0
$\begingroup$

I need to find:

$\int_0^\infty e^{-2t}tcos(t)dt$

Using Laplace transform.

The theorem that I'm supposed to use (I wasn't able to find the correct term in English):

Suppose $L[f(t)]=F(s).$ If $\frac{f(t)}{t}$ is an original, then $L[\frac{f(t)}{t}]=\int_s^\infty F(s)ds$.

I'm not sure how this applies here. The problems I've solved so far (and unfortunately haven't fully understood) had a distinct form of $\frac{f(t)}{t}$.

I'd be grateful if anyone could help point me to the right direction here.

$\endgroup$
3
$\begingroup$

$${\cal L}(\cos t)=\int_0^\infty e^{-st}\cos t\ dt=\dfrac{s}{s^2+1}$$ $${\cal L}(t\cos t)=\int_0^\infty e^{-st}t\cos t\ dt=-\left(\dfrac{s}{s^2+1}\right)'$$ then set $s=2$.

$\endgroup$
  • $\begingroup$ Forgot about that approach, I find it a bit strange that it wasn't mentioned anywhere in my book. $\endgroup$ – frostpad Aug 19 '18 at 15:17
  • 1
    $\begingroup$ With integration by parts prove ${\cal L}(tf)=-F'$. $\endgroup$ – Nosrati Aug 19 '18 at 15:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.