Laplace Heaviside function,how can I continue?

Question

I have to find the Laplace transform of the function $H(t-a)t^{n}$. That's what I have done so far: $$L\{H(t-a)t^{n}\}=\int_{0}^{\infty}e^{-st}H(t-a)t^{n}dt=\int_{a}^{\infty}e^{-st}t^{n}dt=\int_{0}^{\infty}e^{-s(t+a)}(t+a)^{n}dt $$

Could you tell me how I can continue?

Answer 1

I think this formula can help us, besides to @dato's way. You know that $\mathcal{L}\{\mathscr{U}(t-a)\}=\frac{e^{-as}}{s}, a>0$. And $$\mathcal{L}\{t^nf(t)\}=(-1)^n\bigg(\mathcal{L}(f(t))\bigg)^{(n)}$$ Now let think about $\big(\frac{e^{-as}}{s}\big)^{(n)}$. Using Maple, I found:

$$n=1\to \left(\frac{e^{-as}}{s}\right)'=-\exp(-as)\frac{as+1}{s^2}\\ n=2\to \left(\frac{e^{-as}}{s}\right)''=\exp(-as)\frac{a^2s^2+2as+2}{s^3}\\ n=3\to \left(\frac{e^{-as}}{s}\right)'''=-\exp(-as)\frac{a^3s^3+3a^2s^2+6as+6}{s^4} $$ And so I think the possible pattern could be $$(-1)^n\frac{\sum_{k=0}^{n}\frac{n!}{(n-k)!}s^{n-k}a^{n-k}}{s^{n+1}}$$

Answer 2

put $t+\alpha=m$

then $dt=dm$

so we have

$ e^{-s(t+\alpha)}*(t+\alpha)^n=e^{-sm}m^ndm$

you can integrate by part

Answer 3

You can obtain directly from the integral $$\mathcal{L}\{H(t-a)t^{n}\}=\int_{a}^{\infty}e^{-st}t^{n}dt=\frac{1}{s^{n+1}}\int_{as}^{\infty}e^{-z}z^{n}dz= \frac{1}{s^{n+1}}\Gamma(n+1,as)$$ recalling the integral expression of incomplete Gamma function for integer $n$ $$ \Gamma(n,x)=\int_{x}^{\infty}e^{-z}z^{n-1}dz. $$

NOTE Another useful methodology may be the following.

Observing that $$H(t-a)t^n=H(t-a)(t-a+a)^n=H(t-a)\sum_{k=0}^n\binom{n}{k}(t-a)^ka^{n-k}$$ the Laplace transform becomes $$ \mathcal{L}\{H(t-a)t^n\}=\sum_{k=0}^n\binom{n}{k}a^{n-k}\mathcal{L}\{H(t-a)(t-a)^k\} $$ recalling that $\mathcal{L}\{H(t-a)f(t-a)\}=e^{-as}F(s)$. Using $\mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}}$, we obtain $$ \mathcal{L}\{H(t-a)t^n\}=e^{-as}\sum_{k=0}^n\binom{n}{k}a^{n-k}\frac{k!}{s^{k+1}}=\frac{n!e^{-as}}{s^{n+1}}\sum_{k=0}^n\frac{(as)^{n-k}}{(n-k)!}=\frac{1}{s^{n+1}}n!e^{-as}\sum_{\nu=0}^n\frac{(as)^{\nu}}{\nu!} $$ and recalling that $$ \Gamma(n,z)=(n-1)!e^{-z}\sum_{\nu=0}^{n-1}\frac{z^{\nu}}{\nu!} $$ is the incomplete Gamma function for integer $n$, we obtain finally $$ \mathcal{L}\{H(t-a)t^n\}=\frac{1}{s^{n+1}}\Gamma(n+1,as). $$