Dirac delta - sifting

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?

• Look at representations of the delta function, they are all even... – tired Oct 25 '15 at 19:24

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

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

"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. $f(a)$.

• Is that really formal, though? The integral notation is meaningless in the sense that it doesn't really refer to an integral, right? – layman Oct 25 '15 at 19:17
• @user46944 "Formally" in this sense means "only formally", that is, I haven't given the justification for why I can do this manipulation, but this is the manipulation that I "should" be able to do. To really justify this, one should be more careful, for instance by passing to an approximating sequence. – Ian Oct 25 '15 at 19:18
• So by "passing to an approximating sequence of a distribution", do you mean finding a sequence of actual functions $\delta_{n}$ such that the distribution they define is actually integration and the limit of the distributions converges to the dirac delta distribution? – layman Oct 25 '15 at 19:24
• @user46944 Yes. For the Dirac delta this is an especially standard technique; in this case the approximating sequence is called an approximate identity. (The name is because the Dirac delta is the identity element for the convolution operator.) – Ian Oct 25 '15 at 19:24
• @user46944 : "Formally" means it's not logically justified but you do it anyway. – Michael Hardy Oct 25 '15 at 19:24

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.

• I am so confused by this answer. What does it mean for $\delta(x-a)$ to "peak" at an argument when it's not even a defined function? – layman Oct 25 '15 at 19:15
• That's a heuristic argument, I'm editing the answer right now to include a formal one. – Soba noodles Oct 25 '15 at 19:16
• Thanks, I'm looking forward to reading it. My understanding of the dirac delta function is very limited. – layman Oct 25 '15 at 19:17