Find Laplace transform of $f(t)=t^5$

Question

I was double checking some Laplace transforms against Wolfram alpha, and I found this result odd.

Let $f(t)=t^{5}$

Using Laplace integration formula $\int_0^\infty f(t)e^{-st}dt$ we get:

\begin{equation} \int_0^\infty t^5e^{-st}dt=\int_0^\infty te^{5-st}dt \end{equation}

Here we use integration by parts, with $u=t$, $u'=1$, $v'=e^{5-st}$, and $v=\frac{e^{5-st}}{5-s}$, obtaining:

\begin{equation} \int_0^\infty te^{5-st}dt=\bigg[e^{5-st}\bigg(\frac{t}{5-s}-\frac{1}{(5-s)^2}\bigg)\bigg]_0^\infty=0-1\bigg(0-0\bigg)=0 \end{equation}

But Wolframalpha gives this, which is very different.

What is wrong here?

Thanks

Answer 1

Substitute $\tau =st$ so $\frac{d\tau}{dt} = s$ and $$ \int_0^\infty t^5 e^{-st} dt = \int_0^\infty \frac{\tau^5}{s^6} e^{-\tau} d\tau = \frac{1}{s^6}\Gamma(6) = \frac{5!}{s^6} = \frac{120}{s^6}. $$

Answer 2

Another approach is to note that $\mathcal{L} \{t f(t)\} = -f'(s), s>0$ or more generally $\mathcal{L} \{t^n f(t)\} = (-1)^n f^{(n)}(s), s>0$

Doing this with $f(t)=1, f(s)=s^{-1}$ yields the usual Laplace transforms $$\mathcal{L} \{t \} = - \frac{d}{ds} s^{-1} = (1!) s^{-2}$$ $$\mathcal{L} \{t^2 \} = - \frac{d}{ds} s^{-2} = (2!) s^{-3}$$ $$\mathcal{L} \{t^3 \} = -(2!) \frac{d}{ds} s^{-3} = (3!) s^{-4}$$ from which you deduce/prove $$\mathcal{L} \{t^5 \} = (4!) \frac{d}{ds} s^{-5} = (5!) s^{-6}$$ or more generally $$\mathcal{L} \{t^n \} = (n!) s^{-(n+1)}$$