# Complicated double integral

• A
Homework Helper

## 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

PeroK
Homework Helper
Gold Member
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
2019 Award
What does the delta function do when you integrate over s?

PeroK
Homework Helper
Gold Member
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
PeroK
Homework Helper
Gold Member
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.

PeroK
Homework Helper
Gold Member
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##.

PeroK
Homework Helper
Gold Member
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.

PeroK
Homework Helper
Gold Member
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?

PeroK
Homework Helper
Gold Member
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##.

PeroK
Homework Helper
Gold Member
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.

PeroK
Homework Helper
Gold Member
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.

PeroK
Homework Helper
Gold Member
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.

PeroK
Homework Helper
Gold Member
##\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.

PeroK
Homework Helper
Gold Member
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.

• $$\int_{-1}^{q}=\int_{-1}^{a-\varepsilon}+\int_{a-\varepsilon}^{a+\varepsilon}+\int_{a+\varepsilon}^{q}$$