What to do when there is a common factor in the limits of an integral? Let $log : (0,∞) → R$ be defined by $log(x) := \int_{1}^{x} f$ , where $f : (0,∞) → R :x → \frac{1}{x}$
. Show that $log(xy) = log(x) + log(y).$
As log is a continuous function I know that it is regulated. So I need to show that 
$\int_{1}^{xy} f$ = $\int_{1}^{x} f$ + $\int_{1}^{y} f$.
The solutions say to show $\int_{1}^{xy} f$ = $\int_{1}^{x} f$ + $\int_{1}^{xy} f$ = $\int_{1}^{xy} f$ = $\int_{1}^{x} f$ + $\int_{1}^{y} f$.
Can you explain how to get the middle part? 
 A: Note that
$$
\int_{1}^{x}f + \int_{x}^{xy}f = \int_{1}^{xy}f
$$
by basic properties of Riemann integration.
It suffices to prove that
$$
\int_{1}^{y}f = \int_{x}^{xy}f.
$$
Note that if $x > 0$ then
$$
\int_{x}^{xy}\frac{1}{t}dt = \int_{1}^{y}\frac{1}{xu}\cdot xdu = \int_{1}^{y}\frac{1}{u}du
$$
by change-of-variables theorem with the auxiliary function $u \mapsto xu =: t$.
A: So $\int_1^{xy}{f} = \int_1^x{f} + \int_x^{xy}{f}$.
Now it is required to prove that $\int_x^{xy}{f} = \int_1^y{f}$ and the proof will be complete. We can re-write this as $\int_x^{xy}{\frac{1}{u}}du$ due to the definition of the function. Now, substitute $tx=u$ to get $\int_1^y{\frac{1}{tx}}xdt = \int_1^y{\frac{1}{t}}dt = \int_1^y{f}$ and the proof is complete.
I think the answers you have posted are wrong, check the limits and edit your post if it is wrong.
A: Since the variable of integration is $t$, we can rewrite the second integral in the following way:
\begin{align*}
\int_{x}^{xy} \frac{1}{t} dt
&=\int_{t=x}^{t=xy} \frac{1}{t} dt \\
&=\int_{xu=x}^{xu=xy} \frac{1}{xu} d(xu) \\
&=\int_{u=1}^{u=y} \frac{1}{xu} x du \\
&=\int_{u=1}^{u=y} \frac{1}{u} du \\
&=\ln(y)
\end{align*}
where we made the substitution $t=ux$ in the second step.
