Complicated double integral

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hunt_mat
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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
PeroK
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You could try integrating the ##\sinh## function!

What's the difficulty?
 
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  • #3
hunt_mat
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The delta function perhaps??? I get it that it's a complicated integral.
 
  • #4
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What does the delta function do when you integrate over s?
 
  • #5
PeroK
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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
hunt_mat
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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
PeroK
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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
hunt_mat
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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
PeroK
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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
PeroK
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Ps you've got ##\delta(s + a)## in your integral. I assume it should be ##\delta(s -a)##.
 
  • #11
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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
hunt_mat
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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
PeroK
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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
PeroK
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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
PeroK
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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
hunt_mat
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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
hunt_mat
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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
PeroK
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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
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If ##q, s, a## are all negative then ##\delta(s+a)## is always ##0##.
I don't think so.
 
  • #20
hunt_mat
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##\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
PeroK
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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
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##\frac 1 k (\cosh(ka)-1)##
Go figure.
 
  • #23
hunt_mat
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I see where you're confused. Having thought about it [itex]a\in(0,1)[/itex] which is confusing things.
 
  • #24
PeroK
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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
hunt_mat
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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.
 

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