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EDIT: To make what I am asking more clear. I've simplified it and have a more direct question.

Let's say I am writing out an expression, and I want to write: $$\int_0^xF'(y)\,dy$$

However, for reasons outside the scope of this question, I don't want to use the notation $F'(y)$ to represent the first derivative of $F(y)$ with respect to $y$. So, as far as I'm aware (correct me if there is another alternative) I must write:

$$\int_0^x\frac{dF(y)}{dy} \,dy$$

Which, I believe is equivalent to the first formula, but uses different notation. However, in the second form, immediately it looks like the $dy$s should cancel out. But that leaves only:

$$\int_0^x dF(y)$$

My question is: is this third formula in an acceptable and meaningful form? It looks weird to me, but maybe that's just because I'm inexperienced. Would this make sense to other people? Is it the best way for me to express it? Any help in understanding would be super appreciated!!!

Thank you!!


Let's say I want to integrate something like this: $$\int_0^x F_1(x)F_2(y)F_3'(y)\,dy$$ Where $F_1$, $F_2$, and $F_3$ are just placeholders for functions.

Since $F_3'(y) = \frac{dF_3(y)}{dy}$ then this could also be written $$\int_0^x F_1(x)F_2(y)\frac{dF_3(y)}{dy}\,dy$$ Then, wouldn't the $dy$s cancel out, leaving: $$\int_0^x F_1(x)F_2(y)dF_3(y)$$

Is that correct, and how would one interpret that expression, or solve it in terms of $x$?

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From what I see, I believe $x$ is a constant and hence $F_1(x)$ will pull through the integral. What you'll have is then a integrand in $y$ and $dy$ will be born out of $dF_3(y)$. So, the result will be a function in $x$. – user21436 Feb 21 '12 at 23:14
Your last expression could be interpreted as summing up values of $F_2(y)$ multiplied by the corresponding infinitesimal changes in $F_3(y)$. As Kannappan says, then the result is multiplied by $F_1(x)$. – alex.jordan Feb 21 '12 at 23:32
I have integration by parts on the brain due to calculus discussion section this morning. Does it make sense in this context to say: $$\int_0^x F_1(x)F_2(y)F_3'(y)\;dy=F_1(x)F_2(y)F_3(y)|_0^x-\int_0^xF_1(x)F_2'(y)F_3(y)\;dy.‌​$$ – dls Feb 22 '12 at 0:29

Your third equation can be thought of as integrating 1 with respect to F(y) over the interval [0,x].

$$\int_0^x 1 dF(y) $$ Let g=F(y), so we have $$ \int_0^x 1 dg $$ which is trivial to solve: $$\int_0^x 1 dg =(g)|_0^x= g(x)-g(0)$$ by FTC, which is then equivalent to $$F(x)-F(0)$$ by the original definition of g.

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