# Integral dx^2

## Main Question or Discussion Point

Hi pf!

I'm wondering how to evaluate. $$\int_{x_i}^x \int_{x_i}^x (ds)^2$$ I would do it like $$\int_{x_i}^x \int_{x_i}^x (ds)^2 \\ =\int_{x_i}^x ds \int_{x_i}^x ds \\= (x-x_i)^2$$ yet i know this is wrong since the answer should be $(x-x_i)^2/2!$ (taylor series is the application here). It looks like we should evaluate this as $$\int_{x_i}^x \int_{x_i}^x (ds)^2 = \int_{x_i}^x s (ds) = s^2/2$$ and then suddenly place the $x-x_i$ inside the $s$ term (which we obviously don't normally do).

Thanks so much!

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Stephen Tashi
I'm wondering how to evaluate. $$\int_{x_i}^x \int_{x_i}^x (ds)^2$$
How did you arrive at that expression? Perhaps that would give us a clue what it means.

How did you arrive at that expression? Perhaps that would give us a clue what it means.
I arrived at it through a taylor series derivation for a function $f(x)$. Given $$f(x) = f(a) + \int_a^x f'(s) ds \implies f'(x) = f'(a) + \int_a^x f''(s) ds$$. Evidently take this expression for $f'(x)$ as $f'(s) = f'(a) + \int_a^s f''(t) dt$ and substitute this into the above to arrive at $$f(x) = f(a) + \int_a^x \left( f'(a) + \int_a^s f''(t) dt \right) ds \\ = f(a) + f'(a)(x-a) + \int_a^x \int_a^s f''(t) dt ds$$. From the above, $f''(t) = f''(a) + \int_a^t f'''(r) dr$. Substituting this into the previous expression we have $$f(x) = f(a) + f'(a)(x-a) + \int_a^x \int_a^s \left( f''(a) + \int_a^t f'''(r) dr \right) dt ds \\ = f(a) + f'(a)(x-a) + \int_a^x \int_a^s f''(a) dt ds + \int_a^x \int_a^s \int_a^t f'''(r) dr dt ds \\= f(a) + f'(a)(x-a) + \int_a^x f''(a)(s-a) ds + \int_a^x \int_a^s \int_a^t f'''(r) dr dt ds \\ = f(a) + f'(a)(x-a) + f''(a)\left[ x^2/2 - a^2/2 -a(x-a) \right] + \int_a^x \int_a^s \int_a^t f'''(r) dr dt ds \\ = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2 + ...$$

hmmmm, i guess it works after all. at the end the author writes $\int_a^x \cdots \int_a^x M (ds)^{n+1} = M\frac{(x-a)^{n+1}}{(x+1)!}$ and I was confused. they must be abusing notation I guess?

Stephen Tashi
they must be abusing notation I guess?
Their notation using $\int \int dtds$ is clearer than the notation $\int \int (ds)^2$

Their notation using $\int \int dtds$ is clearer than the notation $\int \int (ds)^2$
I agree, but the $dt ds$ notion is mine. theirs was the $(ds)^2$. are they talking about a remain integral?

Stephen Tashi
I agree, but the $dt ds$ notion is mine. theirs was the $(ds)^2$. are they talking about a remain integral?

I'd have to see their work to make a guess about what it means.

mathman
There are obvious errors which make it hard to follow. Specifically he has identities where he has a + rather than an =.

Yes, the paper is not perfect but I like the integration technique to derive the taylor series. but now that we're on the topic, how would one compute an integral with measure $(ds)^2$?

Stephen Tashi
Did the paper use the notation $(ds)^2$? Or did it only use $\int \int ....ds ds$ ?

mathman
Did the paper use the notation $(ds)^2$? Or did it only use $\int \int ....ds ds$ ?
yes $\int\int....dsds$

The author clearly abuses the notation. The integral you asked about should be $$\int_a^x \int_a^s dt ds = \int_a^x (s-a)ds = \int_a^x (s-a)d(s-a)=\frac12 (x-a)^2 ,$$ and you already computed it when you were explaining how did you arrive to your integral.

The notation in the text you are reading is "twice bad": not only $dsds$ or $(ds)^2$ is a bad (and formally wrong) notation, but also the limits of integration in the text are wrong. Your computations are correct, and you can see that in multiple integrals the limits of integrations are different in the inner and outer integrals. In the text they are the same, and that is completely wrong.

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Stephen Tashi
It's confusing to write $\int \int f(s) ds ds$ but it isn't any more wrong than writing $\int \int f(t) dt ds$.

By contrast, the notation $\int \int f(s) (ds)^2$ doesn't have a standard interpretation.

Thanks guys! I was confused with their notation, but I think it makes sense now! I appreciate your help!

WWGD
Formally, you can see this as integrating 2-forms on a 2-manifold , the manifold being $\mathbb R^2$. Then , if $ds^2 =ds \wedge ds$ then it is zero.