Double Integral with Dirac Delta Function and Changing Limits

In summary, the conversation discusses an integral with a Dirac delta function and a sinh function, where a is a positive value. Suggestions are given to split the integral and consider different cases, as well as to use a new variable and set new limits. It is mentioned that the solution may only have one variable left, a.
  • #1
hunt_mat
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TL;DR Summary
Double integral as a result of some work I'm doing
I have an integral:
[tex]
\int_{-1}^{0}\int_{-1}^{q}\delta(s+a)\sinh[k(q-s)]dsdq
[/tex]
where [itex]0<a<1[/itex] and [itex]\delta (s-a)[/itex] is a dirac delta function. Anyone know what to do?
 
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  • #2
You could try integrating the ##\sinh## function!

What's the difficulty?
 
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  • #3
The delta function perhaps? I get it that it's a complicated integral.
 
  • #4
What does the delta function do when you integrate over s?
 
  • #5
hunt_mat said:
The delta function perhaps? I get it that it's a complicated integral.
Okay, there's a ##q## in the integration bounds. I didn't see that.

Try looking at the cases ##q < a## and ##q > a##.
 
  • #6
Vanadium 50 said:
What does the delta function do when you integrate over s?
You don't know a priori if [itex]a\in(-1,q][/itex], that's the crux of the problem.
 
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  • #7
hunt_mat said:
You don't know a priori if [itex]a\in(-1,q]\[/itex], that's the crux of the problem.
Split the integral. Then you do know.

##a \in [-1, q]## when ##q > a##.
 
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  • #8
PeroK said:
Okay, there's a ##q## in the integration bounds. I didn't see that.

Try looking at the cases ##q < a## and ##q > a##.
How's that going to help when you have another integral to do. I was thinking that at one point [itex]a[/itex] would be in the integration range, so you can simply write:
[tex]
\int_{-1}^{0}\int_{-1}^{q}\delta (s+a)\sinh[k(q-s)]dsdq=\int_{-1}^{0}\sinh[k(q+a)]=\frac{1}{k}(\cosh(ka)-\cosh[k(a-1)])
[/tex]
This seems like a sloppy way of arguing however.
 
  • #9
hunt_mat said:
How's that going to help when you have another integral to do. I was thinking that at one point [itex]a[/itex] would be in the integration range, so you can simply write:
[tex]
\int_{-1}^{0}\int_{-1}^{q}\delta (s+a)\sinh[k(q-s)]dsdq=\int_{-1}^{0}\sinh[k(q+a)]=\frac{1}{k}(\cosh(ka)-\cosh[k(a-1)])
[/tex]
This seems like a sloppy way of arguing however.
That doesn't look right. You can split the outer integral in ##dq##.
 
  • #10
Ps you've got ##\delta(s + a)## in your integral. I assume it should be ##\delta(s -a)##.
 
  • #11
PeroK said:
Ps you've got ##\delta(s + a)## in your integral. I assume it should be ##\delta(s -a)##.
You assume wrong. If you look at the interval and the range [itex]a[/itex] can take, then you'll see where you mistake is.
 
  • #12
PeroK said:
That doesn't look right. You can split the outer integral in ##dq##.
How is that going to help? [itex]a\in (-1,0)[/itex] and so it's going to be in there no matter what.
 
  • #13
hunt_mat said:
How is that going to help? [itex]a\in (-1,0)[/itex] and so it's going to be in there no matter what.
Did you come for help or simply to reject any help you are offered?
 
  • #14
hunt_mat said:
You assume wrong. If you look at the interval and the range [itex]a[/itex] can take, then you'll see where you mistake is.

If ##q, s, a## are all negative then ##\delta(s+a)## is always ##0##.
 
  • #15
hunt_mat said:
Summary: Double integral as a result of some work I'm doing

I have an integral:
[tex]
\int_{-1}^{0}\int_{-1}^{q}\delta(s+a)\sinh[k(q-s)]dsdq
[/tex]
where [itex]-1<a<0[/itex] and [itex]\delta (s-a)[/itex] is a dirac delta function. Anyone know what to do?
Note that you have both ##s +a## and ##s-a## in this post.
 
  • #16
PeroK said:
Did you come for help or simply to reject any help you are offered?
No. I'm just questioning your answers. It's an involved integral. I don't have access to a symbolic algebra program. I gave the best thought I had, you rejected that without much justification. I'm open to suggestions, but I do have questions.
 
  • #17
PeroK said:
Note that you have both ##s +a## and ##s-a## in this post.
I wanted to point out that it was a delta function.
 
  • #18
hunt_mat said:
No. I'm just questioning your answers. It's an involved integral. I don't have access to a symbolic algebra program. I gave the best thought I had, you rejected that without much justification. I'm open to suggestions, but I do have questions.
##\int_{-1}^0 dq = \int_{-1}^a dq + \int_a^0 dq##

The first integral vanishes as ##s## is in the range ##[-1, q]## which does not include ##a##.

