Integrating by Partial Fractions

[Electrowonder]

Homework Statement

Integrating by Partial Fractions

Relevant Equations

None

I was doing this problem from Griffith's electrodynamics book and can't figure out how to do this integral. The author suggested partial fractions but the denominator has a fractional exponent which I have never seen for partial fractions, and so, I am unsure how to proceed. The integral I am working on is this:

$$\int \frac{Z-Ru}{(R^2-z^2-2RZu)^{3/2}}du$$

Where u is the variable of integration and all the other ones are constants.

No idea how to get started, please give a hint.

 

[Charles Link]

I think you can split the numerator, with the first term being simple. For the second term, you integrate not by partial fractions, but rather integrate by parts.

 

[vela]

Are you sure the integral is correct?

 

[PeroK]

Problem Statement: Integrating by Partial Fractions
Relevant Equations: None

I was doing this problem from Griffith's electrodynamics book and can't figure out how to do this integral. The author suggested partial fractions but the denominator has a fractional exponent which I have never seen for partial fractions, and so, I am unsure how to proceed. The integral I am working on is this:

$$\int \frac{Z-Ru}{(R^2-z^2-2RZu)^{3/2}}du$$

Where u is the variable of integration and all the other ones are constants.

No idea how to get started, please give a hint.

For these sort of integrals, I prefer to use the substitution $s^2 = R^2 + z^2 - 2Rzu$. I assume that should be a $+z^2$ in the denominator. Or, using the original variable:

$s^2 = R^2 + z^2 - 2Rz \cos \theta$

 

[Electrowonder]

Sorry, that should be a +z^2 in the denominator.

PeroK

The original integral was

$$\int \frac{\sin (\theta) (Z-Rcos\theta)}{(R^2+z^2 -2RZcos\theta)^{3/2}} d\theta$$

If I use the substitution s^2, how do I do find ds?

 

[PeroK]

Sorry, that should be a +z^2 in the denominator.

PeroK

The original integral was

$$\int \frac{\sin (\theta) (Z-Rcos\theta)}{R^2+z^2 -2RZcos\theta} d\theta$$

If I use the substitution s^2, how do I do find ds?

The usual way:

$2s ds = \dots$

 

[Electrowonder]

I think you can split the numerator, with the first term being simple. For the second term, you integrate not by partial fractions, but rather integrate by parts.

$$\int \frac{z}{(R^2 + z^2 - 2Rzu)^{3/2}} du +\int \frac{-Ru}{(R^2 + z^2 - 2Rzu)^{3/2}} du$$

How would I integrate the second term? I have already done a U substitution, am I allowed to just do another substitution?

 

[Electrowonder]

The usual way:

$2s ds = \dots$

Seems quite obvious but I have never done substitution with a quadratic before. Usually, you substitute with something like $s = R^2 + z^2 -2Rzcos\theta$ but I realized that won't go anywhere nice.

Using your way, I did

$$d(s^2) = 2sds = 2 \sqrt{R^2 + z^2 - 2RZcos\theta} d(\sqrt{R^2+z^2 -2RZcos\theta})$$
$$=2RZsin\theta d\theta$$

Plugging it in:
$$\int \frac{(sin\theta)(z-Rcos\theta)}{(s^2)^{3/2}} \frac{d(s^2)}{2Rzsin\theta}$$
$$\int \frac{z(sin\theta)-Rcos\theta sin\theta)}{(s^3} \frac{d(s^2)}{2Rzsin\theta}$$
$$= \frac{1}{2R} \int \frac{d(s^2)}{s^3} - \frac{1}{2z} \int \frac{cos\theta d(s^2)}{s^3} $$
Then I would solve $cos\theta$ from the original substitution to get
$$cos\theta = \frac{s^2 - R^2 - z^2}{2RZ}$$
Then substitute that into the last integral

Am I on the right track?

 

[PeroK]

Seems quite obvious but I have never done substitution with a quadratic before. Usually, you substitute with something like $s = R^2 + z^2 -2Rzcos\theta$ but I realized that won't go anywhere nice.

