Can Differentiating a Simple Fraction Simplify Complex Integration?

[garylau]

Sorry how to integrate this function from step1 to step 2(see the circle inside the picture)

it looks too complicated
i can't help with it
i cannot use partial fraction to do because it looks hard to decompose it
thank

 

[Battlemage!]

I thought partial fractional decomposition required both numerator and denominator to be polynomials?

 

[stevendaryl]

We're trying to integrate
$\frac{z-Ru}{(R^2 + z^2 - 2Rzu)^{\frac{3}{2}}}$

We can rewrite this as:

$\frac{R}{(2Rz)^{\frac{3}{2}}} \frac{\frac{z}{R} - u}{(\frac{R^2 + z^2}{2Rz} - u)^\frac{3}{2}}$

Now $\frac{\frac{z}{R} - u}{(\frac{R^2 + z^2}{2Rz} - u)^\frac{3}{2}}$ is of the form

$(A-u)(C-u)^\alpha$

with $A = \frac{z}{R}$, $C = \frac{R^2 + z^2}{2Rz}$, $\alpha = \frac{-3}{2}$

Here's the general way to integrate such expressions:

Rewrite this as:
$((A-C) + (C-u))(C-u)^\alpha = (A-C) (C-u)^\alpha + (C-u)^{\alpha+1}$

We can easily integrate these terms, to get:

$-\frac{A-C}{\alpha+1} (C-u)^{\alpha+1} - \frac{1}{\alpha+2} (C-u)^{\alpha+2}$
$= -\frac{(C-u)^{\alpha+1}}{(\alpha+1)(\alpha+2)} ((A-C)(\alpha+2) + (C-u)(\alpha+1))$
$= -\frac{(C-u)^{\alpha+1}}{(\alpha+1)(\alpha+2)} (A (\alpha+2) - C - u(\alpha+1))$

So if you plug in our known values for $A, C, \alpha$, you should get the desired result (after some simplification).

 

[garylau]

We're trying to integrate
$\frac{z-Ru}{(R^2 + z^2 - 2Rzu)^{\frac{3}{2}}}$

We can rewrite this as:

$\frac{R}{(2Rz)^{\frac{3}{2}}} \frac{\frac{z}{R} - u}{(\frac{R^2 + z^2}{2Rz} - u)^\frac{3}{2}}$

Now $\frac{\frac{z}{R} - u}{(\frac{R^2 + z^2}{2Rz} - u)^\frac{3}{2}}$ is of the form

$(A-u)(C-u)^\alpha$

with $A = \frac{z}{R}$, $C = \frac{R^2 + z^2}{2Rz}$, $\alpha = \frac{-3}{2}$

Here's the general way to integrate such expressions:

Rewrite this as:
$((A-C) + (C-u))(C-u)^\alpha = (A-C) (C-u)^\alpha + (C-u)^{\alpha+1}$

We can easily integrate these terms, to get:

$-\frac{A-C}{\alpha+1} (C-u)^{\alpha+1} - \frac{1}{\alpha+2} (C-u)^{\alpha+2}$
$= -\frac{(C-u)^{\alpha+1}}{(\alpha+1)(\alpha+2)} ((A-C)(\alpha+2) + (C-u)(\alpha+1))$
$= -\frac{(C-u)^{\alpha+1}}{(\alpha+1)(\alpha+2)} (A (\alpha+2) - C - u(\alpha+1))$

So if you plug in our known values for $A, C, \alpha$, you should get the desired result (after some simplification).

where is this form
$(A-u)(C-u)^\alpha$
coming from

and Where do you got $((A-C) + (C-u))(C-u)^\alpha = (A-C) (C-u)^\alpha + (C-u)^{\alpha+1}$ this equation from?

thank

 

[Battlemage!]

where is this form
$(A-u)(C-u)^\alpha$
coming from

He's just substituting there. A and C are the two terms that have r and z in it. In the first step he just rearranged it to get u by itself in both the numerator and deniminator, so that he could make the substitution.

The second part is just adding and subtracting C, and then using the associstive property. You'll note that A - u = A + C - C - u.

 

[Battlemage!]

Honestly that's a really clever trick and I think I'm going to commit this one to memory.

 

[Integral]

where is this form
$(A-u)(C-u)^\alpha$
coming from

and Where do you got $((A-C) + (C-u))(C-u)^\alpha = (A-C) (C-u)^\alpha + (C-u)^{\alpha+1}$ this equation from?

thank

I suspect both of those relationships came from a book of integrals.

 

[stevendaryl]

where is this form
$(A-u)(C-u)^\alpha$
coming from

The integrand is (after factor out the factor $\frac{2R}{(2Rz)^{3/2}}$):

$\frac{\frac{z}{R} - u}{(\frac{R^2 + z^2}{2Rz} - u)^{3/2}}$

That's the same as:

$(\frac{z}{R} - u)(\frac{R^2 + z^2}{2Rz} - u)^{-3/2}$

Now, let $A = \frac{z}{R}, C = \frac{R^2 + z^2}{2Rz}, \alpha = -3/2$, so you get

$(A - u)(C - u)^{\alpha}$

...and Where do you got $((A-C) + (C-u))(C-u)^\alpha = (A-C) (C-u)^\alpha + (C-u)^{\alpha+1}$ this equation from?

Battlemage! explained it above.

 

[Dragon27]

But the temptation to integrate with respect to $z$ (since $(z^2-Rzu)'=2(z-Ru)$), then integrate with respect to $u$, and then differentiate with respect to $z$ is just irresistible...

 

[Battlemage!]

But the temptation to integrate with respect to $z$ (since $(z^2-Rzu)'=2(z-Ru)$), then integrate with respect to $u$, and then differentiate with respect to $z$ is just irresistible...

Ye gads, is this not cheating? Would that actually work, and if so is it easier?

 

[Dragon27]

Ye gads, is this not cheating?

It feels so good, that it probably is...

Would that actually work, and if so is it easier?

The integrals in the intermediate steps are much easier, in my opinion (or, maybe, more obvious). In the end you just differentiate a simple fraction, it's not too hard.