Evaluation of $\int_{0}^{\infty}t^3e^{-3t}dt$

Question

I have to evaluate the integral $\int_{0}^{\infty}t^3e^{-3t}dt$ using complex analysis techniques (the laplace transform). Can you check my steps, please?

$$\int_{0}^{\infty}t^3e^{-3t}dt =\Rightarrow L(t^{3}e^{-3t})=\frac{6}{(s-3)^{4}}$$ evaluated from 0 to $\infty$ gives $\frac{-6}{3^{4}}$ even if the exact result is its opposite... What's wrong?

Answer 1

$\int_0^\infty t^3 e^{-3t}dt$ is not $L(t^3 e^{-3t})$. Laplace transform of $f$ is defined by \begin{equation} L(f)=\int_0^\infty e^{-st}f(t)dt. \end{equation} Therefore, \begin{equation} \int_0^\infty t^3 e^{-3t}dt=L(t^3)|_{s=3}. \end{equation}

Answer 2

Directly: \begin{eqnarray*} I(a) &=&\int_{0}^{\infty }dte^{-at}=\left[ \frac{e^{-at}}{-a}\right] _{0}^{\infty }=\frac{1}{a} \\ \partial _{a}^{3}I(a) &=&(-1)^{3}\int_{0}^{\infty }dtt^{3}e^{-at}=\partial _{a}^{2}\frac{-1}{a^{2}}=\partial _{a}\frac{2}{a^{3}}=\frac{-6}{a^{4}} \\ \int_{0}^{\infty }dtt^{3}e^{-at} &=&\frac{6}{a^{4}} \\ a &=&3\Rightarrow \int_{0}^{\infty }dtt^{3}e^{-3t}=\frac{6}{3^{4}} \end{eqnarray*}

Answer 3

Notice:

• Laplace Transform of $f(t)$, assume $\Re(s)>0$:

$$\mathcal{L}_t\left[f(t)\right]_{(s)}=\int_{0}^{\infty}f(t)e^{-st}\space\text{d}t$$

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$$\mathcal{L}_t\left[t^n\right]_{(s)}=\int_{0}^{\infty}t^ne^{-st}\space\text{d}t=\lim_{x\to\infty}\int_{0}^{x}t^ne^{-st}\space\text{d}t=\frac{n!}{s^{n+1}}$$ $$\mathcal{L}_t\left[t^3\right]_{(s)}=\int_{0}^{\infty}t^3e^{-st}\space\text{d}t=\lim_{x\to\infty}\int_{0}^{x}t^3e^{-st}\space\text{d}t=\frac{3!}{s^{3+1}}=\frac{3\cdot2\cdot1}{s^4}=\frac{6}{s^4}$$ $$\mathcal{L}_t\left[t^3\right]_{(3)}=\int_{0}^{\infty}t^3e^{-3t}\space\text{d}t=\lim_{x\to\infty}\int_{0}^{x}t^3e^{-3t}\space\text{d}t=\frac{3!}{3^{3+1}}=\frac{3\cdot2\cdot1}{3^4}=\frac{6}{81}=\frac{2}{27}$$