Prove that $\int_{1}^{a} \frac 1t dt + \int_{1}^{b} \frac 1t dt = \int_{1}^{ab} \frac 1t dt$ Prove that
$$\int_{1}^{a} \frac 1t dt + \int_{1}^{b} \frac 1t dt = \int_{1}^{ab} \frac 1t dt$$
Useful facts:  $\int_{1}^{a}  \frac 1t dt$ can be written as $\int_{b}^{ab} \frac 1t dt$


Every partition $P=${$t_0,...,t_n$} on $[1,a]$ gives rise to a partition $P'=${$bt_0,...,bt_n$} on $[b,ab]$, and conversely.
I have to do the question without actually integrating it, since the Fundamental Theorem of Calculus was not proven yet.

Let the partitions $P$ and $P'$ be as given.
Notice that
$$ \frac{1}{b} \cdot \text{inf}\{\frac{1}{t} : t_{i-1} \leq t \leq t_i \} = \text{inf}\{\frac{1}{t} : bt_{i-1} \leq x \leq bt_i \}$$
Now we can show equivalence between the two partitions in the following way :
Let the first inf above be $m_i$ and the second be $m'_i$, then we have
$$\begin{align*} L(f, P') &= \sum\limits_{i = 1}^n m'_i (bt_{i} - bt_{i -1}) \\
                          &= \sum\limits_{i = 1}^n bm'_i (t_{i} - t_{i -1}) \\
                          &= \sum\limits_{i = 1}^n m_i (t_{i} - t_{i -1}) \\
                          &= L(f, P) \end{align*} $$
Since these lower step functions are the same, we can conclude that
$$ \text{sup}\{L(f, P)\} = \text{sup}\{L(f, P')\}$$
and thus that
$$\int_{1}^{a} \frac{1}{t} dt = \int_{a}^{ab} \frac{1}{t} dt $$
Now we need only plug this in to
$$\int_{1}^{a} \frac{1}{t} dt + \int_{1}^{b} \frac{1}{t} dt = \int_{1}^{ab} \frac{1}{t} $$
and remember that
$$\int_{a}^{b} f + \int_{b}^{c} f = \int_{a}^{c} f $$
to complete the proof.
