Problem:
Before I get into the problem, my professor is trying to have us define both the logarithmic and exponential functions, so we are not allowed to use things we know about logs or the exponential function, which is why I'm having trouble.
For $ x \in (0,\infty)$, define $L(x)=\int_{1}^x 1/t dt $
a) Why does the integral exist? By Theorem 30.1 (a theorem in our textbook), since L(x) is monotone, L(x) is integrable.
b) Why is L(x) differentiable and what is its derivative?
This is the one I'm most having trouble with. How exactly do I show that this function is differentiable? I know that since it is integrable, then it is uniformly continuous and that the Fundamental Theorem of Calculus gives me that $L'(x)=l(x)$, but I don't know how to state this in a way that is "proof"
c) Show that $L(1/x)=-L(x)$, in other words $\int_{1}^{1\over x}=-\int_{1}^{x}$
I feel like something that shows the following and then rearranging the terms would be best, but I'm pretty lost. $\int_{1}^{1\over x} -\int_{1}^x = \int_{x}^x; define \int_{x}^x=0$