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Im trying to prove the identity above, what I got until now is:

for

$y = (x-a)(x-b) = x^2 -(a+b)x + ab $

we have

$ dy = [2x - (a+b)] dx $

so

$dx = \frac{dy}{2x - (a+b)}$

from here I used: $0 = x^2 -(a+b)x + (ab -y)$

and now it's just bhaskara, but here it's my problem I can't seem to get the root right,

$ x = \frac{(a+b) \pm \sqrt{(a+b)^2 - 4(ab - y)}}{2}$

I know the root it's supposed to give $(a-b)$ but I'm not getting there cause the 4y factor after i replace the y value it gets confused and it doesn't seem to work (or I'm not seeing what it's the thing)

My teacher also told that we can prove by doing

$\delta c(x-a) = \frac{x-a}{c}$

with $ c = x-b $ and I tried but it didn't seem so obvious to me but maybe I'm wrong, any tips are welcome :)

maria evangelista
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1 Answers1

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In general, when you have a function with a rather complicated argument, it is a good idea to change variables, the way you did. You then also need the relation between the new integration element $dy$ and the old one $dx$. That is all fine.

However, this exercise is about the Dirac delta function. That is a function with special properties. The most important property is that the function only contributes when the argument is zero. For other values of the argument the function is zero. Now it is easy to see that $(x-a)(x-b)$ has two zeros, namely $x=a$ and $x=b$. You can visualize this by drawing an x-axis with two sharp peaks; at $x=a$ and $x=b$.

This property allows us to treat the two peaks separately. Around $x=a$ we can ignore the variation of $x$ in the term $(x-b)$ and treat it as $(a-b)$. And vice versa. Thus, using the property that a constant can be taken outside the delta function (as hinted by your teacher) one obtains indeed:

$$\delta ((x-a)(x-b)) = [\delta (x-a) + \delta (x-b)] / |a-b|$$

Note that this result is only valid when $a$ and $b$ are unequal. When $a = b$ the two peaks merge into one, and consequently the result is different.

M. Wind
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