Calculate the Dirac delta function integral

[PeroK]

so where does the solution lie?

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:

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

 

[PeroK]

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

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.

 

[PRASHANT KUMAR]

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.

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

 

[PeroK]

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

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.

 

[PRASHANT KUMAR]

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.

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?

 

[PeroK]

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?

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

 

[Mark44]

Thread closed, as the OP is now banned.