1

I'm working on an integral that has a Dirac delta, and normally I'm fine with them, but this time I have the Dirac delta paired with a composition of functions.

The integral is,

$$ \int dx \delta(x-x_0)\cos (\phi(x,t)). $$

I might be over thinking this and the answer is just $\cos(\phi(x_0,t))$, but I can't find any reference to whether or not there is some special rule for this situation.

Can someone point me in the right direction?

L.A Rhoads
  • 81

1 Answers1

1

Suppose we have the $2$-variable delta function $\delta(x,y)$ and a $2$-variable function $f(x,y)$, then:

$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y)\delta(x-x_0,y-y_0) \ dx dy = f(x_0,y_0).$$

This gives the same answer whether you integrate with respect to $x$ or $y$ first. Indeed:

$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y)\delta(x-x_0,y-y_0) \ dy dx = \int_{-\infty}^{\infty}f(x,y_0)\delta(x-x_0)dx=f(x_0,y_0)$$

and similarly when you integrate wrt to $x$ first.

Thus we have:

$$\int_{-\infty}^{\infty}\delta(x-x_0)f(x,t)dx=f(x_0,t)$$

for all $x_0 \in \mathbb R.$

Alessio K
  • 10,599
  • Thanks for your answer, but I'm more concerned with how the composition of functions affects the Dirac delta. Does that affect how the integral works? – L.A Rhoads Sep 07 '20 at 15:05
  • 1
    No, it shouldn't affect how the integral works (see here for instance: https://math.stackexchange.com/questions/1580966/double-integral-with-a-delta-function). Of course $f(x,t)=\cos( \phi (x,t))$ is a multivariable composition of functions. – Alessio K Sep 07 '20 at 15:18