Calculate the Dirac delta function integral

[PRASHANT KUMAR]

https://1drv.ms/w/s!Aip12L2Kz8zghV6Cnr8jPcRTpqTX
https://1drv.ms/w/s!Aip12L2Kz8zghV6Cnr8jPcRTpqTX
My question is in the above link

 

[PeroK]

https://1drv.ms/w/s!Aip12L2Kz8zghV6Cnr8jPcRTpqTX

My question is in the above link

Can you draw a graph of $\delta(x^2 -3x+2)$?

 

[PRASHANT KUMAR]

https://1drv.ms/w/s!Aip12L2Kz8zghWCxSwKUHyOsZE-6
here is the the graph of this function

 

[PeroK]

https://1drv.ms/w/s!Aip12L2Kz8zghWCxSwKUHyOsZE-6
here is the the graph of this function

So, it has two spikes?

 

[PRASHANT KUMAR]

yes it has

 

[PeroK]

yes it has

Okay, so does that not give you a clue as to how to evaluate the integral? Hint: trying splitting the integral.

Note: you'll also have to think about a substitution to get the $\delta$ function into a simpler format.

 

[PRASHANT KUMAR]

https://1drv.ms/w/s!Aip12L2Kz8zghWJU31OqLM79se67

 

[PRASHANT KUMAR]

Okay, so does that not give you a clue as to how to evaluate the integral? Hint: trying splitting the integral.

Note: you'll also have to think about a substitution to get the $\delta$ function into a simpler format.

I did not get this Note: you'll also have to think about a substitution to get the $\delta$ function into a simpler format.[/QUOTE]

 

[PeroK]

I did not get this Note: you'll also have to think about a substitution to get the $\delta$ function into a simpler format.

[/QUOTE]

Think about how to calculate, for example:

$\int_{-\infty}^{+\infty} f(x)\delta(x^2)dx$

 

[PRASHANT KUMAR]

it will be f(0)

 

[PeroK]

it will be f(0)

Try the substitution $y = x^2$.

In fact, a simpler example to try first is:

$\int_{-\infty}^{+\infty} f(x)\delta(2x)dx$

 

[PRASHANT KUMAR]

okay i got it
when y =2x then dy = 2 dx so the integral will be 1/2 f(0)

 

[PRASHANT KUMAR]

is the discussion over?

 

[PeroK]

is the discussion over?

I was at the dentist!

So, you have to do something with that quadratic in the delta function.

 

[PRASHANT KUMAR]

what is wrong if i do it like this

$\int_∞^1 (x^2+1) δ(x^2-3x+2) \, dx$+$\int_1^2(x^2+1) δ(x^2-3x+2) \, dx$+$\int_2^\infty(x^2+1) δ(x^2-3x+2) \, dx$ (in the first integral the limits will be from - ∞ to 1)
= 0+2+5=7
since in the first integral the dirac delta function is zero and in the second and third it will be one

 

[PeroK]

what is wrong if i do it like this

$\int_∞^1 (x^2+1) δ(x^2-3x+2) \, dx$+$\int_1^2(x^2+1) δ(x^2-3x+2) \, dx$+$\int_2^\infty(x^2+1) δ(x^2-3x+2) \, dx$ (in the first integral the limits will be from - ∞ to 1)
= 0+2+5=7
since in the first integral the dirac delta function is zero and in the second and third it will be one

You have the same problem that $\delta(x^2 - 3x + 2) \ne \delta(x)$. That first function is not the Dirac Delta function, it's a composition function and doesn't have exactly the same properties as the Dirac Delta function.

 

[PRASHANT KUMAR]

so now if i consider $x^2-3x+2 = y$ then it will give me (2x-3)dx=dy and if i substitute dx=$\frac{dy}{2x-3}$
x will become $\frac{3\pm\sqrt{1+y}}{2}$ and $x^2$= $\frac{(10+y)\pm6\sqrt{1+y}}{4}$then my integral will look like
$\int_{-\infty}^{+\infty}\frac{(\frac{(10+y)\pm6\sqrt{1+y}}{4}+1)δ(y)}{\pm\sqrt{1+y}}\,dy$
here if we take + signs into account then the answer comes out to be 5 and if we consider - signs then the answer is -2 which is not appropriate because the integral of dirac delta function should be a unit step function and which is one if y is greater than or equal to zero and everywhere else it is zero so the positive 5 should be considered as correct answer . i think so
is this correct?

