What would be the answer to this integral?

$\displaystyle\int_0^{\infty} \delta x\; dx$

I can't figure out if it would be zero or infinite? Or maybe 1?

You meant $\displaystyle\int_0^{\infty} \delta y\; dx$

apologies, I did indeed.

What's the answer

Unless I very much miss my guess, you have no sensible definition of what "infinitesimal" means here. So the question has no meaning.

Not from the attachment you've posted it isn't.

When you derive the derivative, you don't use the word infinitesimal. Therefore, "the infinitesimal distance $\delta y$" is a different object from the $\delta y$ you might see in a derivative proof. One of them is infinitesimal (whatever you think that means), and one of them is not.

It might do. But doing so doesn't really make sense here.

Formally, you have two main ways of doing that here:

$\int_0^\infty \lim_{\delta y \to 0^+} \delta y \, dx = \int_0^\infty 0 \,dx = 0$

or

$\lim_{\delta y \to 0^+} \int_0^\infty \delta y \, dx = \lim_{\delta y \to 0^+} \infty = \infty$

neither of which seems very useful.

(or you could not worry about ensuring the limit is from above, in which case the 2nd approach gives an undefined integral. Again not useful).

Well, I suspect there's a fair bit of context you haven't given...

Well, I'm guessing they really *do* mean infinitesimal (so not the same as $\delta y$ in $\lim_{\delta y \to 0}$, where $\delta y$ is just a number).

If you just say the word, then yeah - because what does it actually mean?

But you can do it properly: http://en.wikipedia.org/wiki/Non-standard_analysis

I've not really done any NSA, but from analogy with measure theory I'd guess you can't actually make this integral make sense even under that framework.

On the gripping hand, I suspect all they meant was some vague woolly idea of infinitesimal either. Conceptually the problem is "either it's 0, and the integral's 0, or it isn't 0, and the integral's infinite". Hard to get something inbetween...