Scamuzzig2

2022-09-01

I worked out this far and I am having trouble with the integral
$F\left(s\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt$
$f\left(t\right)={e}^{t+7}$
${\int }_{0}^{\mathrm{\infty }}{e}^{-\left(s-1\right)t}{e}^{7}dt$
we can say that ${e}^{7}={c}_{1}$ and show
$-\frac{{c}_{1}}{s-1}{\int }_{0}^{\mathrm{\infty }}{e}^{-\left(s-1\right)t}dt$
I would surmise that I can take the limit at this point. So
$-\frac{{c}_{1}}{s-1}\underset{b}{\overset{\mathrm{\infty }}{lim}}{\int }_{0}^{b}{e}^{-\left(s-1\right)t}dt$
which is likely evaluates as
$-\frac{{c}_{1}}{s-1}\left(0-1\right)$
yielding
$\frac{{e}^{7}}{s-1}$
What is "good practice" that I have not shown?

Haylie Campbell

The only big problem is that when you've computed the antiderivative the integral becomes an evaluation:
$-\frac{{c}_{1}}{s-1}\underset{u\to \mathrm{\infty }}{lim}\left[{e}^{-\left(s-1\right)t}{|}_{0}^{u}\right]=-\frac{{c}_{1}}{s-1}\left[\underset{u\to \mathrm{\infty }}{lim}\frac{1}{{e}^{\left(s-1\right)u}}-\frac{1}{{e}^{\left(s-1\right)\cdot 0}}\right]=-\frac{{c}_{1}}{s-1}\left[0-1\right]=\frac{{c}_{1}}{s-1}.$

Domianpv

You have computed the antiderivative but kept the integral sign which is a mistake. Namely, we have
${\int }_{0}^{\mathrm{\infty }}{e}^{-\left(s-1\right)t}{e}^{7}\phantom{\rule{thinmathspace}{0ex}}dt={e}^{7}\left(\underset{M\to \mathrm{\infty }}{lim}{\int }_{0}^{M}{e}^{-\left(s-1\right)t}\phantom{\rule{thinmathspace}{0ex}}dt\right)={e}^{7}\left(\underset{M\to \mathrm{\infty }}{lim}{\left[\frac{{e}^{-\left(s-1\right)t}}{1-s}\right]}_{t=0}^{t=M}\right)={e}^{7}\left(\underset{M\to \mathrm{\infty }}{lim}\frac{{e}^{-\left(s-1\right)M}-1}{1-s}\right)=\frac{{e}^{7}}{s-1}.$

Do you have a similar question?