Show integrals are equal and limit of a sequence as a function. Define $L(x)=\int_{1}^x {1\over t} dt $
NOTE: I realize that $L(x)$ is the definition of $ln(x)$, but we aren't allowed to use that.  Our professor is walking us through the definition of $ln(x)$ and $e^x$. 
Part A: Show that $L({1\over x}) = -L(x)$
I've tried several different substitutions for this and even direct proof, but I'm completely stuck after working/thinking about this for the past several hours.  Help is greatly appreciated!
Part B: Using Cauchy Criterion  showing that the sequence $$s_n = 1 + {1 \over 2} + {1 \over 3} + ... + {1 \over n}$$ is divergent if $m>n$, show that $L(x)$ tendsto $\infty$ as $x \to \infty$.
Basically, I'm trying to show that if the limit of the sequence converges to some L, then the function of that sequence also converges to the same L.  I suspect that I need to show that $l(x)=s_n={1 \over x}$, but I don't know how to formally state this idea.  
 A: For the first part, $$L(1/x) = \int_1^{1/x} \dfrac1t dt$$
Let $y=1/t$, then $dt = -\dfrac{dy}{y^2}$.
$$L(1/x) = \int_1^x y\left( -\dfrac{dy}{y^2} \right) = - \int_1^x \dfrac{dy}y = - L(x)$$
For the second part, note that for all $n \in \mathbb{N}$, we have that
$$s_{2n} - s_n = \dfrac1{n+1} + \cdots + \dfrac1{2n} > \dfrac1{2n} + \dfrac1{2n} + \cdots + \dfrac1{2n} = \dfrac12$$
Hence, by Cauchy criteria the sequence diverges. Now note that $$\int_1^x \dfrac{dt}t > \int_1^2 \dfrac{dt}2 + \int_2^3 \dfrac{dt}3 + \cdots + \int_{\lfloor x \rfloor - 1}^{\lfloor x \rfloor} \dfrac{dt}{\lfloor x \rfloor} = \dfrac12 + \dfrac13 + \cdots + \dfrac1{\lfloor x \rfloor} = s_{\lfloor x \rfloor} - 1$$
A: 1 For $x>0$,
$$[L(\frac1x)]^{\prime}=L^{\prime}(\frac1x)(\frac1x)^{\prime}=x\frac{-1}{x^2}=-\frac1x=(-L(x))^{\prime}$$
which implies $L(\frac1x)=-L(x)$
2 The Harmonic series diverges. By the integral test,
$$\sum_{n=1}^{N}\frac{1}{n}\le \int_{1}^{N}\frac{1}{x}dx $$
and so it follows,
$$\lim_{N\to +\infty}L(N)=+\infty$$
A: For part A, I think the substitution $s=1/t$ should see you through.
For part B, think about grouping with powers of 2, or something similar. Then compare the sum to the integral. This might help.
A: $$L\left(\frac{1}{x}\right):=\int_1^{1/x}\frac{dt}{t}\;\;,\;\;u:=\frac{1}{t}\,\,,\,du=-\frac{1}{t^2}dt=-u^2\,dt\Longrightarrow$$
$$L\left(\frac{1}{x}\right)=\int_1^x-\frac{du}{u^2}\cdot u=-\int^x_1\frac{du}{u}=-L(x)$$
Added Part B: For $\,m>n\,$:
$$|s_m-s_n|=\left|\frac{1}{n+1}+\frac{1}{n+2}+\ldots +\frac{1}{m}\right|\geq \frac{m-n}{m}=1-\frac{n}{m}$$
We thus can keep $\,n\,$ fixed and let $\,m\to\infty\,$, getting that $\,|s_m-s_n|\geq 1\,$ , which means the sequence cannot be Cauchy and thus cannot converge. Since it is formed by a sum of positive elements, this means the sequence diverges to $\infty\,$
