Dirac delta - sifting

Question

We know $\int_{-\infty}^\infty \delta(x-a)f(x) \, dx=f(a) $

Is this still true for: $\int_{-\infty}^\infty \delta(a-x)f(x) \, dx=f(a) $

In general, can we call dirac delta even function?

Answer 1

Yes, because $\delta(x-a)$ "activates" when the argument is equal to zero, so it's irrelevant if you write $\delta(x-a)$ or $\delta(a-x)$ because $a-x=0$ is the same statement as $x-a=0$.

A more formal way to show that is to use the formula for when the argument of $\delta$ is a function: $$\delta(f(x))=\sum_{i=1}^n \frac{\delta(x-x_i)}{\left|f'(x_i)\right|},$$ where $x_i$ are the points satisfying $f(x_i)=0$. In this case $f(x)=a-x$ and $|f'(a)|=1$, so $\delta(a-x)=\delta(x-a)$, and so the distribution is, in a way, even.

Substitution also works, as the other answers suggest.

Answer 2

you can do the change of variable, let $u=a-x$,

Also note that: $\int_{-\infty}^\infty dx=\int_{+\infty}^{-\infty}-du=\int_{-\infty}^\infty du$

Answer 3

"Formally", you can substitute $u=a-x$ to get

$$\int_{-\infty}^\infty \delta(a-x) f(x) \, dx = \int_\infty^{-\infty} -\delta(u) f(a-u) \, du=\int_{-\infty}^\infty \delta(u) f(a-u) \, du.$$

Then the last quantity is $f(a-u)$ evaluated at $u=0$, i.e. $f(a)$.