I Griffiths problem 2.7 Messy Integral

[Waxterzz]

Hi all,

"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

 

[Charles Link]

Hi all,

"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.

 

[zaidalyafey]

You can directly use integration by parts.

 

[Waxterzz]

I do think the author must have meant integration by parts.

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?

 

[zaidalyafey]

You can differentiate z - Ru and integrate ( du / (R² + z² - 2 Rz u)^3/2)

 

[mathman]

From first to second line is simply the substitution given. The integration is from second line to third.

 

[Waxterzz]

You can differentiate z - Ru and integrate ( du / (R² + z² - 2 Rz u)^3/2)

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)

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.

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[zaidalyafey]

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