- #1

SisypheanZealot

- 9

- 6

- Homework 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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