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Recently I've been playing around with Feynman's Trick to evaluate integrals. Obviously, one of it's many great features is that it allows derivatives to make expressions simpler. I was wondering whether Laplace Transforms could equally be applied.

I'm not qualified to say the following is proper or rigorous, it was just an experiment.

Consider

$$I = \int_{0}^{\infty}\frac{\sin^2(x)}{x^2(x^2 + 1)}\, \mathrm dx.$$

Let

$$I(t) = \int_{0}^{\infty}\frac{\sin^2(tx)}{x^2(x^2 + 1)} \,\mathrm dx$$

Take the Laplace Transform to yield \begin{align*} \mathscr L[I(t)] &= \int_{0}^{\infty}\frac{\mathscr L[\sin^2(tx)]}{x^2(x^2 + 1)}\,\mathrm dx\\ &= \int_{0}^{\infty}\frac{\mathscr 1}{x^2(x^2 + 1)}\frac{2x^2}{s(s^2 + 4x^2)}\,\mathrm dx\\ &= \frac{2}{s}\int_{0}^{\infty}\frac{1}{(x^2 + 1)(4x^2 + s^2)}\,\mathrm dx. \end{align*} Splitting via Partial Fraction Decomposition we arrive at \begin{align*} \mathscr L[I(t)] &= \frac{2}{s(s^2 - 4)}\int_{0}^{\infty}\left[ \frac{1}{x^{2} + 1} - \frac{4}{4x^{2} + s^2}\right] \,\mathrm dx\\ &= \frac{2}{s(s^2 - 4)}\left[\arctan(x) - \frac{2}{s}\arctan\left(\frac{2x}{s}\right)\right]_{0}^{\infty}\\ &= \frac{2}{s(s^2 - 4)}\left[\frac{\pi}{2} - \frac{2}{s}\frac{\pi}{2} \right]\\ &= \frac{\pi}{s^2(s + 2)} \end{align*} And so $$I(t) = \mathscr L^{-1}\left[\frac{\pi}{s^2(s + 2)}\right] = \pi\left[\frac{t}{2} + \frac{e^{-2t}}{4} - \frac{1}{4}\right]$$ Hence, $$I(1) = \pi\left[\frac{1}{2} + \frac{e^{-2}}{4} - \frac{1}{4} \right] = \frac{\pi}{4}\left[1 + e^{-2}\right]$$ which is correct. I'm unsure if this is mere luck or whether this is a viable method.

Has anyone used this method before?

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    $\begingroup$ This is correct. You tacitly appealed to Fubini's Theorem to interchange the order of the Laplace Transform and the integral representation of $I(t)$. $\endgroup$ – Mark Viola Oct 11 '18 at 0:50
  • $\begingroup$ @MarkViola - Much appreciated! Will have to read into that. Thanks for your comment. $\endgroup$ – user150203 Oct 11 '18 at 1:03
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$ – Mark Viola Oct 11 '18 at 1:13
  • $\begingroup$ @MarkViola - Do you want to put your response as an official answer, so I can mark it 'answered' ? $\endgroup$ – user150203 Oct 15 '18 at 0:37
  • $\begingroup$ I'd be happy to do so, but I will have to wait unti tomorrow. $\endgroup$ – Mark Viola Oct 15 '18 at 3:54
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Noting that the iterated integral

$$I_1=\int_0^\infty \left(\int_0^\infty e^{-st}\frac{\sin^2(tx)}{x^2(x^2+1)}\,dt\right)\,dx$$

is finite, the Fubini-Tonelli Theorem guarantees that the iterated integral formed by interchanging the order of integration,

$$I_2=\int_0^\infty \left(\int_0^\infty e^{-st}\frac{\sin^2(tx)}{x^2(x^2+1)}\,dx\right)\,dt$$

is also finite with

$$I_1=I_2$$.

But $I_2$ is the Laplace Transform of $\int_0^\infty \frac{\sin^2(tx)}{x^2(x^2+1)}\,dx$. Hence, we assert that

$$\begin{align} \mathscr{L}\left(\int_0^\infty \frac{\sin^2(tx)}{x^2(x^2+1)}\,dx\right)(s)&=\int_0^\infty \left(\int_0^\infty e^{-st}\frac{\sin^2(tx)}{x^2(x^2+1)}\,dt\right)\,dx\\\\ &=\int_0^\infty\left(\frac2s \frac{1}{(x^2+1)(4x^2+s^2)}\right)\,dx \end{align}$$

which agrees with the development in the OP!

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