I have to evaluate the following integral by using the Laplace transform:

$$\int_0^\infty \frac{\sin^4 (tx)}{x^3}\,\mathrm{d}x.$$

How am I supposed to approach this question by using the Laplace transform?

  • $\begingroup$ So is it just $t$ or $tx$ inside the sin? $\endgroup$ – mathreadler Apr 17 '17 at 19:20

Such integral is just $t^2\int_{0}^{+\infty}\frac{\sin^4(x)}{x^3}\,dx$, or $t^2$ times:

$$ \int_{0}^{+\infty}\left(\mathcal{L}^{-1}\frac{1}{x^3}\right)(s)\cdot\left(\mathcal{L}\sin^4 x\right)(s)\,ds = \int_{0}^{+\infty}\frac{12 s}{(s^2+4)(s^2+16)}\,ds $$ i.e. $\color{red}{t^2\log(2)}$, by partial fraction decomposition.
Have a look at this Wikipedia entry about the key property exploited here.

  • $\begingroup$ how did t^2 come outside? $\endgroup$ – Aayushman Mishra Apr 17 '17 at 19:33
  • $\begingroup$ (+1) Oh, I'm happy I just started to write my answer where one would integrate the Laplace transform of $\sin^4(x)$ three times. It was horrible to do, but now at least I don't have to write it down. This is the way to go! $\endgroup$ – mickep Apr 17 '17 at 19:33
  • 1
    $\begingroup$ @AayushmanMishra Rewrite as an integral over tx and replace tx by x (since it's just a dummy variable) to get this expression. $\endgroup$ – Anamay Chaturvedi Apr 17 '17 at 19:40

You must evaluate the Laplace transform

$$ \int_{0}^{\infty}dt \sin^{4} (xt) e^{-st}=F(s) $$

This can be evaluated by using the Euler's identity

After you have evaluated $ F(s) $ you must integrate F(s) three times and set ·$ s=0$


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