Evaluate ##\int_{-\infty}^{\infty} e^{-|x|}\delta(x^2 +2x -3) dx##

[DragonBlight]

Homework Statement

evaluate $\int_{-\infty}^{\infty} e^{-|x|}\delta(x^2 +2x -3) dx$

Relevant Equations

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

Hi,

Is it correct to say that the dirac delta function is equal to 0 except if the argument is 0?
Thus, $x^2 +2x -3$ must be equal to 0.

Then, we have x = 1 or -3. What does that means?

$\int_{-\infty}^{\infty} e^{-|x|}\delta(x^2 +2x -3) dx = e^{-1}$ and/or $e^{-3}$ ?

Thank you

 

[Orodruin]

The delta function argument will be zero at two points. Both will contribute to the integral. You also need to consider that $\delta(ax) = \delta(x)/|a|$ with the appropriate generalisation to non-linear arguments.

 

[DragonBlight]

I'm not sure to fully understand.
What should be the result? In a case where the argument is 0 only if x = 1, the result would have been $e^{-1}$

 

[Orodruin]

I'm not sure to fully understand.
What should be the result? In a case where the argument is 0 only if x = 1, the result would have been $e^{-1}$

No, this is incorrect. That is only true if you have something like $\delta(x-1)$. Whether or not there are additional constants depend also on the derivative of the argument. See the Wikipedia entry https://en.wikipedia.org/wiki/Dirac_delta_function particularly under ”properties”.

 

[vela]

Then, we have x = 1 or -3. What does that means?

$\int_{-\infty}^{\infty} e^{-|x|}\delta(x^2 +2x -3) dx = e^{-1}$ and/or $e^{-3}$ ?

Naively, you might expect the result to be the sum of those two because the integral is essentially a sum, but it's a little more complicated.

I'm not sure to fully understand.
What should be the result? In a case where the argument is 0 only if x = 1, the result would have been $e^{-1}$

The one thing you know about the delta function is
$$\int f(x)\delta(x)\,dx = f(0).$$ Now consider an integral like
$$\int f(x)\delta(2x)\,dx.$$ Because of the $2x$ inside the delta function, it's not exactly in the same form as the first integral. Hence, it would be wrong to conclude the second integral is equal to $f(0)$ even though that's where the argument of the delta function is 0. If you use the substitution $u=2x$, however, you can transform the integral into the form above so that you can evaluate it:
$$\int f(x)\delta(2x)\,dx = \int\underbrace{ \frac 12 f(u/2)}_{g(u)} \delta(u) \,du = \int g(u)\delta(u)\,du = g(0) = \frac 12 f(0).$$

 

[DragonBlight]

So, can I replace the argument with u to find a "form" that I know for the delta function in every situation?

For example, $u = x^2 +2x -3$ and $x = -1 \pm \sqrt{u + 4}$
Thus, I have $e^{- 1 - \sqrt{u + 4}}$ and $\delta(u)$

So we have
$\int_{-\infty}^{\infty} e^{-1-\sqrt{u+4} \delta(u) \frac{du}{2\sqrt{u+4}}}$
For $\delta(0)$
$=e^{-3}/4$

 

[vela]

Something like that. You should read the Wikipedia page @Orodruin linked to, particularly the part about composition with a function.

 

[Orodruin]

So, can I replace the argument with u to find a "form" that I know for the delta function in every situation?

For example, $u = x^2 +2x -3$ and $x = -1 \pm \sqrt{u + 4}$
Thus, I have $e^{- 1 - \sqrt{u + 4}}$ and $\delta(u)$

So we have
$\int_{-\infty}^{\infty} e^{-1-\sqrt{u+4} \delta(u) \frac{du}{2\sqrt{u+4}}}$
For $\delta(0)$
$=e^{-3}/4$

You must be much more careful with your charge of variables in the integral. What are the bounds of the new integral?