# Finding an integral using the Laplace transform

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?

• So is it just $t$ or $tx$ inside the sin? – 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.

• how did t^2 come outside? – Aayushman Mishra Apr 17 '17 at 19:33
• (+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! – mickep Apr 17 '17 at 19:33
• @AayushmanMishra Rewrite as an integral over tx and replace tx by x (since it's just a dummy variable) to get this expression. – 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 ·$s=0$