 

[PRASHANT KUMAR]

You have the same problem that $\delta(x^2 - 3x + 2) \ne \delta(x)$. That first function is not the Dirac Delta function, it's a composition function and doesn't have exactly the same properties as the Dirac Delta function.

$δ(x^2-3x+2)≠δ(x)$ because δ(x) has only one spike at zero while $δ(x^2-3x+2)$ has two spikes at 1 and 2 .
But i have a doubt what is the need of equating these two?

 

[PeroK]

what is wrong if i do it like this

$\int_∞^1 (x^2+1) δ(x^2-3x+2) \, dx$+$\int_1^2(x^2+1) δ(x^2-3x+2) \, dx$+$\int_2^\infty(x^2+1) δ(x^2-3x+2) \, dx$ (in the first integral the limits will be from - ∞ to 1)
= 0+2+5=7
since in the first integral the dirac delta function is zero and in the second and third it will be one

This was better. You need to split the integral before you look at substitutions. I would do the first integral from $0$ to $3/2$ and the second from $3/2$ to $3$, say. That seems a lot neater.

Then you can use the $y$ substitution. Although, there is also a trick to leave the x-function as it is. Since you are not going to integrate it, it won't change. This is what I mean:

With $y = x^2 - 3x + 2 = (x - 3/2)^2 - 1/4$ and $dy = (2x - 3)dx$

$\int_0^{3/2}(x^2 + 1) \delta(x^2 - 3x + 2)dx = \int_2^{-1/4} \frac{x^2 +1}{2x-3} \delta(y) dy = -\int_{-1/4}^2 \frac{x^2 +1}{2x-3} \delta(y) dy = -\frac{x^2 +1}{2x-3}|_{y=0} = -\frac{x^2 +1}{2x-3}|_{x=1} = 2$

Notice that because $2x-3 = -1$ you get same answer,. This is a coincidence. If you change the quadratic it won't be the same.

 

[PRASHANT KUMAR]

This was better. You need to split the integral before you look at substitutions. I would do the first integral from $0$ to $3/2$ and the second from $3/2$ to $3$, say. That seems a lot neater.

Then you can use the $y$ substitution. Although, there is also a trick to leave the x-function as it is. Since you are not going to integrate it, it won't change. This is what I mean:

With $y = x^2 - 3x + 2 = (x - 3/2)^2 - 1/4$ and $dy = (2x - 3)dx$

$\int_0^{3/2}(x^2 + 1) \delta(x^2 - 3x + 2)dx = \int_2^{-1/4} \frac{x^2 +1}{2x-3} \delta(y) dy = -\int_{-1/4}^2 \frac{x^2 +1}{2x-3} \delta(y) dy = -\frac{x^2 +1}{2x-3}|_{y=0} = -\frac{x^2 +1}{2x-3}|_{x=1} = 2$

Notice that because $2x-3 = -1$ you get same answer,. This is a coincidence. If you change the quadratic it won't be the same.

how did you chose the limits of integration from 0 to 3/2 and 3/2 to3

 

[PeroK]

how did you chose the limits of integration from 0 to 3/2 and 3/2 to3

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?

 

[PRASHANT KUMAR]

This was better. You need to split the integral before you look at substitutions. I would do the first integral from $0$ to $3/2$ and the second from $3/2$ to $3$, say. That seems a lot neater.

Then you can use the $y$ substitution. Although, there is also a trick to leave the x-function as it is. Since you are not going to integrate it, it won't change. This is what I mean:

With $y = x^2 - 3x + 2 = (x - 3/2)^2 - 1/4$ and $dy = (2x - 3)dx$

$\int_0^{3/2}(x^2 + 1) \delta(x^2 - 3x + 2)dx = \int_2^{-1/4} \frac{x^2 +1}{2x-3} \delta(y) dy = -\int_{-1/4}^2 \frac{x^2 +1}{2x-3} \delta(y) dy = -\frac{x^2 +1}{2x-3}|_{y=0} = -\frac{x^2 +1}{2x-3}|_{x=1} = 2$

Notice that because $2x-3 = -1$ you get same answer,. This is a coincidence. If you change the quadratic it won't be the same.

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

 

[PRASHANT KUMAR]

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?

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.

 

[PeroK]

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

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 $dy$.

 

[PRASHANT KUMAR]

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 $dy$.

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

 

[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.

 

[PeroK]

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

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

 

[PRASHANT KUMAR]

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

so where does the solution lie?

 

[PRASHANT KUMAR]

@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.

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

 

[PRASHANT KUMAR]

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 $dy$.

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

 

[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?