Find the Laplace transform of f(t) = t*e^(2t), and state its region of convergence.

Step-by-step solution

1. 1

   Start from the transform you already know. The basic power rule gives

   $\mathcal{L}\{t^{n}\}=\frac{n!}{s^{n+1}},\qquad\text{so}\qquad \mathcal{L}\{t\}=\frac{1}{s^{2}}\quad (s>0).$

   Everything else in this problem is the exponential factor, and there is a rule made exactly for that.

2. 2

   Apply the first shifting theorem. The theorem says that multiplying a function by $e^{at}$ shifts its transform:

   $\mathcal{L}\{e^{at}f(t)\}=F(s-a),\qquad F(s)=\mathcal{L}\{f(t)\}.$

   With $f(t)=t$, $F(s)=\tfrac{1}{s^{2}}$ and $a=2$, so replace every $s$ by $s-2$:

   $\mathcal{L}\{t\,e^{2t}\}=\frac{1}{(s-2)^{2}}.$

3. 3

   Cross-check with the multiplication-by-$t$ rule. An independent route is $\mathcal{L}\{t\,g(t)\}=-\frac{d}{ds}\mathcal{L}\{g(t)\}$ with $g(t)=e^{2t}$, whose transform is $\tfrac{1}{s-2}$:

   $-\frac{d}{ds}\left(\frac{1}{s-2}\right)=\frac{1}{(s-2)^{2}}.$

   Two different theorems giving the same expression is strong evidence the answer is right.

4. 4

   Confirm straight from the definition. Integrating by parts with $u=t$, $dv=e^{-(s-2)t}dt$:

   $\int_{0}^{\infty}t\,e^{2t}e^{-st}\,dt=\int_{0}^{\infty}t\,e^{-(s-2)t}\,dt=\left[\frac{-t\,e^{-(s-2)t}}{s-2}\right]_{0}^{\infty}+\frac{1}{s-2}\int_{0}^{\infty}e^{-(s-2)t}dt=\frac{1}{(s-2)^{2}}.$

5. 5

   State the region of convergence. The boundary term vanishes and the final integral converges only when the exponent $-(s-2)$ is negative, i.e.

   $s-2>0\;\Longleftrightarrow\;s>2.$

   That is also visible in the answer itself: the transform has a double pole at $s=2$, and the region of convergence always lies to the right of the rightmost pole.