# A Complicated double integral

#### hunt_mat

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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 $0<a<1$ and $\delta (s-a)$ is a dirac delta function. Anyone know what to do?

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• HallsofIvy

#### PeroK

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You could try integrating the $\sinh$ function!

What's the difficulty?

• hunt_mat

#### 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?

#### 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 $q > a$.

#### hunt_mat

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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

#### PeroK

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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 $q > a$.

• • #### 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 $a$ 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

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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 $dq$.

#### PeroK

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

#### hunt_mat

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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.

#### 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? $a\in (-1,0)$ and so it's going to be in there no matter what.

#### PeroK

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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?

#### PeroK

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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 $0$.

#### PeroK

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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.

#### 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.

#### 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.

#### 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 $a$.

• #### hunt_mat

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

#### 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.

#### 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.

#### PeroK

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

#### hunt_mat

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

#### PeroK

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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.

• #### hunt_mat

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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.