Griffiths problem 2.7 Messy Integral

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    Griffiths Integral
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Discussion Overview

The discussion revolves around solving a specific integral from Griffiths' textbook, focusing on techniques such as partial fractions and integration by parts. Participants explore various substitution methods and express their experiences with non-integer exponents in integrals.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant seeks advice on how to approach an integral involving non-integer exponents and mentions the possibility of using partial fractions.
  • Another participant suggests that the author likely intended for integration by parts to be used instead of partial fractions.
  • Some participants propose using integration by parts directly, indicating that it can lead to a messy result.
  • A participant shares their work on the integral, detailing the steps taken and expressing uncertainty about the correctness of their approach.
  • There are mentions of differentiating and integrating specific terms related to the integral, with some participants providing their interpretations of the substitution process.
  • One participant humorously expresses the need to learn LaTeX for clearer communication of their mathematical work.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to the integral, with multiple competing views on whether to use partial fractions or integration by parts. The discussion remains unresolved regarding the most effective method.

Contextual Notes

Some participants note that the integral becomes complex and messy, indicating potential limitations in their current understanding or approach. There are also references to specific substitutions and integration techniques that may depend on the context of the problem.

Waxterzz
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Hi all,
twRFdMD.png


"Integral can be done by partial fractions - or look it up" So second line, that's what I want to do.

How to deal with this? What substitution can I use? Never encountered partial fractions with non-integer exponents.

Someone give me a tip?

Thanks in advance
 
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Waxterzz said:
Hi all,
twRFdMD.png


"Integral can be done by partial fractions - or look it up" So second line, that's what I want to do.

How to deal with this? What substitution can I use? Never encountered partial fractions with non-integer exponents.

Someone give me a tip?

Thanks in advance
I do think the author must have meant integration by parts. The term without a ## u ## in the numerator that has a ## (A+Bu)^{3/2} ## in the denominator is a straightforward integration and the ## u \, du/(A+Bu)^{3/2} ## can be readily integrated by parts.
 
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You can directly use integration by parts.
 
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Charles Link said:
I do think the author must have meant integration by parts.

zaidalyafey said:
You can directly use integration by parts.

I get something extremely messy. Here is the first part, the second part with integration by parts,I'll do it again tomorrow.

So the first term

integral (z du / (R² + z² - 2 Rz u) ^3/2 )

d(R² + z² - 2 Rz u) = -2 R z du

The integral becomes

-1/2R * integral ( d(R² + z² - 2 Rz u) / (R² + z² - 2 Rz u) ^3/2) )

Then I get1/R * (1/ (R² + z² -2Rzu) ^ 1/2)

The second part I'm going to do again tomorrow, and I'm going to try to put it in latex (have to learn latex quick) because posting like this is unreadable. But the first part is correct right?
 
You can differentiate z - Ru and integrate ( du / (R² + z² - 2 Rz u)^3/2)
 
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From first to second line is simply the substitution given. The integration is from second line to third.
 
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zaidalyafey said:
You can differentiate z - Ru and integrate ( du / (R² + z² - 2 Rz u)^3/2)

mathman said:
From first to second line is simply the substitution given. The integration is from second line to third.
Ok, so partial integration. integral (fdg) = fg - integral (gdf)

f = z-Ru
dg = du / (R² +z² -2zRu) ^ 3/2

For g I get
1 / ( (Rz) * (R² + z² -2*R*z*u)^1/2)

fg = (z-Ru) / ( (Rz) * (R² + z² -2*R*z*u)^1/2)

now the integral part - integral ( g d f)

df = -R du

So

The integral to solve

Rdu / ((Rz) * (R² + z² -2*R*z*u)^1/2))

Becomes

1/z * (2/-2*R*z) * (R² + z² -2*R*z*u)^1/2

Combining two terms, and rearranging

I get solutionDEAR GOD

hhahaha

(Okay I really need to learn latex now haha)
Charles Link said:
I do think the author must have meant integration by parts. The term without a ## u ## in the numerator that has a ## (A+Bu)^{3/2} ## in the denominator is a straightforward integration and the ## u \, du/(A+Bu)^{3/2} ## can be readily integrated by parts.

Thank u also.

How to give everyone upvotes in this thread?
 
Waxterzz said:
How to give everyone upvotes in this thread?

You can hit the like button in the bottom right corner of each post.
 
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