Laplace Transformation $f(t) = e^{t+7}$

Question

I worked out this far and I am having trouble with the integral

$$F(s) = \int_0^\infty{e^{-st}f(t)dt}$$

$$f(t) = e^{t+7}$$

$$\int_0^\infty{e^{-(s-1)t}e^7dt}$$

we can say that $e^7=c_1$ and show

$$-\frac{c_1}{s-1}\int_0^\infty{e^{-(s-1)t}dt}$$

I would surmise that I can take the limit at this point. So

$$-\frac{c_1}{s-1}\lim_b^\infty\int_0^b{e^{-(s-1)t}dt}$$

which is likely evaluates as

$$-\frac{c_1}{s-1}(0-1)$$

yielding

$$\frac{e^7}{s-1}$$

What is "good practice" that I have not shown?

Answer 1

The only big problem is that when you've computed the antiderivative the integral becomes an evaluation: $$-\frac{c_1}{s-1}\lim_{u\rightarrow\infty}\left[e^{-(s-1)t}\Big|_{0}^{u}\right]=-\frac{c_1}{s-1}\left[\lim_{u\rightarrow \infty}\frac{1}{e^{(s-1)u}}-\frac{1}{e^{(s-1)\cdot0}}\right]=-\frac{c_1}{s-1}[0-1]=\frac{c_1}{s-1}.$$

Answer 2

You have computed the antiderivative but kept the integral sign which is a mistake. Namely, we have

$$ \int_0^{\infty} e^{-(s-1)t} e^7 \, dt = e^7 \left( \lim_{M \to \infty} \int _0^M e^{-(s-1)t} \, dt \right) = e^7 \left( \lim_{M \to \infty} \left[ \frac{e^{-(s-1)t}}{1-s}\right]_{t=0}^{t=M} \right) = e^7 \left( \lim_{M \to \infty} \frac{e^{-(s-1)M}-1}{1-s} \right) = \frac{e^7}{s-1}. $$