# Solving Integral Limit Trouble: A_x Calculation

• latentcorpse
So, for the first region we have \xi=x-x' and dx'=-d\xi whereas for the second region we have \xi=x'-x and dx'=d\xi. So, basically, you did do what I suggested.Now, it seems easier to me to use the second definition of \xi (i.e. \xi=x'-x) for both integrals. Then we have:A_{x,1}=\frac{-\mu_0 I}{4 \pi} \int_{\infty}^{0} \frac{d\xi}{((x'-x)^2+y^2+z^2)^{\frac{1}{2}}}=\frac{\mu_0 I}{4
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?

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

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?

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

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

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

Why is that? What is the entire problem?

Last edited:
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 traveling 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!

hi I've managed that. and I've found A_y and A_z=0 as no current in that direction

i'll post my working just now

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 I am 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?

Looks good to me

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?

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!

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?

latentcorpse said:

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

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?

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

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.

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?

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:

so in first region $$d\xi =dl'$$ and in second region $$d\xi =-dl'$$

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?

latentcorpse said:
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!

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

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

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

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

Try writing your general expression for B in terms of the dimensionless variable $$w=\frac{x^2+y^2}{z^2}[/itex] then Taylor expand it about w=0. latentcorpse said: oh and the substitution you recommend in post 12, how do u show $du=d \xi$? You don't because it's not true. If $u=\xi+\sqrt{\xi^2+y^2+z^2}$ , then [tex]du=\left(1+\frac{\xi}{\sqrt{\xi^2+y^2+z^2}}\right)d\xi=\frac{\sqrt{\xi^2+y^2+z^2}+\xi}{\sqrt{\xi^2+y^2+z^2}}d\xi=\frac{u}{\sqrt{\xi^2+y^2+z^2}}d\xi$$

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?

latentcorpse said:
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?

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.

## 1. What is the purpose of solving integral limit trouble?

The purpose of solving integral limit trouble is to accurately calculate the value of a given integral. This can be useful in many scientific fields, such as physics and engineering, where integrals are used to calculate important quantities.

## 2. What is the meaning of A_x in the context of integral limit trouble?

A_x is a variable used to represent the limit of integration in the context of integral limit trouble. It is typically a constant or a function of other variables, and it determines the range over which the integral is evaluated.

## 3. What are some common challenges in solving integral limit trouble?

Some common challenges in solving integral limit trouble include determining the appropriate limits of integration, dealing with discontinuities or singularities in the integrand, and choosing the right integration method for the given problem.

## 4. How can I check if my solution to an integral limit trouble problem is correct?

To check the correctness of your solution, you can use numerical methods such as approximation or integration software to evaluate the integral. Additionally, you can also check your solution by taking the derivative of the antiderivative and comparing it to the original integrand.

## 5. Are there any tips for solving integral limit trouble more efficiently?

Some tips for solving integral limit trouble more efficiently include breaking up the integral into smaller, more manageable pieces, using appropriate substitution or integration techniques, and familiarizing yourself with common integrals and their solutions. It can also be helpful to double-check your work and avoid careless mistakes.

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