# Homework Help: Calculate the Dirac delta function integral

1. Jan 3, 2018

### PRASHANT KUMAR

Last edited by a moderator: Jan 3, 2018
2. Jan 3, 2018

3. Jan 3, 2018

4. Jan 3, 2018

5. Jan 3, 2018

yes it has

6. Jan 3, 2018

### PeroK

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.

7. Jan 3, 2018

8. Jan 3, 2018

### PRASHANT KUMAR

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]

9. Jan 3, 2018

### PeroK

[/QUOTE]

Think about how to calculate, for example:

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

10. Jan 3, 2018

### PRASHANT KUMAR

it will be f(0)

11. Jan 3, 2018

### PeroK

Try the substitution $y = x^2$.

In fact, a simpler example to try first is:

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

12. Jan 3, 2018

### PRASHANT KUMAR

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

13. Jan 3, 2018

### PRASHANT KUMAR

is the discussion over?

14. Jan 3, 2018

### PeroK

I was at the dentist!

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

15. Jan 3, 2018

### 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

16. Jan 3, 2018

### PeroK

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.

17. Jan 3, 2018

### 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?

18. Jan 3, 2018

### PRASHANT KUMAR

$δ(x^2-3x+2)≠δ(x)$ 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?

19. Jan 3, 2018

### PeroK

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.

20. Jan 3, 2018

### PRASHANT KUMAR

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