- #1

signalcarries

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## Homework Statement

I'm trying to prove that [tex]\delta'(y)=-\delta'(-y)[/tex].

## Homework Equations

## The Attempt at a Solution

I'm having trouble getting the LHS and the RHS to agree. I've used a test function [tex]f(y)[/tex] and I am integrating by parts.

For the LHS, I have

[tex]\int_{-\infty}^{\infty} f(y)\delta'(y)dy = \int_{-\infty}^{\infty} \frac{d}{dy}[f(y)\delta(y)]dy - \int_{-\infty}^{\infty} \delta(y)\frac{df(y)}{dy}dy = 0 - f'(0) = -f'(0)[/tex]

For the RHS, I have

[tex]-\int_{-\infty}^{\infty} f(y)\delta'(-y)dy = \int_{\infty}^{-\infty} f(-t)\delta'(t)dt = -\int_{-\infty}^{\infty} f(-t)\delta'(t)dt = -\int_{-\infty}^{\infty} \frac{d}{dt} [f(-t)\delta(t)]dt + \int_{-\infty}^{\infty} \frac{df(-t)}{dt} \delta(t)dt = 0 + \int_{-\infty}^{\infty} \frac{df(-t)}{dt} \delta(t)dt = f'(0)[/tex].

I seem to be off by a minus sign, but I can't figure out where. Any help would be appreciated.