# Integral limit trouble

1. Feb 18, 2009

### latentcorpse

we have $\mathbf{A(r)}=\frac{\mu_0 I}{4 \pi} \int dV' \mathbf{\frac{dl'}{|r-r'|}}$

i need to show that $A_x=-\frac{\mu_0 I}{4 \pi} \int_{-x}^x \frac{d \xi}{(\xi^2 + y^2 + z^2)^{-\frac{1}{2}}}$

i said $|\mathbf{r-r'}|=\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}$

then if we pick $\mathbf{r'}$ on the x axis, y'=z'=0

then we let $\xi=x-x' \Rightarrow dl'=-d \xi$

so everything's looking good up till now but i can't get the limits on the integration to come out right.

x' goes between $-\infty$ and $+\infty$ btw

any ideas?

2. Feb 19, 2009

### gabbagabbahey

The way this is writen, it makes absolutely no sense. The LHS of the equation is a vector, and yet every term on the LHS is a scalar. How can a vector equal a scalar? And you seem to be integrating both over a volume dV' and along a curve dl'....how is that possible?

Surely you meant to write:

$$\mathbf{A}(\mathbf{r})=\frac{\mu_0 I}{4 \pi} \int \frac{\mathbf{dl'}}{|\mathbf{r}-\mathbf{r'}|}$$

right?

How do you justify picking $\mathbf{r'}$ to be on the x axis? What physical problem is this in regards to?

Why is that? What is the entire problem?

Last edited: Feb 19, 2009
3. Feb 23, 2009

### latentcorpse

4. Feb 23, 2009

### gabbagabbahey

The figure isn't very clear, but I think you are supposed to assume that the current moving in the positive x-direction goes from x'=-x too x'=0; and the current travelling in the negative x-direction goes from x'=x to x=0. That gives you two integrals (two sections of current) which you can combine to get the desired form.

However, looking only at the figure I would have assumed that the current go from -infinity to zero and infinity to zero; so I recommend you clarify this point with your professor first!

5. Feb 23, 2009

### latentcorpse

hi ive managed that. and ive found A_y and A_z=0 as no current in that direction

i'll post my working just now

6. Feb 23, 2009

### latentcorpse

you have a region with i in positive x direction and x' from -infinity to 0. here xi goes from +infinity to x. and a region in which i is in negative x direction but with x' from 0 to +infinity. here xi goes from x to -infinity.

xi=x-x'

so in first region $d \xi = dl'$ as $x' <0$ and in second region $d \xi=-dl'$. here dl' is a scalar as im about to take x component of A!

$A_x=\frac{\mu_0 I}{4 \pi} \int_{+\infty}^x \frac{d \xi}{(\xi^2+y^2+z^2)^{\frac{1}{2}}} + \frac{\mu_0 (-I)}{4 \pi} \int_x^{-\infty} \frac{-d \xi}{(\xi^2+y^2+z^2)^{\frac{1}{2}}}=-\frac{\mu_0 I}{4 \pi} \int_{-x}^{-\infty} \frac{d \xi}{(\xi^2+y^2+z^2)^{\frac{1}{2}}} - \frac{\mu_0 (-I)}{4 \pi} \int_{-\infty}^x \frac{-d \xi}{(\xi^2+y^2+z^2)^{\frac{1}{2}}}$
so
$A_x=-\frac{\mu_0 I}{4 \pi} \int_{-x}^x \frac{d \xi}{(\xi^2+y^2+z^2)^{\frac{1}{2}}}$

does this look acceptable?

7. Feb 23, 2009

### gabbagabbahey

Looks good to me

8. Feb 23, 2009

### latentcorpse

1, is it ok to say $d \xi=dl'$ in region 1 and $d \xi=-dl'$ in region 2?

2,in the first integral i pull out a negative sign by changing the limits from i.e. is it true that $\int_{\infty}^x=-\int_{-x}^{-\infty}$

3, in the second integral i go from $\int_{x}^{-\infty}=-\int_{-\infty}^{x}$
is this ok? it seems to disagree with my question 2?

