So, I've come across the following integral (and it's expansion) many times and in my study so far, Complex Residues have been used to evaluate it. I was hoping to find an alternative approach using Laplace Transforms. I believe the method I've taken is correct, but I'm concerned there may certain theorems/tests I should have applied first
$$I = \int_{0}^{\infty} \frac{\sin\left(\frac{x}{1}\right)\sin\left(\frac{x}{3}\right)}{\left(\frac{x}{1}\right)\left(\frac{x}{3}\right)}\:dx$$
The first step is to make a slight change of variable $x = 3u$, which gives us
$$I = \int_{0}^{\infty} \frac{\sin\left(3u\right)\sin\left(u\right)}{u^2} \: du $$
Here I employ the Feynman Trick but with two variables, i.e.
$$I(a,b) = \int_{0}^{\infty} \frac{\sin\left(3ua\right)\sin\left(ub\right)}{u^2}\: du$$
Take the Laplace Transform w.r.t '$a$'
$$\mathscr{L}_{a} \left[I(a,b)\right] = \int_{0}^{\infty} \frac{\mathscr{L}_{a}\left[\sin\left(3ua\right)\right]\sin\left(ub\right)}{u^2}\: du = \int_{0}^{\infty} \frac{3u\sin\left(ub\right)}{\left(s^2 + 9u^2\right)u^2}\: du $$
Or
$$ \overline{I}(s,b) = \int_{0}^{\infty} \frac{3\sin\left(ub\right)}{\left(s^2 + 9u^2\right)u}$$
Now apply the Laplace Transform w.r.t '$b$'. Here $\omega$ will be used as the alternate '$s$' variable. Hence we arrive at
$$ \mathscr{L}_{b}\left[\overline{I}(s,b)\right] = \int_{0}^{\infty} \frac{3\mathscr{L}_{b}\left[\sin\left(ub\right)\right]}{\left(s^2 + 9u^2\right)u}\:du = \int_{0}^{\infty} \frac{3u}{\left(s^2 + 9u^2\right)u\left(\omega^2 + u^2\right)}\:du $$
Or
$$\overline{\overline{I}}\left(s,\omega\right) = \int_{0}^{\infty} \frac{3}{\left(s^2 + 9u^2\right)\left(\omega^2 + u^2\right)}\:du = \frac{3\pi}{2s\omega}\left(\frac{1}{s + 3\omega} \right)$$
In no specific order we now take the Inverse Laplace Transform w.r.t. '$\omega$'
$$\overline{I}\left(s,b\right) = \mathscr{L}_{\omega}^{-1}\left[\frac{3\pi}{2s\omega}\left(\frac{1}{s + 3\omega} \right) \right] = \frac{3\pi}{2}\left[\frac{1}{s^2} – \frac{e^{\frac{sb}{3}}}{s^2}\right]$$
We now take the Inverse Laplace Transform w.r.t. '$s$'
$$I(a,b) = \frac{3\pi}{2}\mathscr{L}_{s}^{-1}\left[ \frac{1}{s^2} – \frac{e^{\frac{sb}{3}}}{s^2}\right] = \frac{3\pi}{2}\left[a – \left(a – \frac{b}{3} \right)\mathcal{H}\left(a – \frac{b}{3}\right) \right]$$
And so,
$$I = I(1,1) = \frac{3\pi}{2}\left[1 – \left(1 – \frac{1}{3} \right)\mathcal{H}\left(1 – \frac{1}{3}\right) \right] = \frac{\pi}{2}$$
As required.
Is this a stroke of luck? or is it just employing the Dominated Convergence Theorem and Fubini's Theorm (as I believe is valid here).
Best Answer
Not sure what you are asking with your "Is this a stroke of luck?" question...
Just in case, here's a different approach to the integral:
$$I=\int_0^\infty \frac{\sin (3x) \sin(x)}{x^2} dx=\frac{1}{2}\int_0^\infty \frac{\cos (2x)-\cos(4x)}{x^2} dx$$
$$I=\int_0^\infty \frac{\cos (x)-\cos(2x)}{x^2} dx$$
Now we also introduce a parameter, though we only need one:
$$I(a)=\int_0^\infty \frac{\cos (ax)-\cos(2ax)}{x^2} dx$$
$$I'(a)=\int_0^\infty \frac{2\sin (2ax)-\sin(ax)}{x} dx$$
The integral can now be safely separated into two terms, and each has a well known value:
$$\int_0^\infty \frac{\sin(x)}{x} dx=\frac{\pi}{2}$$
So:
$$I'(a)=\int_0^\infty \frac{2\sin (2ax)-\sin(ax)}{x} dx=\frac{\pi}{2}$$
Integrating (the constant of integration is determined by $I(0)$), we have:
$$I(a)=\frac{\pi}{2}a$$
$$I(1)=\frac{\pi}{2}$$
The proofs of the $\text{sinc}$ integral can be found elsewhere, including this site. Evaluating the integral $\int_0^\infty \frac{\sin x} x \ dx = \frac \pi 2$?