Solve $$\int_{∞}^{∞}dxf(x)\delta((x-x_1))$$: Dirac Delta Function

[Arman777]

Homework Statement

Dirac Delta Properties

Relevant Equations

$$ I = \int_{∞}^{∞}dxf(x)δ((x - x_1)(x-x_2)) = ?$$

If the question was
$$ \int_{∞}^{∞}dxf(x)δ((x - x_1)) = ? $$ The answer would be $f(x_1)$

So the delta function has two roots, I searched the web and some books but I am not sure what approach should I use here. I guess there's sometihng happens when $x_1 = -x_2$.

So I am not sure what to do at this point.

 

[WWGD]

Notice that I is not the delta function ( which is usually considered a distribution and not a function) but an expression involving the delta function or distribution. What are the arguments , if any, given for the expression I?

 

[Arman777]

Notice that I is not the delta function ( which is usually considered a distribution and not a function) but an expression involving the delta function or distribution. What are the arguments , if any, given for the expression I?

There arent any. We need to find I

 

[George Jones]

$$\int_{-∞}^{∞}dxf(x)δ((x - x_1)(x-x_2)) = ?$$

This is a special case of
$$\int_{-\infty}^{\infty} dx f \left(x\right) \delta \left( g \left(x \right) \right).$$

Have you seen this before?

 

[Arman777]

This is a special case of
$$\int_{-\infty}^{\infty} dx f \left(x\right) \delta \left( g \left(x \right) \right).$$

Have you seen this before?

Yes I have seen it.
$$\int_{-\infty}^{\infty} dx f \left(x\right) \delta \left( g \left(x \right) \right) = \Sigma f(x_i)/g'(x_i)$$ ? But I am not sure why this is the case

 

[George Jones]

Yes I have seen it.
$$\int_{-\infty}^{\infty} dx f \left(x\right) \delta \left( g \left(x \right) \right) = \Sigma f(x_i)/g'(x_i)$$ ? But I am not sure why this is the case

Then, it might be good to do your original question as an illustrative example. Shift your coordinates such that the new coordinate system has its zero halfway between $a_1$ and $a_2$.

 

[PeroK]

Yes I have seen it.
$$\int_{-\infty}^{\infty} dx f \left(x\right) \delta \left( g \left(x \right) \right) = \Sigma f(x_i)/g'(x_i)$$ ? But I am not sure why this is the case

Perhaps that should be $|g'(x_i)|$?

 

[Arman777]

Perhaps that should be $|g'(x_i)|$?

Yes I was being lazy to put those sign.