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Let $y: \mathbb{R} \rightarrow \mathbb{R}$.

Claim: If $y' = \frac{1}{x} $ then $y(ab) = y(a) + y(b)$

Proof: \begin{align} y' &= \frac{1}{x}\\ \implies y(x) &= \int_1^x \frac{1}{z} \ dz &\text{(fundamental theorem)}\\ \implies y(ab) &= \int_1^{ab} \frac{1}{z} \ dz \\ &= \int_1^a \frac{1}{z} dz + \int_a^{ab} \frac{1}{z} dz\\ &= y(a) + \int_1^b \frac{1}{t} \ dt &\text{(change of variables $z = at$)} \\ = y(a) + y(b) \end{align}

However, $y = \ln(4x)$ is a counterexample to the above proof, since

$$y(ab) = \ln(4ab) = \ln(4) + \ln(a) + \ln(b) \ne \ln(16) + \ln(a) + \ln(b) = y(a) + y(b)$$

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  • $\begingroup$ you assume $y(1)=0$ else the formula become $y(ab)-y(1)=y(a)-y(1)+y(b)-y(1)$ $\endgroup$ – zwim Apr 26 '18 at 19:56
  • $\begingroup$ In your first step, why do you assume the starting point of your integral is $1$? We could also think about $\int_{17}^x{1\over z}dz$ ... $\endgroup$ – Noah Schweber Apr 26 '18 at 19:59
  • $\begingroup$ starting point could be anything $\endgroup$ – David Warren Katz Apr 26 '18 at 20:00
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If $y^{\prime}(x) = \dfrac{1}{x}$, it follows that $$\int_{1}^{x}y^{\prime}(t)\text{ d}t=y(x)-y(1)$$ You cannot drop the $y(1)$, since you don't have the assumption that $y(1) = 0$.

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  • $\begingroup$ thanks, i see it now. I think that is what the other comments were hinting at too. $\endgroup$ – David Warren Katz Apr 26 '18 at 20:14
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The error is in your very first step. The claim should be If $y' = \frac{1}{x}$, there exists a $y$ such that $y(ab) = y(a) + y(b)$. The claim as is does not imply $y(x) = \int_{1}^{x} \frac{1}{z} \, dz$.

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  • $\begingroup$ doesn't that follow from the fundamental theorem? $\endgroup$ – David Warren Katz Apr 26 '18 at 19:58
  • $\begingroup$ @Auburn It does not. $\endgroup$ – Jeffery Opoku-Mensah Apr 26 '18 at 20:00

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