Could you please explain what happens in the second last line, where they had the integral larger or equal to the e^-s integral of 1/t. I am confused as it seems that they just pulled the e^-s out and claimed that it was smaller than the integral itself. What law are they using here? sorry I did not type in Latex, but please refer to the image below. thank you.
First, it is a general fact that, if $f(x) \geq g(x)$ for all $x \in [a,b]$, then $$ \int_a^b f(x) \,\mathrm{d}x \geq \int_a^b g(x) \,\mathrm{d}x \text{.} $$ (To prove this, observe that $f(x) - g(x) \geq 0$. Then subtract the right-hand side of the display form the left-hand side, observe that the new (single) integral on the left is an integral of an everywhere nonnegative function, so is nonnegative.)
Then, from $\mathrm{e}^{-st} \geq \mathrm{e}^{-s}$ on the interval of integration, we have \begin{align*} \mathrm{e}^{-st} &\geq \mathrm{e}^{-s}, \\ \frac{\mathrm{e}^{-st}}{t} &\geq \frac{\mathrm{e}^{-s}}{t} \qquad t \neq 0, \\ \int_0^1 \; \frac{\mathrm{e}^{-st}}{t} \,\mathrm{d}t &\geq \int_0^1 \; \frac{\mathrm{e}^{-s}}{t} \,\mathrm{d}t \\ &= \mathrm{e}^{-s}\int_0^1 \; \frac{1}{t} \,\mathrm{d}t \text{,} \end{align*} because $\mathrm{e}^{-s}$ is independent of $t$ so is constant while evaluating the integral.