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Let's say I have a function $$ f(x) = \int_{0}^{1} \frac{\partial f}{\partial x}(\sigma x)\cdot x d\sigma$$

with $$ \frac{\partial f}{\partial x}(\sigma x)\cdot x d\sigma $$ that satisfy all the hypothesis for the Fundamental theorem of calculus.

Let's also define $$ g(\sigma) = \sigma x $$

Now my question Is it mathematically correct the following implication?

$$ f(x) = \int_{0}^{1} \frac{\partial f}{\partial x}(\sigma x)\cdot x d\sigma = \int_{0}^{1} \frac{\partial f}{\partial x}(g(\sigma))\cdot x d\sigma \Rightarrow f(x,g(\sigma)) = \int_{0}^{\sigma} \frac{\partial f}{\partial x}(g(\sigma))\cdot x d\sigma $$ for $$ \sigma = [0,1] $$

and thus

$$ \frac{\text{d}f}{\text{d}\sigma}(x,g(\sigma)) = \frac{\partial f}{\partial x}(g(\sigma))\cdot x d\sigma $$

again for $$ \sigma = [0,1] $$

ALL FROM HERE IS EDITED

Thanks the answering I realized that my implication was wrong.

But what about the following

$$ f(x) = \int_{0}^{1} \frac{\partial f}{\partial x}(\sigma x)\cdot x d\sigma = \int_{0}^{1} \frac{\partial f}{\partial x}(g(\sigma))\cdot x d\sigma \Rightarrow f(g(\sigma)) = \int_{0}^{\sigma} \frac{\partial f}{\partial x}(g(\sigma))\cdot x d\sigma $$ for $$ \sigma = [0,1] $$

Now makes more sense because if for instance we set sigma = 1 $$ f(g(\sigma)) = f(g(1)) = f(x) $$.

But I'm not completely convinced. Is it wright my (edited) implication or still wrong?

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Lets compare the beginning and the end of your question.

What is $f$ ?

-(In the beginning)A function from $\mathbb{R} \to \mathbb{R}$.

-(In the end) A function from $\mathbb{R}^2 \to \mathbb{R}$.

So do you think such maps can be equal to each other ?

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  • $\begingroup$ Thank You, that makes a lot of sense! Although it's a trivial observation, it's crucial. My implication is wrong, but what if a edit it in the following way (look at the edit part in the question) $\endgroup$ – Tommaso Bendinelli Jul 21 '18 at 11:29
  • $\begingroup$ Do you realise that $g(\sigma) = \sigma x$, but you still write $f(g(\sigma))$ in the LHS of $$f(g(\sigma)) = \int_{0}^{\sigma} \frac{\partial f}{\partial x}(g(\sigma))\cdot x d\sigma$$ $\endgroup$ – onurcanbektas Jul 21 '18 at 13:20

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