I want to calculate the following improper integral using Laplace and transforms (and laplace transforms only).

$$\int_0^{\infty} x e^{-3x} \sin{x}\, dx$$

I propose the following method. I plan to use $\mathcal{L}(\int_0^x f(t) dt)=\frac{\mathcal{L}(f)(s)}{s}$.

So, first I need to find out the laplace transform of the integrand, which can be done using $\mathcal{L}({xf})=-\frac{dL(f)}{ds}=-\frac{d}{ds}(e^{-3x} \sin{x})=-\frac{d}{ds}(\frac{1}{(s+3)^2+1})=\frac{2(s+3)}{((s+3)^2)+1)^2}$

So, my first formula for the transform of an integral means I have to calculate the inverse transform of $\frac{2(s+3)}{s((s+3)^2)+1)^2}=2[\frac{1}{((s+3)^2)+1)^2}+\frac{3}{s((s+3)^2)+1)^2}]$. I am stuck after this. How do I calculate this inverse transform?

Is this the right method I am using? If yes, how do I calculate the inverse transform w/o using the bromwich integral.


1 Answer 1


Hint: Consider the more general integral

$$ \int_{0}^{\infty} x\sin(x) e^{-sx} dx $$

which is the Laplace transform of $x\sin(x)$. You need to use the following property

$$ L(xf(x))= -F'(s). $$

So, basically, you need to find only the Laplace transform of $\sin(x)$

  • 1
    $\begingroup$ Awesome! Thanks. $\endgroup$
    – user23238
    Commented Apr 30, 2013 at 6:05
  • $\begingroup$ @ramanujan_dirac: You are welcome. By the way, do not forget to subs $s=3$ at the end. $\endgroup$ Commented Apr 30, 2013 at 6:58

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