# How do we integrate with respect to (dx)^2 (not dx)?

1. Nov 4, 2013

### AakashPandita

$$∫x(dx)^2=?$$

and what is the difference between dx^2 and (dx)^2?

2. Nov 4, 2013

### pwsnafu

Where's this from? I've seen $dx^2$ but not $(dx)^2$. The only way I can parse the latter is with wedge products

Last edited: Nov 4, 2013
3. Nov 4, 2013

### Simon Bridge

The integral is zero.
i.e. it has no contribution to the overall sum.

But lets be careful: where have you seen this come up?

4. Nov 5, 2013

### jackmell

Isn't that just short-hand notation for a double integral, i.e.,

$$\iint x dx dx$$

5. Nov 5, 2013

### pwsnafu

I don't think you are allowed to have the same variable for multiple integrals. A double integral is integrating with respect to the area element.

6. Nov 5, 2013

### jackmell

Ok, but the reason I suggested this is because I recall in one of my books on integral equations, the author, when converting from IVP to Volterra Equations, use the syntax such as:

$$\int_0^x \int_0^x f(t) dt dt=\int_0^x (x-t) f(t)dt$$

however, to be fair, he never used the syntax $(dt)^2$ so if I am in error, my apologies.

7. Nov 5, 2013

### AakashPandita

I was trying to find the moment of inertia of a disk rotating at the axis through its center and perpendicular to it.

i got an integral for some function with (dx)^2 in it.

8. Nov 5, 2013

### HallsofIvy

I strongly suspect that you have misread
$$\int_0^x\int_0^x f(t)d\tau dt$$

Neither "$dtdt$" nor "$(dx)^2$" has any meaning.

9. Nov 5, 2013

### jackmell

With all due respect Hall, I have not. It's from "A First Course in Integral Equations" by Abdul-Majid Wazwaz. It's on p. 20 and includes also, the expression:

$$\int_0^x \int_0^x \int_0^x f(t)dt dt dt=1/2 \int_0^x (x-t)^2 f(t)dt$$

Last edited: Nov 5, 2013
10. Nov 5, 2013

### HallsofIvy

Then it is just bad notation.

11. Nov 5, 2013

### qbert

Dear lord, that's awful. I'd read the lhs as
$$\int_0^x \int_0^x \int_0^x f(t)dt dt dt = \int_0^x \int_0^x \left( \int_0^x f(t)dt \right) dt_1 \; dt_2 = \int_0^x \int_0^x H(x) dt_1 \; dt_2 = x^2 H(x) = x^2 \int_0^x f(t) dt$$

while, what was probably meant is
$$\int_0^x \int_0^{t_2} \int_0^{t_1} f(t) dt \; dt_1 \; dt_2 = \frac{1}{2} \int_0^x (x-t)^2 f(t) dt.$$

12. Nov 5, 2013

### economicsnerd

Yikes, that's awful notation.

13. Nov 5, 2013

### pwsnafu

That's the Cauchy formula for repeated integration. And you are not allowed to write it as dtdtdt. You must keep the variables separate.
$\int_{a}^x \int_a^{t_1}\ldots \int_a^{t_{n-1}}f(t_n) \, dt_n \, dt_{n-1} \ldots dt_1 = \frac{1}{(n-1)!} \int_a^x (x-t)^{n-1} f(t) \, dt.$

Last edited: Nov 5, 2013
14. Nov 5, 2013

### jackmell

Ok. Thanks! That makes sense to me now. Sorry for causing trouble.

15. Nov 5, 2013

### Simon Bridge

Please show how you derived that.

16. Nov 5, 2013

### AakashPandita

$$I=\int_0^R dmr^2 dm=σπ[(r+dr)^2-r^2]$$
and then I put $$dm$$ in first.

Last edited: Nov 6, 2013
17. Nov 5, 2013

### AakashPandita

$$I=\int_0^R dmr^2 ;dm=σπ[(r+dr)^2-r^2]$$
and then I put $$dm$$ in first.

Last edited: Nov 6, 2013
18. Nov 6, 2013

### pwsnafu

What does $(r+dr)^2$ mean?

Edit: I think it should read $dm = 2\sigma \pi r \, dr$, which is compatible with what you wrote if we take $(dr)^2 = 0$.

Was there a textbook or a teacher who said you could square differentials? What made you think $(r+dr)^2$ is well defined?

Last edited: Nov 6, 2013
19. Nov 6, 2013

### AakashPandita

i tried to find the area of a ring by subtracting the area of smaller circle from larger circle.
The difference in the radii is dr.

area of ring $$=π[(r+dr)^2 - r^2]$$

20. Nov 6, 2013