lx) - \sin(-Lx) \big) - \frac{i}{x} \big( \cos(Lx) - \cos(-Lx) \big) \bigg) \\ &= \frac{1}{2 \pi} \big( \frac{1}{x} 2 \sin(Lx) \big) \\ &= \frac{1}{\pi x} \sin(Lx) \end{align*} since cosine is an even function and sine odd.

Now we must show $\delta_L(x)$ behaves like Dirac delta. This means that if we take $L \rightarrow \infty$, $\delta_L(x)$ must be zero for all $x$ except for $x = 0$, which is infinity. For the second part, I would just do: \begin{align*} \delta_L(x) = \frac{1}{2 \pi} \int_{- \infty}^\infty e^{ikx} \, dk = \frac{1}{\pi} \int_0^\infty dk = \infty \end{align*} if we let $x = 0$. Is this kind of reasoning correct? Also, how to show the other property?

## 1 Answer

The $\delta$-function, despite its name, is not really a function, but a distribution. Very roughly speaking this means that it's only meant to be 'under an integral sign'.

To show that $\delta_L$ converges (in distribution) to a Dirac delta function we need to show that it satisfies the defining relation

$$\lim_{L\to\infty}\int_{-\infty}^\infty f(x)\delta_L(x)\,{\rm d}x = f(0)$$

for any suitable test-function $f$. A proof for test-functions $f(x)\in C_c^\infty(\mathbb{R})$ (the class of smooth functions with compact support) is given in this answer.

• I think I understand better now. The 'test function' you're speaking of, this is not equal to $\delta_L (x)$? It can be an arbitrary function? – Kamil Oct 17 '15 at 21:24
• @Kamil Yes that is correct, $f$ is not $\delta_L$, but an arbitrary function (up to some requirements). – Winther Oct 17 '15 at 21:32
• Hans, how did you arrive at the inequality $$\left|\int_{-\sqrt L}^\sqrt L \frac{\sin(y)}{\pi y} (f(y/L)-f(0))\right|<\int_{-\sqrt L}^{\sqrt L} \frac{\sin(y)}{\pi y}\epsilon\,dy\,\,?$$Application of the triangle inequality would have included the absolute value of $\frac{\sin(y)}{y}$ on the right hand side of the inequality. I must be missing something here. When you have a chance, please explain. – Mark Viola Jul 17 '17 at 14:42
• @MarkViola Yes there should have been an absolute value there. Thanks for pointing it out, tried to stitch it up. Was done in a hurry here so there is probably some holes still needs fixing. Will go through it properly later tonight. – Winther Jul 17 '17 at 15:24
• Hans, have you tried using the second mean value theorem. That way, we don't need to use the triangle inequality and the absolute value doesn't appear inside the integral. Since $f$ is smooth and of compact support, all should flow nicely. Let me know if this works. -Mark – Mark Viola Jul 17 '17 at 23:55