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April 21st, 2012, 07:50 PM   #1
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Are there issues with this?

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April 21st, 2012, 08:12 PM   #2
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Re: Are there issues with this?

I simplified your result some, then differentiated and did not get back the original integrand.

That is a real bear of an integral , at least by factoring the denominator:



and then using partial fraction decomposition, substitutions, etc.
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April 21st, 2012, 09:04 PM   #3
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Re: Are there issues with this?

Quote:
Originally Posted by MarkFL
I simplified your result some, then differentiated and did not get back the original integrand.

That is a real bear of an integral
I'm aware. That is rather crazy. Can you post your work? Also, is there an error algebraically in what I wrote? It wasn't really easy to compose, but the idea seemed simple.

EDIT: I made a stupid mistake... Namely,



Redoing and reposting, one moment. . .
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April 21st, 2012, 09:14 PM   #4
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Re: Are there issues with this?

I simply looked at the suggested method given by wolframalpha.com.

Your factorization of ...nevermind, I see you caught it.
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April 21st, 2012, 09:30 PM   #5
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Re: Are there issues with this?



How's this?
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April 21st, 2012, 09:37 PM   #6
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Re: Are there issues with this?

You don't seem to actually be integrating...you should have natural logarithms in your result, i.e.,:

where
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April 21st, 2012, 09:41 PM   #7
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Re: Are there issues with this?

Woooooooow.

-facepalm- This is why you never try to integrate and multitask.

It works out to, instead,

\ln|x-it|+(B-3B&#39\ln|x+it|+(C-3C&#39\ln|x+t|+(D-3D&#39\ln|x-t|+C" />

Thank you for the patience, Mark. The nuances and little things are what I always miss.

EDIT: The big reason I asked this topic was because I was uncertain if extending the method of partial fraction decomposition to complex numbers was a bad idea. Wolfram Alpha does not seem to like it, but it makes sense to me.
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April 21st, 2012, 09:52 PM   #8
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Re: Are there issues with this?

Another issue:

You cannot state:



You need to use partial fraction decomposition. For example:



With 4 linear factors in the denominator it will be more complicated. You want:

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April 21st, 2012, 10:01 PM   #9
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Re: Are there issues with this?

Quote:
Originally Posted by MarkFL
Another issue:

You cannot state:



You need to use partial fraction decomposition. For example:



With 4 linear factors in the denominator it will be more complicated. You want:

I don't understand this issue. I am using partial fraction decomposition...

Here are the steps:



This gives us an equation:



I evaluated this equation at the zeros. (it, -it, t, -t) This gave me the expressions for A, B, C, and D. Take A as an example:

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April 21st, 2012, 10:15 PM   #10
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Re: Are there issues with this?

My apologies, I see that now. You are even using the cover-up method.

You can do a lot of simplification of your constants. For example:



I find that that coefficient should be (using W|A):

which is simply the negative of what you have.
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