# How to discretize an integral

1. Mar 24, 2015

### VictorVictor5

Greetings all,

I looked this up in the forums but there didn't seem to be a response.

I have the following equation:

$i\sum\limits_{n = 1}^N {{\sigma _n}} \left[ {\int\limits_{{\zeta _{n - 1}}}^\infty {\phi \left( \zeta \right)d\zeta - \int\limits_{{\zeta _{n}}}^\infty {\phi \left( \zeta \right)d\zeta } } } \right]$

where φ represents two cases:

Vertical Case:

${\phi _V}\left( \zeta \right) = \frac{{4\zeta }}{{{{\left( {4{\zeta ^2} + 1} \right)}^{3/2}}}}$

And horizontal case:

${\phi _H}\left( \zeta \right) =2 - \frac{{4\zeta }}{{{{\left( {4{\zeta ^2} + 1} \right)}^{1/2}}}}$

and ζ = z/r.

I am trying to express these integrals in summation form, but have never attempted discrete calculus. I apologize if the question is not complete, but I can try and provide more details as the post proceeds.

I am trying to learn.

Thanks!
VV5

2. Mar 24, 2015

### mathman

It is not clear what you are trying to do . You can simplify the integrand with x=ζ2, while the two integrals can be collapsed into one. The integral can then be explicitly evaluated.

3. Mar 25, 2015

### VictorVictor5

Mathman,

Sorry about the confusion - trust me I'm new at discretization too. But first, thanks for the response.

Basically, I'm seeing if there's a way to represent these integrals in summation form via the Phi equations.

What these "Phi" equations represent is the relative responses for vertical and horizontal dipoles for use with electromagnetic induction instruments that collect apparent conductivity (I can provide the reference if we get that far).

So, in my current effort, discretizing will allow me to see what the conductivity response is and how to compare that answer to an integral.

If we consider 4 layers, with the first one being air and the rest water, and the conductivity of the layer of air being essentially negligible, so we'll turn our attention starting at layer 2 (water)

I was told, very confusingly from the start of this problem, that substitution of variables can occur, such that:

${\sigma _a}= i\sum\limits_{n = N}^1 {{\sigma _n}\left[ {\sum\limits_{p = n - 1}^1 {{\phi _p} - \sum\limits_{p = n}^1 {{\phi _p}} } } \right]}$

I should have stated that the first equation I ever introduced is also equal to sigma_a (sorry about that one, the person who is tasking me to solve this problem said that the details don't matter, just the variables. Right.). Now, as stated before, we start at layer 2 to define apparent conductivity (sigma_a) since layer 1 is essentially 0 in terms of electrical conductivity:
So for N=1:
${\sigma _a}(1) = - i{\sigma _1}{\phi _1}$

For N=2:
${\sigma _a} = i\left[ {{\sigma _2}\left[ {{\phi _1} - \left[ {\left( {{\phi _1} + {\phi _2}} \right)} \right]} \right]} \right]$

For N=3:
${\sigma _a} = i{\sigma _3}\left[ {\left( {{\phi _1} + {\phi _2}} \right) - \left( {{\phi _1} + {\phi _2} + {\phi _3}} \right)} \right] = i{\sigma _3}{\phi _3}$

And for N=4:
${\sigma _a} = -i{\sigma _4}{\phi _4}$

And therefore, the discretization can be reported as

${\sigma _a}(n) = - i\sum\limits_{n = 1}^N {{\sigma _n}{\phi _n}}$

Now, I don't know how discretization works as an expert does, but I'm seeing if this approach is even legit with the above to integrals aforementioned.

Apologies if this is confusing as all get out. It is to me, so I figured I'd strike up a conversation with the experts here.

Thanks!
VV5

Last edited: Mar 25, 2015
4. Mar 26, 2015

### mathman

$\sum_{p=n-1}^1 \phi_p - \sum_{p=n}^1 \phi_p = -\phi_n$
Usually in summation formulas the lower limit is first. Why do you have it reversed?

I can't comment on the physics (not my thing).

5. Mar 26, 2015

### VictorVictor5

Why do I have it reversed?

Good question - not too sure! That' why I threw the question on the boards.

So is the discretization I stated correct?

I'm still a bit confused.

Thanks!

Last edited: Mar 26, 2015
6. Mar 27, 2015

### mathman

Your notation is very sloppy. You sometimes mix n and N. Also you use $\sigma_a$ sometimes for individual terms and sometimes for the sum. However the last line seems correct if you use $\sigma_a (N) \ not\ \sigma_a(n)$.

7. Apr 6, 2015