Laplace Transform of a complex valued function with power argument

Question

I'm getting confused with the Laplace transform of a complex valued function.

For example, I know that: \begin{equation} \int_0^\infty t^n e^{-st} dt = \frac{n!}{s^{n+1}}. \end{equation}

However what if I have: \begin{align} \int_0^\infty (t+i)^n e^{-st} dt &= \int_0^\infty (t+i)^n e^{-s(t+i)}e^{i s} dt \\ &=e^{i s} \frac{n!}{s^{n+1}}, \end{align} is this correct?

Thankks in advance.

Answer 1

No, it is not correct.

Note that for $i>0$, $$\int_0^\infty (t+i)^n e^{-st} dt =\int_0^\infty (t+i)^n e^{-s(t+i)}e^{i s} dt=e^{i s}\int_{i}^\infty \tau^n e^{-s\tau}d\tau\not=e^{i s}\int_{0}^\infty \tau^n e^{-s\tau}d\tau,$$ so your last equality does not hold.

Try also for example the case where $n=1$ and $i=1$. Then $$\int_0^\infty (t+1) e^{-st} dt =\int_0^\infty te^{-st} dt+\int_0^\infty e^{-st} dt=\frac{1}{s^{2}}+\frac{1}{s}\not=\frac{e^{s}}{s^{2}}.$$

Answer 2

We can use the binomial theorem to write

$$\begin{align} \int_0^\infty (t+i)^n e^{-st}\,dt&=\sum_{k=0}^n\binom{n}{k}(i)^{n-k}\int_0^\infty t^ke^{-st}\,dt\\\\ &=\sum_{k=0}^n\binom{n}{k}(i)^{n-k}\frac{k!}{s^{k+1}} \end{align}$$