Laplace transforms please help

Question

I really need help to find to Laplace transforms of $f(x)=x+e^{-x}$, and $g(x)=xe^x$. I'm having big troubles on the calculations. Thanks.

Answer 1

Recall $F(s)=\int_0^{\infty}{f(t)e^{-st}dt}=\int_0^{\infty}{(t+e^{-t})e^{-st}dt}=\int_0^{\infty}{te^{-st}dt}+\int_0^{\infty}{e^{-t}e^{-st}dt}=\int_0^{\infty}{te^{-st}dt}+\int_0^{\infty}{e^{-t(s+1)}dt}$ And $G(s)=\int_0^{\infty}{te^{t}e^{-st}dt}=\int_0^{\infty}{te^{-t(s-1)}dt}$

Use integration by parts for both.

Also remember since we have improper integrals we will end up with limits that are undefined, so we need L'hopital's rule to solve them (the exponential always wins!....).

EDIT: First I'll do $\int_0^{\infty}{te^{-st}dt}$ (and then we can do the same thing for $s+1$).
Let $$u=t\to{du}=dt,$$ $$dv=e^{-st}dt\to{v}=\frac{e^{-st}}{-s}$$ So using integration by parts,
$$\int_0^{\infty}{te^{-st}dt}=\lim_{b\to \infty}\int_0^b{te^{-st}dt}=\lim_{b\to \infty}\frac{te^{-st}}{-s}|_0^b-\lim_{b\to \infty}\frac{1}{-s}\int_0^b{e^{-st}dt}$$ $$=\lim_{b\to \infty}\frac{t}{-se^{st}}|_0^b-\lim_{b\to \infty}\frac{e^{-st}}{(-s)^2}|_0^b=\lim_{b\to \infty}\left(\frac{b}{-se^{sb}}-\frac{0}{s}\right)-\lim_{b\to \infty}\left({\frac{1}{s^2e^{sb}}}-\frac{1}{s^2}\right)$$ $$=\lim_{b\to \infty}\frac{b}{se^{sb}}-0-0+\frac{1}{s^2}$$ Because $$\lim_{b\to \infty}{\frac{b}{-se^{sb}}}=\frac{\infty}{\infty},$$ We can use L'Hopital's rule to get it in the form $$\lim_{b\to \infty}\frac{1}{s^2e^{st}}=0$$ Finally we can say that $$\int_0^{\infty}{te^{-st}dt}=\frac{1}{s^2}$$ Since $s$ is just a constant when we integrate with respect to $t$, we can also say that $$\int_0^{\infty}{te^{-(s-1)t}dt}=\frac{1}{(s-1)^2}$$

Thus we can see that $$F(s)=\frac{1}{s^2}+\frac{1}{s+1}$$ $$G(s)=\frac{1}{(s-1)^2}.$$

Answer 2

We use the properties of Laplace transformation, you can find them here

http://en.wikipedia.org/wiki/Laplace_transform

$\displaystyle{ \bullet \quad \mathcal{L} \{x \} = \int_{0}^{+ \infty} e^{-st}t dt = \lim_{a \to + \infty} - \frac{e^{-st}}{s^2} (st+1) \big|_0^a = \cdots = \frac{1}{s^2} }$

$\displaystyle { \bullet \quad \mathcal{ L} \{ e^x \} = \int_{0}^{+ \infty} e^{-st} e^t dt= \lim_{ a \to + \infty} \frac{ e^{(1-s)t}}{1-s} |_0^a = \frac{1}{s-1} }$

Using the properties now we get:

$\displaystyle{ \bullet \quad \mathcal{L} \{f \} = \mathcal{ L} \{x+e^{-x} \} =\mathcal{L} \{x\} +\mathcal{L} \{e^{-x} \cdot 1 \} = \frac{1}{s^2} + F(s+1) = \frac{1}{s^2} + \frac{1}{s+1} }$ ,where $\displaystyle{ F(s)= \mathcal{L} \{1\} }$

$\displaystyle{ \bullet \mathcal{ L} \{xe^x \} = F(s-1) =\frac{1}{(s-1)^2} }$, where we used the property that $\displaystyle{ \mathcal{L} \{ e^{at} f(t) \} = F(s-a) ; F(s)= \mathcal{L} \{f\} }$. In your case is$\displaystyle{ F(s)=\mathcal{L} \{t\} } $