1.I tried to use u substitution for the first part of the problem, but just ended up with $1/2$($n^2$-1). Looking on stack exchange, I've seen people making multiple integrals by splitting them based on the upper and lower integrals, but got confused on how to do this with an upper limit like ln n.
I also looked at this, but got confused: https://www.youtube.com/watch?v=bruvOqbMcM4 (Solving for $\int_{0}^{\infty }$$[x]e^{-x}dx$)
I'm working on a proof by induction for the second part of the problem, but I'm not sure this is the right approach.
All help is appreciated.
The first one can be shown by induction on $n \in \mathbb{N}$.
I'll leave you the base case to check it, and will show how to prove de inductive case.
Hypothesis. $ \displaystyle \int_0^{\ln(n)} \left \lfloor e^x \right \rfloor dx = n \ln n - \ln n! $
Thesis. $ \displaystyle \int_0^{\ln(n+1)} \left \lfloor e^x \right \rfloor dx = (n+1) \ln (n+1) - \ln (n+1)! $
Proof. We notice that: $$ \begin{align*} &\displaystyle \int_0^{n+1} \frac{\left \lfloor t\right \rfloor }{t} dt = \displaystyle \int_0^{n} \frac{\left \lfloor t\right \rfloor }{t} dt + \displaystyle \int_{n}^{n+1} \frac{\left \lfloor t\right \rfloor }{t} dt \\ &\Leftrightarrow (t=e^{x}) \\ &\displaystyle \int_0^{\ln(n+1)} \left \lfloor e^x \right \rfloor dx = \displaystyle \int_0^{\ln(n)} \left \lfloor e^x \right \rfloor dx + \displaystyle \int_{\ln(n)}^{\ln(n+1)} \left \lfloor e^x \right \rfloor dx \end{align*} \tag{*} $$ The left side is the thesis meanwhile on the right side, the first term is the hypothesis. We want to know what the second term is: let's study it.
$$ \displaystyle \int_{\ln(n)}^{\ln(n+1)} \left \lfloor e^x \right \rfloor dx \overset{t=e^x}{=} \displaystyle \int_{n}^{n+1} \frac{\left \lfloor t \right \rfloor}{t} dt = \displaystyle \int_{n}^{n+1} \frac{n}{t} dt = n(\ln(n+1) - \ln(n)) $$
Now, you have the three terms of the equality at $(*)$ and it's not hard to check that it holds.