# Complicated double integral

• A
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Double integral as a result of some work I'm doing
I have an integral:
$$\int_{-1}^{0}\int_{-1}^{q}\delta(s+a)\sinh[k(q-s)]dsdq$$
where $0<a<1$ and $\delta (s-a)$ is a dirac delta function. Anyone know what to do?

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• HallsofIvy

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

What's the difficulty?

• hunt_mat
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The delta function perhaps? I get it that it's a complicated integral.

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What does the delta function do when you integrate over s?

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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##.

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What does the delta function do when you integrate over s?
You don't know a priori if $a\in(-1,q]$, that's the crux of the problem.

• Delta2
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You don't know a priori if $a\in(-1,q]\$, 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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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 $a$ would be in the integration range, so you can simply write:
$$\int_{-1}^{0}\int_{-1}^{q}\delta (s+a)\sinh[k(q-s)]dsdq=\int_{-1}^{0}\sinh[k(q+a)]=\frac{1}{k}(\cosh(ka)-\cosh[k(a-1)])$$
This seems like a sloppy way of arguing however.

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How's that going to help when you have another integral to do. I was thinking that at one point $a$ would be in the integration range, so you can simply write:
$$\int_{-1}^{0}\int_{-1}^{q}\delta (s+a)\sinh[k(q-s)]dsdq=\int_{-1}^{0}\sinh[k(q+a)]=\frac{1}{k}(\cosh(ka)-\cosh[k(a-1)])$$
This seems like a sloppy way of arguing however.
That doesn't look right. You can split the outer integral in ##dq##.

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Ps you've got ##\delta(s + a)## in your integral. I assume it should be ##\delta(s -a)##.

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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 $a$ can take, then you'll see where you mistake is.

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That doesn't look right. You can split the outer integral in ##dq##.
How is that going to help? $a\in (-1,0)$ and so it's going to be in there no matter what.

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How is that going to help? $a\in (-1,0)$ 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?

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You assume wrong. If you look at the interval and the range $a$ can take, then you'll see where you mistake is.

If ##q, s, a## are all negative then ##\delta(s+a)## is always ##0##.

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Summary: Double integral as a result of some work I'm doing

I have an integral:
$$\int_{-1}^{0}\int_{-1}^{q}\delta(s+a)\sinh[k(q-s)]dsdq$$
where $-1<a<0$ and $\delta (s-a)$ is a dirac delta function. Anyone know what to do?
Note that you have both ##s +a## and ##s-a## in this post.

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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.

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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.

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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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If ##q, s, a## are all negative then ##\delta(s+a)## is always ##0##.
I don't think so.

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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.

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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.

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##\frac 1 k (\cosh(ka)-1)##
Go figure.

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I see where you're confused. Having thought about it $a\in(0,1)$ which is confusing things.

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I see where you're confused. Having thought about it $a\in(0,1)$ 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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If you split up the inner integral as:
$$\int_{-1}^{q}=\int_{-1}^{a-\varepsilon}+\int_{a-\varepsilon}^{a+\varepsilon}+\int_{a+\varepsilon}^{q}$$
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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##\frac 1 k (\cosh(ka)-1)##
Go figure.
This solution doesn't look right to me.

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This solution doesn't look right to me.
I figured as much. DIY

• hunt_mat
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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.

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Thank you for your help thought PeroK.

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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.

• • hunt_mat and PeroK
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Thank you.

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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?

HomogenousCow
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".