<s>[INST] Solving $\int_{0}^{\infty} \frac{\sin(x)}{x^3}dx$ 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...
 [/INST] As the other answers have pointed out, the integral does indeed diverge. But if want to assign a finite value to it, there are a couple ways to see that in fact $-\pi/4$ is the "right" value.
One is to take the integral not quite down to zero, but instead to $\epsilon$. If we do, then expand in a series in $\epsilon$, we get
$$\int_\epsilon^\infty\frac{\sin x}{x^3}\,dx=\epsilon^{-1}-\frac{\pi}{4}+O(\epsilon).$$
The leading term is the divergent $\epsilon^{-1}$, but if we ignore that then the next term is $-\pi/4$.
Another way to get the same value is to first extend the integral to $-\infty$. Since the integrand is even, we would expect
$$\int_0^\infty\frac{\sin x}{x^3}\,dx=\frac12\int_{-\infty}^\infty\frac{\sin x}{x^3}\,dx.$$
Of course, the right-hand integral diverges, as well. But the only problem point is when $x=0$. If we imagine $x$ as being in the complex plane, travelling from $-\infty$ to $\infty$, then we can just "go around" $0$ by curving $x$ slightly out into the complex plane, for example:

If we do, then we end up with the same answer of $-\pi/4$. I'm not sure what this sort of regularization is called, but it's frequently used in quantum field theory where divergent integrals abound.
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