# Dirac Delta using periodic functions

#### SisypheanZealot

Problem Statement
Exercise1.3: Using the periodic form (1.14) for the Dirac delta, show
$$\int^{\infty}_{-\infty}dx\:f(x)\delta(x-y)=\sum^{\infty}_{n=-\infty}f(y+2nL)$$
Relevant Equations
$$\delta(x-y)=\frac{1}{2L}\sum^{\infty}_{n=-\infty}\lbrace sin(\frac{n\pi x}{L})sin(\frac{n\pi y}{L})+cos(\frac{n\pi x}{L})cos(\frac{n\pi y}{L})\rbrace$$
I know it is something simple that I am missing, but for the life of me I am stuck. So, I used the identity $sin(a)sin(b)+cos(a)cos(b)=cos(a-b)$ which gives me $$\int^{\infty}_{-\infty}dx\:f(x)\delta(x-y)=\int^{\infty}_{-\infty}dx\:f(x)\frac{1}{2L}\sum^{\infty}_{n=-\infty}\lbrace sin(\frac{n\pi x}{L})sin(\frac{n\pi y}{L})+cos(\frac{n\pi x}{L})cos(\frac{n\pi y}{L})\rbrace$$ $$=\int^{\infty}_{-\infty}dx\:f(x)\frac{1}{2L}\sum^{\infty}_{n=-\infty}cos(\frac{n\pi}{L}(x-y))$$ Then I performed the transformation $u=\frac{\pi}{L}(x-y)$ giving $$=\frac{1}{2L}\sum^{\infty}_{n=-\infty}\frac{L}{\pi}\int^{\infty}_{-\infty}du\:f(u\frac{L}{\pi}+y)cos(u\cdot n)$$ $$=\int^{\infty}_{-\infty}du\:f(u\frac{L}{\pi}+y)\frac{1}{2\pi}\sum^{\infty}_{n=-\infty}cos(u\cdot n)$$ $$\int^{\infty}_{-\infty}du\:f(u\frac{L}{\pi}+y)\lbrace\frac{1}{2\pi}+\frac{1}{\pi}\sum^{\infty}_{n=1}cos(u\cdot n)\rbrace$$ Now we have $u=2\pi$ as the solutions for $cos(u\cdot n)=1$ which to me looks like it gives $f(2L+y)$. What am I not seeing?

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#### SisypheanZealot

New thought. What if I work backwards on this. so start with $$\sum^{\infty}_{n=-\infty}f(2nL+y)$$ Then set this equal to $$\int dx f(x)\sum^{\infty}_{n=-\infty}\delta(x-(2nL+y))$$ Now look at the delta term as a periodic term, and look at the Fourier transform definition of the Dirac Delta $$\delta(x-y)=\int\frac{dz}{2\pi}e^{iz(x-y)}$$ The Fourier will be invariant under rotation of multiples of $2\pi$. So then, I can rewrite the integral as $$\int\frac{dz}{2\pi}e^{iz(x-y)}e^{-i2n\pi}$$ This will then give the new definition of the delta function $$\sum^{\infty}_{n=-\infty}\delta(x-(2n\pi-y))=\sum^{\infty}_{n=-\infty}\int\frac{dz}{2\pi}e^{iz(x-y)}e^{-i2n\pi}$$. But, then I need to worry about the modes of the system. I don't know if this moved me forward, or just in circles.

#### SisypheanZealot

So, for the interested. I think I finally got this thing figured out. The first thing that we are going to do is transition to the complex Fourier series definitions of $f(x)$ and $\delta(x-y)$ and set $L=\pi$. i.e. $$f(x)=\sum_{n=-\infty}^{\infty}c_{n}e^{inx}$$ and $$\delta(x-y)=\int\frac{dz}{2\pi}e^{iz(x-y)}$$ So that the formula takes the form $$\int dx f(x)\delta(x-y)=\int dx \sum_{n=-\infty}^{\infty}c_{n}e^{inx}\int\frac{dz}{2\pi}e^{iz(x-y)}$$ then doing some algebra we get $$\int dx \sum_{n=-\infty}^{\infty}c_{n}\int\frac{dz}{2\pi}e^{i[z(x-y)+nx]}$$ performing the integration we then get $$\sum_{n=-\infty}^{\infty}c_{n}\int dx\frac{e^{i[z(x-y)+nx]}}{2\pi i(x-y)}$$ Now we apply Cauchy's formula $\oint\frac{g(w)}{w-w_0}dz=2\pi ig(w_0)$ to this. Doing so we notice that the translations are $w=x,w_0=y$, and that there is a pole at $x=y$. So, the problem becomes $$\sum_{n=-\infty}^{\infty}c_{n}\int dx\frac{e^{i[z(x-y)+nx]}}{2\pi i(x-y)}=\sum_{n=-\infty}^{\infty}c_{n}e^{i[z(y-y)+ny]}$$ Which then gives us the common result of the original problem $$\int dx f(x)\delta(x-y)=\sum_{n=-\infty}^{\infty}c_{n}e^{iny}=f(y)$$ Now, to progress from here we need to introduce yet another definition for the Dirac delta $\delta(n)=\sum^{\infty}_{m=-\infty}e^{i2\pi nm}$ we now multiple this by the new Fourier series giving $$\sum_{n=-\infty}^{\infty}c_{n}e^{iny}\sum^{\infty}_{m=-\infty}e^{i2\pi nm}=\sum^{\infty}_{m=-\infty}\sum_{n=-\infty}^{\infty}c_{n}e^{in(y+2\pi m)}=\sum^{\infty}_{m=-\infty}f(y+2\pi m)$$

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