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