# Complicated double integral

• A
Homework Helper
TL;DR 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 $0<a<1$ and $\delta (s-a)$ is a dirac delta function. Anyone know what to do?

Last edited:
HallsofIvy

Homework Helper
Gold Member
2022 Award
You could try integrating the ##\sinh## function!

What's the difficulty?

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

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

Homework Helper
Gold Member
2022 Award
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##.

Homework Helper
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
Homework Helper
Gold Member
2022 Award
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##.

Homework Helper
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.

Homework Helper
Gold Member
2022 Award
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##.

Homework Helper
Gold Member
2022 Award
Ps you've got ##\delta(s + a)## in your integral. I assume it should be ##\delta(s -a)##.

Homework Helper
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.

Homework Helper
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.

Homework Helper
Gold Member
2022 Award
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?

Homework Helper
Gold Member
2022 Award
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##.

Homework Helper
Gold Member
2022 Award
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.

Homework Helper
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.

Homework Helper
Note that you have both ##s +a## and ##s-a## in this post.
I wanted to point out that it was a delta function.

Homework Helper
Gold Member
2022 Award
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##.

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

Homework Helper
##\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.

Homework Helper
Gold Member
2022 Award
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.

Homework Helper
Gold Member
2022 Award
##\frac 1 k (\cosh(ka)-1)##
Go figure.

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

Homework Helper
Gold Member
2022 Award
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.

Homework Helper
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.

Homework Helper
##\frac 1 k (\cosh(ka)-1)##
Go figure.
This solution doesn't look right to me.

Homework Helper
Gold Member
2022 Award
This solution doesn't look right to me.
I figured as much. DIY

hunt_mat
Homework Helper
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.

Homework Helper
Thank you for your help thought PeroK.

Staff Emeritus
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
Homework Helper
Thank you.

Homework Helper
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".