# Integrating by Partial Fractions

• Electrowonder
In summary, the author suggests using substitution s^2 = R^2 + z^2 - 2Rzcos\theta which allows for the integrals to be solved easily.
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.

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.

Are you sure the integral is correct?

Electrowonder said:
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##

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?

Last edited:
Electrowonder said:
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 ##

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?

PeroK said:
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.

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

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?

Electrowonder said:
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.

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

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}##

Electrowonder
PeroK said:
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}##

PeroK said:
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?

Electrowonder said:
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".

PeroK said:
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.

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

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 ##.

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?

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

## 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.

## 1. What is the purpose of integrating by partial fractions?

Integrating by partial fractions is a technique used to simplify and solve integrals that involve rational functions. It breaks down a complex rational function into simpler fractions that are easier to integrate.

## 2. When is integrating by partial fractions necessary?

Integrals involving rational functions with a degree of 2 or higher in the denominator require the use of partial fractions. This is because they cannot be solved using basic integration techniques.

## 3. How do you identify which method to use for integrating by partial fractions?

To determine the appropriate method for integrating by partial fractions, you must first factor the denominator of the rational function. If it can be factored into linear and/or quadratic terms, then partial fractions can be used.

## 4. What are the steps for integrating by partial fractions?

The steps for integrating by partial fractions are as follows: 1) Factor the denominator of the rational function. 2) Write the partial fraction decomposition, using the factors from the denominator. 3) Equate the coefficients of the terms on both sides of the equation. 4) Solve the resulting system of equations to find the values of the unknown coefficients. 5) Substitute the values of the coefficients back into the partial fraction decomposition. 6) Integrate each term separately.

## 5. Are there any limitations to integrating by partial fractions?

Integrating by partial fractions is limited to rational functions with a degree of 2 or higher in the denominator. It also cannot be used for improper rational functions, where the degree of the numerator is greater than or equal to the degree of the denominator.

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