Finding Laplace Transform of $te^{-t}$

Question

I started with this integral:

$$ \int_{0}^{\infty} e^{-st}\cdot te^{-t}dt$$

= $$\int_{0}^{\infty} te^{-(s+1)t}dt$$

let $dv=e^{-(s+1)t}dt, u=t$ and thus $v=-\frac{1}{s+1}e^{-(s+1)t}dt, du=dt$

$\rightarrow$

$-\frac{t}{s+1}e^{-(s+1)t}|_0^\infty + \frac{1}{s+1}\int_{0}^{\infty}e^{-(s+1)t}dt$

= $-\frac{1}{(s+1)^2}e^{-(s+1)t}|_0^\infty$

What mistakes have I made or do I just use L'Hopital's rule to finish?

Answer 1

If you see an integral of the form $$\int t f(t) \,dt$$ then try partial integration! Assuming $s\neq 1$ \begin{align}\int_0^{\infty} t f(t) \,dt &= \left[t\frac{-1}{s+1}e^{-(s+1)t} \right]_{t=0}^\infty-\int_0^{\infty} \frac{-1}{s+1}e^{-(s+1)t} \,dt \\&= 0 - 0 + \left[ \frac{-1}{(s+1)^2}e^{-(s+1)t} \right]_{t=0}^\infty \\ &= \frac{1}{(s+1)^2}\end{align}

Answer 2

$$\mathcal{L\{te^{-t}\}}=\int^{\infty}_{0} e^{-st}te^{−t}dt $$

$$=\int^{\infty}_{0} e^{-t(s+1)}tdt $$ substitute $$st=x;dt=\frac{dx}{s}$$ $$=\frac{1}{s^2}\int^{\infty}_{0} xe^{-\frac{x}{s}(s+1)}dx $$ $\frac{x(s+1)}{s}=m;dx\frac{(s+1)}{s}=dm; $ $$=\frac{1}{(s+1)^2}\int^{\infty}_{0}m \ e^{-m}dm $$

$$=\frac{\Gamma (2)}{(s+1)^2}$$

$$=\frac{1}{(s+1)^2}$$

Answer 3

As an exercise I would have my DE students see how many of the elementary Laplace transform rules they could derive using only linearity plus $L\{y^\prime\}=sL\{y\}-y(0)$.

It is simple then to find $L\{e^{at}\}=\frac{1}{s-a}$. Having that one can find $L\{te^{at}\}$ as follows:

Let $y=te^{at}$. Then $y^\prime=e^{at}+ate^{at}$. \begin{eqnarray} L\{y^\prime\}&=sL\{y\}-y(0)\\ L\{e^{at}+ate^{at}\}&=sL\{te^{at}\}\\ L\{e^{at}\}+aL\{te^{at}\}&=sL\{te^{at}\}\\ \frac{1}{s-a}&=(s-a)L\{te^{at}\}\\ L\{te^{at}\}&=\frac{1}{(s-a)^2} \end{eqnarray}