Using your way, I did

$$d(s^2) = 2sds = 2 \sqrt{R^2 + z^2 - 2RZcos\theta} d(\sqrt{R^2+z^2 -2RZcos\theta})$$
$$=2RZsin\theta d\theta$$

Plugging it in:
$$\int \frac{(sin\theta)(z-Rcos\theta)}{(s^2)^{3/2}} \frac{d(s^2)}{2Rzsin\theta}$$
$$\int \frac{z(sin\theta)-Rcos\theta sin\theta)}{(s^3} \frac{d(s^2)}{2Rzsin\theta}$$
$$= \frac{1}{2R} \int \frac{d(s^2)}{s^3} - \frac{1}{2z} \int \frac{cos\theta d(s^2)}{s^3} $$
Then I would solve $cos\theta$ from the original substitution to get
$$cos\theta = \frac{s^2 - R^2 - z^2}{2RZ}$$
Then substitute that into the last integral

Am I on the right track?

That looks too complicated to me. Instead, simply:

$s^2 = R^2 + z^2 - 2Rz \cos \theta$

$2sds = 2Rz \sin \theta d\theta$

$\sin \theta d\theta = \frac{sds}{Rz}$

 

[PeroK]

That looks too complicated to me. Instead, simply:

$s^2 = R^2 + z^2 - 2Rz \cos \theta$

$2sds = 2Rz \sin \theta d\theta$

$\sin \theta d\theta = \frac{sds}{Rz}$

PS

$\Rightarrow \ \cos \theta = \frac{R^2 + z^2 - s^2}{2Rz}$

 

[Electrowonder]

That looks too complicated to me. Instead, simply:

$s^2 = R^2 + z^2 - 2Rz \cos \theta$

$2sds = 2Rz \sin \theta d\theta$

$\sin \theta d\theta = \frac{sds}{Rz}$

Thank you, I was able to fully solve it! That was a brilliant substitution, and now it seems almost obvious after the fact. But what thought process do I need to adopt to do smart substitutions like that in the future?

Can someone advise me how to also do it by partial fractions as the author suggested?

 

[PeroK]

Thank you, I was able to fully solve it! That was a brilliant substitution, and now it seems almost obvious after the fact. But what thought process do I need to adopt to do smart substitutions like that in the future?

It's not hard to remember, because $s$ is the distance, which Griffiths denotes as the squiggly "r".

 

[Electrowonder]

It's not hard to remember, because $s$ is the distance, which Griffiths denotes as the squiggly "r".

Oh that's right, got it, thank you. That should prove helpful for other problems.

 

[Charles Link]

$$\int \frac{z}{(R^2 + z^2 - 2Rzu)^{3/2}} du +\int \frac{-Ru}{(R^2 + z^2 - 2Rzu)^{3/2}} du$$

How would I integrate the second term? I have already done a U substitution, am I allowed to just do another substitution?

The second term becomes $\int -Ru \,(\frac{1}{RZ}) d(R^2+z^2-2RZu)^{-1/2}$.$\\$ You should be able to do this by parts routinely. Remember $\int\limits_{a}^{b} u \, dv=uv|_a^b-\int\limits_{a}^{b} v \, du$.

 

[Electrowonder]

The second term becomes $\int -Ru \,(\frac{1}{RZ}) d(R^2+z^2-2RZu)^{-1/2}$.$\\$ You should be able to do this by parts routinely. Remember $\int\limits_{a}^{b} u \, dv=uv|_a^b-\int\limits_{a}^{b} v \, du$.

I'm not sure what to do with the $\int v du$ term, what is v going to be?

 

[Charles Link]

$v=(R^2+Z^2-2RZu)^{-1/2}$.
The integral $\int v \, du$ is straightforward.

 

[Electrowonder]

$v=(R^2+Z^2-2RZu)^{-1/2}$.
The integral $\int v \, du$ is straightforward.

Oh, I see you meant $d[(R^2 +z^2 -2Rzu)^{-1/2}]$, I was a bit confused with the exponent outside as I haven't done this for 3 years.