Finding original function of $1/x$ by using a step function Is it technically possible to show, that the area under $1/x$ in the interval $[1,a]$ equals $\log a$ without using any kind of differentiation, but only step functions.
I tried this, but without success.
Thank you
 A: One can get a closed form for an upper (by that meaning right endpoint) Riemann sum $R_n$ by using, instead of the partition of $[1,a]$ into $n$ equal parts, the partition 
$$x_0=a^{0/n}=1,\ x_1=a^{1/n},\ \ldots \ x_n=a^{n/n}=a.$$
Then the $k$th subinterval $[x_{k-1},x_k]$ has length $a^{(k/n)}-a^{(k-1)/n},$ and for a right endpoint sample point at $x_k$ of the function $y=1/x$ this interval length is to be multiplied by $1/x_k=1/a^{k/n}$ for its contribution to the sum $R_n$. The "nice" thing here is that the $k$ cancels and the $k$th sum contribution is $1-\frac{1}{a^{1/n}}.$ This is then summed over $k,$ and since here are no remaining $k$ in the expression the right sum $R_n$ for this partition (into $n$ unequal parts) of $[1,a]$ for the integral $\int_0^a (1/x) dx$ is
$$R_n=n\cdot (1-\frac{1}{a^\frac1n}).$$
The latter can be shown to approach $\ln a$ by various means such as L'Hospital, though without knowing what is an "acceptable" definition of $\ln a$ it needs some more work, since the question said "without using any kind of differentiation" which would seem to rule out L'Hospital.
Added later: If one puts $h=1/n$ then the above expression for $R_n$ becomes $(a^0-a^{-h})/h,$ which is the left-hand difference quotient for the derivative of $g(x)=a^x$ at $x=0.$ Then  $g'(x)=\ln(a)\cdot a^x,$ which is $\ln a$ at $x=0.$ So no need for LHospital, only the derivative formula for exponentials.
I do find it interesting that, using this unequal partition, a closed form could be obtained for a Riemann sum. [Note: it needs to be checked that, as $n \to \infty,$ the mesh size of this partition goes to $0,$ but that's straightforward.]
