How do you integrate dirac delta functions?

[KeithKp]

Homework Statement

∫δ(x³ - 4x²- 7x +10)dx. Between ±∞.

Homework Equations

The Attempt at a Solution

Well I don't really know how to attempt this. In the case where inside the delta function there is simply 2x, or 5x, I know the answer would be 1/2 or 1/5. Or for say δ(x^2-5), the answer would be 1/√5.
But I'm unsure how to go about this one. Wolfram Alpha states the answer is 1/6. Tried to factor the polynomial which results in (x-1)(x+2)(x-5). But that doesn't seem to help either.

Any guidance would be great thanks!

 

[Orodruin]

You are on the right track. You now know what points give a contribution (the roots of the polynomial) if they are inside your integration interval.

Now, do you remember where the factor 1/sqrt5 comes from in the case of x^2-5?

 

[KeithKp]

Are you saying that I should only consider the (x-5) and (x-1) roots? Would that then result in the answer being 1/6? Why would I not consider the negative root? :S
I know this relation exists:

∫dx⋅f(x)⋅δ(ax+b) = 1/|a|⋅f(-a^-1b). (sorry for the formatting) (limits ±∞ again)

Which I understand makes sense for say δ(2x). Not quite sure how √5 is a in δ(x^2-5). But I get the system. Then I'm really confused for the 3 order polynomial above.

https://www.physicsforums.com/file:///page36image21544 https://www.physicsforums.com/file:///page36image21704

 

[Ray Vickson]

Are you saying that I should only consider the (x-5) and (x-1) roots? Would that then result in the answer being 1/6? Why would I not consider the negative root? :S
I know this relation exists:

∫dx⋅f(x)⋅δ(ax+b) = 1/|a|⋅f(-a^-1b). (sorry for the formatting) (limits ±∞ again)

Which I understand makes sense for say δ(2x). Not quite sure how √5 is a in δ(x^2-5). But I get the system. Then I'm really confused for the 3 order polynomial above.

I can see i take the positive root for x^2-5, is that the same thinking for taking (x-5) and (x-1) above? But why is that?https://www.physicsforums.com/file:///page36image21544 https://www.physicsforums.com/file:///page36image21704

$$\delta(x^2-5) = \delta((x-\sqrt{5})(x + \sqrt{5})) = \delta(2 \sqrt{5} (x - \sqrt{5}) + \delta(-2 \sqrt{5}(x + \sqrt{5}))$$

 

[KeithKp]

$$\delta(x^2-5) = \delta((x-\sqrt{5})(x + \sqrt{5})) = \delta(2 \sqrt{5} (x - \sqrt{5}) + \delta(-2 \sqrt{5}(x + \sqrt{5}))$$

...Thank you, I'm an idiot.