Graduate Double Integral with Dirac Delta Function and Changing Limits

[hunt_mat]

TL;DR

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

I have an integral:
<br /> \int_{-1}^{0}\int_{-1}^{q}\delta(s+a)\sinh[k(q-s)]dsdq<br />
where 0&lt;a&lt;1 and \delta (s-a) is a dirac delta function. Anyone know what to do?

 

[PeroK]

You could try integrating the $\sinh$ function!

What's the difficulty?

 

[hunt_mat]

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

 

[Vanadium 50]

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

 

[PeroK]

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 $q > a$.

 

[hunt_mat]

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.

 

[PeroK]

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 $q > a$.

 

[hunt_mat]

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:
<br /> \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)]) <br />
This seems like a sloppy way of arguing however.

 

[PeroK]

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:
<br /> \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)]) <br />
This seems like a sloppy way of arguing however.

That doesn't look right. You can split the outer integral in $dq$.

 

[PeroK]

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

 

[hunt_mat]

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.

 

[hunt_mat]

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]

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]

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]

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

I have an integral:
<br /> \int_{-1}^{0}\int_{-1}^{q}\delta(s+a)\sinh[k(q-s)]dsdq<br />
where -1&lt;a&lt;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.

 

[hunt_mat]

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.

 

[hunt_mat]

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]

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 $a$.

 

[hunt_mat]

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

I don't think so.

 

[hunt_mat]

$\int_{-1}^0 dq = \int_{-1}^a dq + \int_a^0 dq$

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]

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]

Spoiler

$\frac 1 k (\cosh(ka)-1)$

Go figure.

 

[hunt_mat]

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

 

[PeroK]

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.

 

[hunt_mat]

If you split up the inner integral as:
<br /> \int_{-1}^{q}=\int_{-1}^{a-\varepsilon}+\int_{a-\varepsilon}^{a+\varepsilon}+\int_{a+\varepsilon}^{q}<br />
Then you're going to get the function I posted. Splitting up the outer integral isn't going to do much in my opinion.

 

[hunt_mat]

Spoiler

$\frac 1 k (\cosh(ka)-1)$

Go figure.

This solution doesn't look right to me.

 

[PeroK]

This solution doesn't look right to me.

I figured as much. DIY

 

[hunt_mat]

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.

 

[hunt_mat]

Thank you for your help thought PeroK.

 

[Vanadium 50]

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.