In my attempt to solve the this improper integral, I employed a well known improper integral (part of the Borwein family of integrals):
$$ \int_{0}^{\infty} \frac{\sin\left(\frac{x}{1}\right)\sin\left(\frac{x}{3}\right)\sin\left(\frac{x}{5}\right)}{\left(\frac{x}{1}\right)\left(\frac{x}{3}\right)\left(\frac{x}{5}\right)} \: dx = \frac{\pi}{2}$$
To begin with, I made a simple rearrangement
$$ \int_{0}^{\infty} \frac{\sin\left(\frac{x}{1}\right)\sin\left(\frac{x}{3}\right)\sin\left(\frac{x}{5}\right)}{x^3} \: dx = \frac{\pi}{30}$$
From here I used the Sine/Cosine Identities
$$ \int_{0}^{\infty} \frac{\frac{1}{4}\left(-\sin\left(\frac{7}{15}x\right)+ \sin\left(\frac{13}{15}x\right) + \sin\left(\frac{17}{15}x\right) -\sin\left(\frac{23}{15}x\right) \right)}{x^3} \: dx = \frac{\pi}{30}$$
Which when expanded becomes
$$ -\int_{0}^{\infty} \frac{\sin\left(\frac{7}{15}x\right)}{x^3}\:dx + \int_{0}^{\infty} \frac{\sin\left(\frac{13}{15}x\right)}{x^3}\:dx +
\int_{0}^{\infty} \frac{\sin\left(\frac{17}{15}x\right)}{x^3}\:dx –
\int_{0}^{\infty} \frac{\sin\left(\frac{23}{15}x\right)}{x^3}\:dx
= \frac{2\pi}{15}$$
Using the property
$$\int_{0}^{\infty}\frac{\sin(ax)}{x^3}\:dx = a^2 \int_{0}^{\infty}\frac{\sin(x)}{x^3}\:dx$$
We can reduce our expression to
$$\left[ -\left(\frac{7}{15}\right)^2 + \left(\frac{13}{15}\right)^2 + \left(\frac{17}{15}\right)^2 – \left(\frac{23}{15}\right)^2\right] \int_{0}^{\infty} \frac{\sin(x)}{x^3}\:dx = \frac{2\pi}{15}$$
Which simplifies to
$$ -\frac{120}{15^2}\int_{0}^{\infty} \frac{\sin(x)}{x^3}\:dx = \frac{2\pi}{15}$$
And from which we arrive at
$$\int_{0}^{\infty} \frac{\sin(x)}{x^3}\:dx = -\frac{\pi}{4}$$
Is this correct? I'm not sure but when I plug into Wolframalpha it keeps timing out…
Best Answer
You cannot expand the integrals since they are not convergent.
Moreover, given that $\int_a^b f(x)+g(x)dx$ converges, $\int_a^b f(x)+g(x)dx=\int_a^b f(x)dx+\int_a^b g(x)dx$ only if $\int_a^b f(x)dx$ and $\int_a^b g(x)dx$ converge.