Logical mistake in a proof of Dirichlet's test While writing a proof of Dirichlet's Test, [ in which, a required condition is that $\displaystyle{\int_a^B\phi(x)dx}$ is bounded $\forall B>a]$, I made a logical mistake, which I noticed upon revision. The mistake is as follows:
Let $\displaystyle{\vert\int_a^B\phi(x)dx \vert}\leq M, \ [ \ \forall B>a] \implies \displaystyle{\vert\int_{x_1}^\xi\phi(x)dx \vert} \leq M$
, $[x_1,\xi] \subset (a,B)$.
To support the fact that I was mistaken, I try to construct a function $\phi(x)$, such that :
$\phi(x)=-1$, when $x \in [1,2]$
$=2$, when $x\in [2,3]$
$=1/x^2$ in $(3, B]$
Clearly, $\displaystyle{ \vert \int_1^B\phi(x)dx \vert}\leq1$, but $\displaystyle{ \vert \int_2^3\phi(x)dx \vert}=2\nleq1$. 
Is this example valid enough to show that I was mistaken?
 A: Your counterexample does not show that for all $B >1$
$$\left|\int_1^B f(x) \, dx \right| \leqslant 1$$
If $B > 3$, then we have
$$\begin{align}\left|\int_1^B f(x) \, dx \right| &= \left|\int_1^2 (-1) \, dx +  \int_2^3 (2) \, dx + \int_3^B \frac{dx}{x^2} \right| \\ &= -1 + 2 + \frac{1}{3} - \frac{1}{B} \\ &= \frac{4}{3} - \frac{1}{B} \\ &> \frac{4}{3} - \frac{1}{3} \\ &= 1\end{align}$$
You can easily find other constant values $f(x) = C$ on $[1,2]$  where this works. Using $C = -\frac{3}{2}$, for example, we have 
$$-\frac{3}{2} + 2 + \frac{1}{3} - \frac{1}{B} = \frac{5}{6} - \frac{1}{B} < \frac{5}{6} < 1$$
Of course to prove the Dirichlet's test where it is given that $\left|\int_a^B f(x) \, dx\right|< M$ is uniformly bounded for all $B > a$ it is not necessary to have $\left|\int_{x_1}^\xi f(x) \, dx\right|< M$ for any $[x_1,\xi] \subset [a,B]$.
All that will matter is that we have the bound
$$\left|\int_{x_1}^\xi f(x) \, dx\right| = \left|\int_{a}^\xi f(x) \, dx - \int_{a}^{x_1} f(x) \, dx \right| \leqslant \left|\int_{a}^\xi f(x) \, dx \right| +\left|\int_{a}^{x_1} f(x) \, dx \right| \leqslant M+M = 2M $$
Note that $\left|\int_{a}^\xi f(x) \, dx \right|, \,\,\left|\int_{a}^{x_1} f(x) \, dx \right| \leqslant M$ when $\xi > a$.
