Calculate the Dirac delta function integral

1. Jan 3, 2018

PeroK

They can be anything, as long as they cover the two zeroes of the quadratic. $0.9 - 1.1$ and $1.9 - 2.1$ would have done just as well.

Can you see the problem trying to do the quadratic substitution across the whole range? And why you should split the integral first?

2. Jan 3, 2018

PRASHANT KUMAR

$-\int_{-1/4}^2 \frac{x^2 +1}{2x-3} \delta(y) dy$ why did you put a negative sign here?

3. Jan 3, 2018

PRASHANT KUMAR

yes i think the problem in this case arises because of the two spikes of the function so there is a need to split the integral so that we can solve it separately for each one.

4. Jan 3, 2018

PeroK

The integral with respect to $y$ came out the "wrong" way. The lower limit was greater than the upper limit. In general, you need to look out for this when integrating:

$\int_a^b = - \int_b^a$

So, for example:

$\int_1^{-1}\delta(x)dx = -1$

I could have avoided this by having $y = -(x^2 -3x +2)$ in that integral. Then the limits would have been the right way round (with respect to $y$) and the negative sign would have come from $dy$.

5. Jan 3, 2018

PRASHANT KUMAR

But if we consider $y = -(x^2 -3x +2)$then our dirac delta function will look like δ(-y) is that okay?

6. Jan 3, 2018

PeroK

@PRASHANT KUMAR even at this level graphs help. I have a nice graph of $x^2-3x +2$, so I can see clearly what the function is doing about the zeros and what happens when I substitute $y$ It is easy to miss that the function has a negative gradient around $x=1$ so the natural integral substitution comes out the wrong way.

7. Jan 3, 2018

PeroK

Good point. Best to stick with the way I did it first time!

8. Jan 3, 2018

PRASHANT KUMAR

so where does the solution lie?

9. Jan 3, 2018

PRASHANT KUMAR

yes it has negative gradient but i could not link these two : natural integral and negative gradient

10. Jan 3, 2018

PRASHANT KUMAR

$\int_1^{-1}\delta(x)dx = -1$ i think this should be 1

11. Jan 3, 2018

PeroK

Okay. Let's look at

$I = \int f(x) \delta((x-a)(x-b))dx$ where $a < b$

Your problem is a special case of this, but it may be easier to look at the general case to see what is going on.

If we draw a graph of this quadratic, we get two roots at $x = a$ and $x = b$. So, we can see that:

$I = \int_{a-\epsilon}^{a+\epsilon} f(x) \delta((x-a)(x-b))dx + \int_{b-\epsilon}^{b+\epsilon} f(x) \delta((x-a)(x-b))dx$

Where we have focused on the two zeroes and taken a small interval around each, using a small $\epsilon$

Now, if we do the substitution $y = (x-a)(x-b)$ and $dy = (2x - a -b) dx$ and

$I = \int_{y_1}^{y_2} \frac{f(x)}{2x-a-b} \delta(y)dy + \int_{y_3}^{y_4} \frac{f(x)}{2x-a-b} \delta(y)dy$

Where $y_1, y_2$ etc. are the y-values for $a-\epsilon, a+\epsilon$ etc.

Now, from the graph, we have to note that $y_1 > 0$ and $y_2 < 0$. So, we need to get these limits in the correct order. Also, as we have this in the Dirac function for $y$, we are simply evaluating at y=0, which equates to $x =a$ and $x=b$ respectively. So:

$I = -\frac{f(x)}{2x-a-b}|_{x=a} + \frac{f(x)}{2x-a-b}|_{x=b} = \frac{f(a) +f(b)}{b-a}$

12. Jan 3, 2018

PeroK

No.

$\int_1^{-1}\delta(x)dx = -\int_{-1}^{1}\delta(x)dx = -1$

As I said, this is generally true for any integral on any interval. If you swap the limits, you change the sign of the integral.

13. Jan 3, 2018

PRASHANT KUMAR

yes that is true if we change the sign then the limits are interchanged but how can it imply that the lower limit should always be less than the upper limit

14. Jan 3, 2018

PeroK

It doesn't, but if you use a substitution to solve an integral sometimes the limits come out the wrong way. You might get something like:

$\int_{\pi}^{0} \sin(x) dx = -\cos(x)|_{\pi}^0 = -1 -1 = -2$

And, we know that $\int_0^{\pi} \sin(x) = 2$

This is a basic property of the integral we are talking about here.

15. Jan 3, 2018

PRASHANT KUMAR

while solving the problem if we do such type of substitution then it is necessary to put a negative sign otherwise it will be wrong?

16. Jan 3, 2018

PeroK

No. But, you were going to use:

(X) $\int_{-1}^{1}\delta(x)dx = 1$

And, you were going to assume that the order of the limits doesn't matter. So, you were going to assume that:

$\int_1^{-1}\delta(x)dx = 1$

Which would be wrong. So, you have to get the limits in the right order before you apply equation (X).

Take a look at the first property of the definite integral here (half way down the page).

http://tutorial.math.lamar.edu/Classes/CalcI/DefnOfDefiniteIntegral.aspx

17. Jan 3, 2018

Staff: Mentor

Thread closed, as the OP is now banned.