The integral I want to solve is:
$$ \int_{-\infty}^0 \frac{1-e^{-t}}{t} dt$$
By substitution:
$$ -t =u$$
$$ \int_{0}^{\infty} \frac{ 1-e^t}{t} dt$$
Maclaurin expanding the numerator,
$$ \int_0^{\infty} - \sum_{k=1}^n \frac{t^{k-1} }{k!}dt$$
Integrating termwise:
$$ -\sum_{k=1}^n \frac{ t^k}{k k!} \biggm|_0^{\infty}$$
Now the tricky part is upper bound.. I'm thinking of using squeeze to figure out limit. My attempt is as follows:
$$ \sum_{k=1}^n \frac{ t^k}{ k!} \leq \sum_{k=1}^n \frac{ t^k}{k k!} \leq \sum_{k=1}^n \frac{ t^k}{(k+1)!} $$
This doesn't seem to work.. any hints?
By the substitution $-t=u$ we have $$\int_{-\infty}^0 \frac{1-e^{-t}}{t} dt$$ $$=\int_{\infty}^{0}\frac{1-e^{u}}{u}du=\int_{0}^{\infty}\frac{e^{t}-1}{t}dt$$
and $\frac{e^{t}-1}{t}\geq1$ for all $t>0$ (where $\lim_{t\rightarrow 0}\frac{e^t-1}{t}=1)$, which follows from Bernoulli's inequality.