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.


1 Answer 1


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

  • $\begingroup$ Forgot about that approach, I find it a bit strange that it wasn't mentioned anywhere in my book. $\endgroup$
    – frostpad
    Aug 19, 2018 at 15:17
  • 1
    $\begingroup$ With integration by parts prove ${\cal L}(tf)=-F'$. $\endgroup$
    – Nosrati
    Aug 19, 2018 at 15:20

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