Laplace transform of $t \ \cos(t) \ e^{-2t}$

Question

$$ \mathscr{L} \{ t \ \cos(t) \}(s)=\int_0^{+\infty} t \cos(t) \ e^{-st} \ dt=\frac{1}{2} \ \int_0^{+\infty} t (e^{it-st}+ e^{-it-st} ) \ dt=$$

$$=\frac{1}{2} \Big( \Big[ t \Big( \frac{e^{it-st}}{i-s}+\frac{e^{-it-st}}{-i-s} \Big) \Big]_0^{+\infty} - \int_0^{+\infty} \frac{e^{it-st}}{i-s}+\frac{e^{-it-st}}{-i-s} \ dt \Big)=$$

$$=\frac{1}{2} \Big( - \Big[ \frac{e^{it-st}}{(i-s)^2}+\frac{e^{-it-st}}{(-i-s)^2} \Big]_0^{+\infty} \Big)=$$

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

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

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

$$\mathscr{L} \{ t \ \cos(t) \ e^{-2t} \}(s)=\frac{(s+2)^2-1}{( \ (s+2)^2+1)^2}$$

Is it correct?

Thanks!

Answer 1

Well, you can use the 'frequency shifting' property of the Laplace transform:

$$\mathscr{L}_t\left[\text{f}\left(t\right)e^{\alpha t}\right]_{\left(\text{s}\right)}=\text{F}\left(\text{s}-\alpha\right)\tag1$$

And we can use the 'frequency-domain derivative' property of the Laplace transform:

$$\mathscr{L}_t\left[t\text{f}\left(t\right)\right]_{\left(\text{s}\right)}=-\text{F}\space'\left(\text{s}\right)\tag2$$

So, when we know that (when $\Re\left(\text{s}\right)>0$):

$$\mathscr{L}_t\left[\cos\left(t\right)\right]_{\left(\text{s}\right)}:=\int_0^\infty\cos\left(t\right)e^{-\text{s}t}\space\text{d}t=\frac{\text{s}}{1+\text{s}^2}\tag3$$

So, we also get that (when $\Re\left(\text{s}\right)>0$):

$$\mathscr{L}_t\left[t\cos\left(t\right)\right]_{\left(\text{s}\right)}:=\int_0^\infty t\cos\left(t\right)e^{-\text{s}t}\space\text{d}t=-\frac{\text{d}}{\text{d}\text{s}}\left(\frac{\text{s}}{1+\text{s}^2}\right)=\frac{\text{s}^2-1}{\left(1+\text{s}^2\right)^2}\tag4$$

And using the frequency shifting, we get (when $\Re\left(\text{s}\right)>-2$):

$$\mathscr{L}_t\left[te^{-2t}\cos\left(t\right)\right]_{\left(\text{s}\right)}:=\int_0^\infty te^{-2t}\cos\left(t\right)e^{-\text{s}t}\space\text{d}t=\frac{\left(2+\text{s}\right)^2-1}{\left(1+\left(2+\text{s}\right)^2\right)^2}\tag5$$

So, yes you're right! But you can use the frequency-domain derivative, instead of the integral.