Integrating a delta function of a function

[Milsomonk]

Homework Statement

Evaluate the integral:
$$\int_{-\infty}^{\infty} dx *\dfrac {\delta (x^2-2ax)} {x+b}$$

Homework Equations

$$ x^2-2ax=0 $$

The Attempt at a Solution

I know that the delta function can only be none zero when $$ x=2a$$ so then I have the following integral:
$$\int_{-\infty}^{\infty} dx \dfrac {1} {2a+b} $$
But I'm not sure how to evaluate this integral, or even if this is the correct approach to the problem. Any guidance would be appreciated :)

 

[Orodruin]

I know that the delta function can only be none zero when $$ x=2a$$ so then I have the following integral:
$$\int_{-\infty}^{\infty} dx \dfrac {1} {2a+b} $$
But I'm not sure how to evaluate this integral, or even if this is the correct approach to the problem. Any guidance would be appreciated :)

The delta function is non-zero whenever its argument is zero. Is $x = 2a$ the only point where this happens?

In addition, your integral is formally infinite. This indicates that you have not used the delta function appropriately, please show how you arrived at this conclusion.

 

[Milsomonk]

I suppose also when $$ x=0$$ it would be non zero. I just set it equal to 1 for the value of $$x$$ that I found. Thanks for the response :)

 

[Orodruin]

I suppose also when $$ x=0$$ it would be non zero. I just set it equal to 1 for the value of $$x$$ that I found. Thanks for the response :)

That is not how the delta function works. You cannot just replace it with a one in the integral. What is the defining property of the delta function?

 

[Milsomonk]

Ah, whoops. That it has area 1, so only makes sense once integrated over?

 

[Milsomonk]

Ok, so it would make sense though to substitute my value for x into the denominator and then take it outside of the integrand, then I can integrate the delta function on its own?
$$\int_{-\infty}^{\infty} dx *\dfrac {\delta (x^2-2ax)} {2a+b} = \dfrac {1}{2a+b} \int_{-\infty}^{\infty} dx \delta (x^2-2ax) = \dfrac {1}{2a+b}$$

 

[Orodruin]

First of all, you need to account for the fact that the delta function is non-zero in two points, which you are not doing at the moment. Second, you need to take into account that the argument of the delta function is a function of x and not just $x - x_0$.

 

[Milsomonk]

Thanks very much for your help and patience :)

Ok, I'm really not sure how to proceed with regards to the first point, perhaps split it into two separate delta functions but I don't know how I'd go about that. The second point I think I need to use this formula:
$$\delta (f(x))=\dfrac {\delta (x-x_0)} {\vert f^\prime (x_0) \vert}$$
So then I have:
$$\int_{-\infty}^{\infty} dx \dfrac {\delta (x-x_0)} {\vert 2x_0-2a \vert} \dfrac {1}{x+b}$$
Then if $$ x_0=2a$$ I have:
$$\dfrac {1} {2a}\int_{-\infty}^{\infty} dx \dfrac {\delta (x-x_0)}{x+b}$$
I can then I think us the sifting property to evaluate this but I'm not even sure I'm right up to this point.

 

[Milsomonk]

Ok so I have followed your advice and found a way to account for the two places at which the delta function of zero, I can't see any issue with my method but I do not get the same answer as wolfram alpha:
$$\int_{-\infty}^{\infty} dx *\dfrac {\delta (x^2-2ax)} {x+b}$$
Then using:
$$\delta (f_{(x)}) =\dfrac {\delta (x-x_1)} {\vert f^\prime_{(x_1)} \vert} + \dfrac {\delta (x-x_2)} {\vert f^\prime_{(x_2)} \vert}$$
I can now use this to rewrite the integral. I also know the roots:
$$x_1=2a , x_2=0$$ and $$ f^\prime_{(x_1)} =4a , f^\prime_{(x_2)} =0$$
so
$$ \int_{-\infty}^{\infty} dx *\dfrac {\delta (x^2-2ax)} {x+b}=\dfrac {1} {4a}\int_{-\infty}^{\infty} dx *\dfrac {\delta (x-2a)} {x+b}+\dfrac {1} {2a} \int_{-\infty}^{\infty} dx *\dfrac {\delta (x)} {x+b}$$
Then I use the sifting property:
$$\int_{-\infty}^{\infty} dx * \delta (x-x_0) f(x) = f(x_0)$$
$$f(x)=\dfrac {1} {x+b}$$
my integral becomes:
$$ \dfrac {1} {4a(2a+b)} +\dfrac {1} {2ab}=\dfrac {1} {8a^2+6ab} $$
However Mathematica and wolfram tell me the answer is:
$$\dfrac {1} {2ab} $$
Any guidance on where my method or calculation could be flawed would be appreciated :)

 

[Orodruin]

Overall your methodology looks fine. Don't be overconfident in what Mathematica spits out. It is clearly missing the contribution from $x = 2a$.

However, note that
$$
f'(x) = 2x - 2a
$$
and therefore $f'(0) = -2a \neq 0$ and $f'(2a) = 2a \neq 4a$.

 

[Milsomonk]

Hi, thanks :) that's good to hear. I'd hoped that was the issue with mathematica. whoops that was a typo in my notes for x_1 and a silly mistake for x_2 ! thanks very much for your help it :)

 

[Orodruin]

So after some testing, for some reason Mathematica 11 seems to handle the delta function properly when you put an actual number instead of $a$ but seems unable to handle it symbolically. I find this weird as it is perfectly able to solve $x^2 - 2ax = 0$ and find the derivative of $f$ ...

 

[Milsomonk]

That's pretty surprising, I can't imagine why that would be, it surely couldn't be hard for it to evaluate symbolically if it can handle it numerically. Wolfram alpha appears to be the same...