Observe that when we only substitute $\frac{1}{x} = y$
$$I = \int_1^\infty \frac{dy}{\lfloor y\rfloor y^2} = \sum_{n=1}^\infty \frac{1}{n}\int_n^{n+1}\frac{dy}{y^2}$$
$$ = \sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n=1}^\infty \frac{1}{n} - \frac{1}{n+1} = \frac{\pi^2}{6} - 1$$
To prove : for $A,B \in \Bbb{R} $
$\lfloor(A + B)\rfloor = \lfloor A\rfloor
+ \lfloor B\rfloor + \lfloor \{A\} + \{B\}\rfloor.$
By definition:
$A = \lfloor A\rfloor + \{A\}.$
$B = \lfloor B\rfloor + \{B\}.$
Therefore,
$$(A + B) = \lfloor A\rfloor + \lfloor B\rfloor
+ \{A\} + \{B\}. \tag1$$
Also, since $0 \leq \{A\},\{B\} < 1$
$0 \leq \{A\} + \{B\} < 2.$
Also, if $P \in \Bbb{R}$ and $z \in \Bbb{Z}$ such that
$0 \leq (P - z) < 1$ then $\lfloor P\rfloor = z.$
Consider two cases separately.
$\underline{\text{Case 1:}~~0 \leq \{A\} + \{B\} < 1}$
Then
$\lfloor \{A\} + \{B\}\rfloor = 0$
Using equation (1) above,
$0 \leq (A + B) - (\lfloor A\rfloor + \lfloor B\rfloor) < 1. $
Therefore,
$\lfloor (A + B)\rfloor = \lfloor A\rfloor + \lfloor B\rfloor =
\lfloor A\rfloor + \lfloor B\rfloor + \lfloor \{A\} + \{B\}\rfloor.$
Therefore, the assertion holds in Case 1.
$\underline{\text{Case 2:}~~1 \leq \{A\} + \{B\} < 2}$
Then
$\lfloor \{A\} + \{B\}\rfloor = 1$
Using equation (1) above,
$0 \leq (A + B) - [(\lfloor A\rfloor + \lfloor B\rfloor) + 1]< 1. $
Therefore,
$\lfloor (A + B)\rfloor = \lfloor A\rfloor + \lfloor B\rfloor + 1 =
\lfloor A\rfloor + \lfloor B\rfloor + \lfloor \{A\} + \{B\}\rfloor.$
Therefore, the assertion holds in Case 2.
Best Answer
This is a plot of floor(ln(x)) along x axis.Find these points of discontinuity and break your integral at these points. The the numerator in all of these sub parts will be a constant and you will have just $\frac{some-constant}{x}$ as the integrand.
You can find these points of discontinuity using desmos.com or a scientific calculator (which is easily available on smartphones or computers these days).