For the second integral tge range of ##s## always includes ##a##.
 
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  • #19
PeroK said:
If ##q, s, a## are all negative then ##\delta(s+a)## is always ##0##.
I don't think so.
 
  • #20
PeroK said:
##\int_{-1}^0 dq = \int_{-1}^a dq + \int_a^0 dq##

The first integral vanishes as ##s## is in the range ##[-1, q]## which does not include ##a##.

For the second integral tge range of ##s## always includes ##a##.
I don't understand why this is relevant.
 
  • #21
hunt_mat said:
I don't think so.
Why not? ##\delta(x) =0## except when ##x=0##. You can't get zero by adding two negative numbers. But, you can by subtracting them.
 
  • #22
##\frac 1 k (\cosh(ka)-1)##
Go figure.
 
  • #23
I see where you're confused. Having thought about it [itex]a\in(0,1)[/itex] which is confusing things.
 
  • #24
hunt_mat said:
I see where you're confused. Having thought about it [itex]a\in(0,1)[/itex] which is confusing things.
You see where I'm confused! That's a bit rich!

It's your post. If ##a## should be positive and you stated it was negative that's your confusion. Not mine.
 
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  • #25
If you split up the inner integral as:
[tex]
\int_{-1}^{q}=\int_{-1}^{a-\varepsilon}+\int_{a-\varepsilon}^{a+\varepsilon}+\int_{a+\varepsilon}^{q}
[/tex]
Then you're going to get the function I posted. Splitting up the outer integral isn't going to do much in my opinion.
 
  • #26
PeroK said:
##\frac 1 k (\cosh(ka)-1)##
Go figure.
This solution doesn't look right to me.
 
  • #27
hunt_mat said:
This solution doesn't look right to me.
I figured as much. DIY
 
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  • #28
PeroK said:
I figured as much. DIY
I did, or what I thought looked sensible. My solution doesn't look right to you, and yours doesn't look right to me.
 
  • #29
Thank you for your help thought PeroK.
 
  • #30
This has been painful to watch. PeroK has been giving excellent advice.

  • First, split the integral into two pieces, one where the delta function is zero everywhere and one where it is not.
  • Do the inner integral. The first part (above) is zero and the second part (above) sets s = -a. The only q left should be inside the sinh.
  • Set a new variable r = q + a. Set you limits in terms of r.
  • Do the outer (and only remaining) integral. I believe you will have only one a left.
 
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  • #31
Thank you.
 
  • #32
Vanadium 50 said:
This has been painful to watch. PeroK has been giving excellent advice.

  • First, split the integral into two pieces, one where the delta function is zero everywhere and one where it is not.
  • Do the inner integral. The first part (above) is zero and the second part (above) sets s = -a. The only q left should be inside the sinh.
  • Set a new variable r = q + a. Set you limits in terms of r.
  • Do the outer (and only remaining) integral. I believe you will have only one a left.
Actually I did that, and I posted the function. Why was that incorrect?
 
  • #33
You can do it pretty easily just be inspecting the integral limits and looking at the interval on which the delta function is nonzero. More specifically, changing the lower bound on the outer integral to ##-a## projects out the integration interval on which the delta function is "satisfied".
 

Related to Double Integral with Dirac Delta Function and Changing Limits

1. What is a double integral with Dirac delta function and changing limits?

A double integral with Dirac delta function and changing limits is a mathematical concept used in the field of calculus to calculate the area under a surface or volume between two surfaces. It involves the use of the Dirac delta function, which is a mathematical tool used to represent a point mass or impulse, and changing limits, which allows for the integration to be performed over a varying range of values.

2. How is a double integral with Dirac delta function and changing limits different from a regular double integral?

A regular double integral involves integrating over a fixed range of values, while a double integral with Dirac delta function and changing limits allows for the integration to be performed over a varying range of values. Additionally, the Dirac delta function introduces a point mass or impulse into the calculation, which can change the overall result of the integral.

3. What are some real-world applications of a double integral with Dirac delta function and changing limits?

A double integral with Dirac delta function and changing limits has various applications in physics, engineering, and other scientific fields. Some examples include calculating the electric potential of a charged particle, determining the moment of inertia of a rotating object, and analyzing the distribution of stress in a material.

4. How do you solve a double integral with Dirac delta function and changing limits?

Solving a double integral with Dirac delta function and changing limits involves breaking down the problem into smaller, more manageable parts and applying the appropriate mathematical techniques, such as substitution or integration by parts. It is also important to carefully consider the properties of the Dirac delta function and the changing limits to accurately evaluate the integral.

5. Are there any limitations or restrictions when using a double integral with Dirac delta function and changing limits?

Yes, there are some limitations and restrictions when using a double integral with Dirac delta function and changing limits. One limitation is that the Dirac delta function is only defined for real numbers, so the integration limits must also be real numbers. Additionally, the changing limits must be carefully chosen to ensure that the integral is well-defined and converges to a finite value.

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