<s>[INST] Prove that $F(xy) =F(x) + F(y)$ when $F(x)$ is not $\ln(x)$ So we have this function for $x > 0$. 
$$\int_{1}^{x} \frac{1}{t}\text{d}t$$
Show that $F(xy) = F(x) + F(y)$ without assuming $F(x) = \ln(x)$.
I came so far as this point, but I can't crack the last step.
$$\int_{1}^{x} \frac{1}{t}\text{d}t + \int_{x}^{xy} \frac{1}{t}\text{d}t$$
which is:
$$F(x) + \int_{x}^{xy} \frac{1}{t}\text{d}t$$
However, I am having difficulties proving that: 
$$\int_{x}^{xy} \frac{1}{t}\text{d}t = F(y)$$
Help would be very very appreciated!
 [/INST] Spoiler below... Don't forget to change both upper and bottom integral bounds when using an integral sustitution
This is the full answer... 
$$ F(xy):=\int_{1}^{xy} \frac{1}{t}dt $$ 
$$ let \,\,\,\begin{matrix}  t = x \cdot u    &  x \cdot u = xy \\ dt = x \cdot du  & x \cdot u = 1\end{matrix} 
 \,\,\,OR\,\,\,
\begin{matrix}  u = y \\ u = \frac{1}{x} \end{matrix} $$ 
$$ then \,\,\,\,F(xy)=\int_{1}^{xy} \frac{1}{t}dt = \int_{\frac{1}{x}}^{y} \frac{1}{x \cdot u} (x \cdot du) = \int_{\frac{1}{x}}^{y} \frac{1}{u} du$$
$$ splitting \int_{\frac{1}{x}}^{y} \frac{1}{u} du = \int_{\frac{1}{x}}^{1} \frac{1}{u} du + \int_{1}^{y} \frac{1}{u} du$$ 
creating one last u-substitition
$$ let \,\,\,\begin{matrix}  v = x \cdot u    &  x \cdot 1 = x =  v \\ \frac{dv}{x} = du  & x\cdot (1/x) = 1 = v \end{matrix}  \,\,\, which\,\, becomes 
$$ 
$$ F(xy)= \int_{1}^{x} \frac{1}{v/x} (dv/x) + \int_{1}^{y} \frac{1}{u} du = \int_{1}^{x} \frac{1}{v} dv + \int_{1}^{y} \frac{1}{u} du = F(x) + F(y)$$
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