$$\int_{-\infty}^\infty (x^2 + 1) \delta(x^2 - 3x + 2) dx =\ ?$$
Which property of Dirac Delta function can be used to solve this type of problems? Or can we call it a Dirac Delta function having two centres?
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2Would [math.se] be a better home for this question? – Qmechanic Mar 29 '18 at 08:30
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Just take a look of its properties, link. – secavara Mar 29 '18 at 09:57
3 Answers
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As already pointed out you should take a look of the properties of the Dirac delta for the general case.
Nonetheless in this case your intuition gives the correct answer:
you can regard $\delta(x^2-3x+2)$ as $\delta(x-1)+\delta(x-2)$, where $1$ and $2$ are the roots of the polynomial $x^2-3x+2$.
The result is then simply achieved, and the original integral becomes:
$$\int_{-\infty}^{+\infty}(x^2+1)\delta(x-2) + \int_{-\infty}^{+\infty}(x^2+1)\delta(x-1) = (2^2+1)+(1^2+1) = 7$$
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1This is generally not correct: $\delta(x^2-3x+2)=|2x-3|^{-1} \left ( \delta(x-1) + \delta(x-2) \right )$. It is a coincidence here that $|2(1)-3|=|2(2)-3|=1$. – Ian Mar 29 '18 at 13:10
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Correct, I guess I should have put more emphasis on this, thanks. – Raffaele d'Amelio Mar 29 '18 at 13:17
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$$\int_{-\infty}^\infty (x^2 + 1) \delta(x^2 - 3x + 2) dx=\int_{-\infty}^\infty (x^2 + 1) \delta((x-1)(x-2)) dx$$ $$=\int_{2-\epsilon}^{2+\epsilon} (x^2 + 1) \delta((x-1)(x-2)) dx+\int_{1-\epsilon}^{1+\epsilon} (x^2 + 1) \delta((x-1)(x-2)) dx$$ $$=2^2+1 + 1^2+1 = 7$$
Jim Haddocc
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Use the property that $$\delta [g(x)]= \sum_i \frac{\delta(x-x_i)}{\left| g'(x_i)\right|}$$ then, $$\delta(x^2-3x+2)=\frac{\delta(x-1)}{1}+\frac{\delta(x-2)}{1}$$ For more information check this post: Dirac Delta Function of a Function
oyster
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