I am stumped on this problem. I think my procedure is incorrect. If someone could take a look at guide me through, I would be most appreciative.
The problem states: Use to definition of Laplace Transform to find $L{f(t)}$.
We know the defintion of the Laplace Transform is:
$F(s) = \int_0^{\infty}e^{-st}f(t)dt$
so $f(t) = e^{t+7}$ and I began to simplify the integrand to find:
$e^{-st+t+7}$ if I let $u=-st + t+ 7$ then $du=(-s+1)dt$ Integrating that is
$\int_0^{\infty}e^u(-s+1)du$
I don't think this is the way to do it..I need to use a limit from some variable, say $T$ that approaches infiity..
$\int e^{t+7} e^{-s t} \, dt=\int e^{-s t+t+7} \, dt$
Let $-s t+t+7=u\to t-st=u-7\to t=\dfrac{u-7}{1-s}$
so we have $dt=\dfrac{du}{1-s}$ and the integral becomes
$\dfrac{1}{1-s}\int e^u \, du=\dfrac{e^u}{1-s}=\dfrac{e^{-s t+t+7}}{1-s}=F(t)$
Now perform the Laplace transform evaluating the integral function at $t=m\to\infty$ and at $t=0$. We get: $$\int_0^{\infty } e^{t+7} e^{-s t} \, dt=\lim_{m\to +\infty } \, \frac{e^{-ms +m+7}}{1-s} - F(0)=0-\left(-\frac{e^7}{s-1}\right)=\dfrac{e^7}{s-1}$$ Remark
The term $e^7$ is a constant and could have been brought out of the integral from the beginning. I mean $\int e^{t+7} e^{-s t} \, dt=e^7\int e^{t} e^{-s t} \, dt$
For the sake of simplicity I left everything as you started it
hope this helps