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.
 A: Since $\int_1^a \frac{1}{t}\, dt = \int_b^{ab} \frac{1}{u}\, du$ (using the $u$-sub $u = tb$), we have $$\int_1^a \frac{1}{t}\, dt + \int_1^b \frac{dt}{t} = \int_b^{ab} \frac{1}{u}\, du + \int_1^b \frac{1}{t}\, dt = \int_1^b \frac{1}{t}\, dt + \int_b^{ab} \frac{1}{t}\, dt = \int_1^{ab} \frac{1}{t}\, dt.$$
A: 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.
A: Simply, use the change change of variable $u = b\cdot t$ in the first integral, or $u = a \cdot t$ in the second integral, then use Chasles's relation.
A: Abuse your useful fact:
$$\int_{1}^{a} \frac 1t dt + \int_{1}^{b} \frac 1t dt=\int_b^{ab}\frac1t dt+ \int_{1}^{b} \frac 1t dt =-\int_{ab}^{b}\frac1t dt- \int_{b}^{1} \frac 1t dt = -\left(\underbrace{\int_{ab}^{b}\frac1t dt+ \int_{b}^{1} \frac 1t dt}_{\int_{ab}^1\frac1{t}dt}\right)=\;\therefore \int_1^{ab}\frac1{t}dt$$
Other useful facts used:
$$\int_a^bf(t)dt=-\int_b^af(t)dt$$
$$\int_a^bf(t)dt+\int_b^cf(t)dt=\int_a^cf(t)dt$$
A: Following the hint, you want to show that
\begin{align}
U(f,P) &= U(f,P') \\  L(f,P) &= L(f,P')
\end{align}
Therefore
$$
\int_1^a \frac{1}{t}\,dt = \int_b^{ab} \frac{1}{t}\,dt
$$