9. Feb 23, 2009

### latentcorpse

i then evaluate the integral and get $A_x=-\frac{\mu_0 I}{4 \pi} [\ln{(x+(x^2+y^2+z^2)^{-\frac{1}{2}})}-\ln{(-x+(x^2+y^2+z^2)^{-\frac{1}{2}})}]$
i need to show that $\frac{\partial{A_x}}{\partial{y}}=-\frac{\mu_0 I}{4 \pi}[-\frac{2xy}{\sqrt{x^2+y^2+z^2}(y^2+z^2)}]$

ive pretty much been trying to show this for the best part of eight hours and am nearing the stage of wanting to kill myself!

10. Feb 23, 2009

### gabbagabbahey

Let's start by writing your original integrals as:

$$A_{x,1}=\frac{\mu_0 I}{4 \pi} \int_{-\infty}^0 \frac{dx'}{((x-x')^2+y^2+z^2)^{\frac{1}{2}}}$$

and

$$A_{x,2}=\frac{-\mu_0 I}{4 \pi} \int_{0}^{\infty} \frac{dx'}{((x-x')^2+y^2+z^2)^{\frac{1}{2}}}$$

Where dl'=dx'i

Okay with this part?

if $\xi=x-x'$, then $d\xi=-dx'$ and $\xi$ goes from $x-(-\infty)=+\infty$ to $x-0=x$

$$\implies A_{x,1}=\frac{-\mu_0 I}{4 \pi} \int_{\infty}^x \frac{d\xi}{(\xi^2+y^2+z^2)^{\frac{1}{2}}}$$

On the other hand, you could define $\xi=x'-x$ instead,and then $d\xi=+dx'$ and $\xi$ goes from $(-\infty)-x=-\infty$ to $0-x=-x$

$$\implies A_{x,1}=\frac{\mu_0 I}{4 \pi} \int_{-\infty}^{-x} \frac{d\xi}{(\xi^2+y^2+z^2)^{\frac{1}{2}}}$$

Follow?

No, this is not alright. What is true is that $\int_{\infty}^x=-\int_{x}^{\infty}$

I had assumed that you had gone from $x\to -x$ and $-\infty \to \infty$ by making the transformation (substitution) $\xi \to -\xi$ which would have been acceptable.

Yes, this part is acceptable.

You need to realize that $\xi$ is essentially a dummy variable snce it is being integrated over and so you don't have to have it represent the exact same quantity in both integrals. It is easiest if you define $\xi=x-x'$ for one of the integrals and $\xi=x'-x$ for the other ----which is what I had thought you did when you said:

11. Feb 23, 2009

### latentcorpse

yeah that's what i was meaning but just couuldnt figure out how to put it in maths

do you have any ideas for the very last bit of the question?

12. Feb 23, 2009

### gabbagabbahey

This is just straight integration and differentiation. For the integration, try the substituion $u=\xi+\sqrt{\xi^2+y^2+z^2}$

13. Feb 23, 2009

### latentcorpse

yeah got that now cheers. what about the last part as in what happens if $z^2 >>x^2+y^2$ bit?

14. Feb 23, 2009

### latentcorpse

oh and the substitution you recommend in post 12, how do u show $du=d \xi$?

15. Feb 23, 2009

Try writing your general expression for B in terms of the dimensionless variable $$w=\frac{x^2+y^2}{z^2}[/itex] then Taylor expand it about w=0. 16. Feb 23, 2009 ### gabbagabbahey You don't because it's not true. If $u=\xi+\sqrt{\xi^2+y^2+z^2}$ , then [tex]du=\left(1+\frac{\xi}{\sqrt{\xi^2+y^2+z^2}}\right)d\xi=\frac{\sqrt{\xi^2+y^2+z^2}+\xi}{\sqrt{\xi^2+y^2+z^2}}d\xi=\frac{u}{\sqrt{\xi^2+y^2+z^2}}d\xi$$

17. Feb 24, 2009

### latentcorpse

ok is there some physical explanation of why you chose xi=x-x' in one region and xi=x'-x in the other - i.e. something other than "to make the maths work"?

how did you know to use this trick?

18. Feb 24, 2009

### gabbagabbahey

No physical reason; but it's not really a 'trick' either. Its a simple mathematical substitution. You treat each integral separately, and so you can use different substitutions for different integrals. It turns out that if you use xi=x-x' for one and xi=x'-x for the other, both your integrals end up in the same form and can therefor be combined in the desired way.