Solving $\int_{0}^{\infty} \frac{\sin(x)}{x^3}dx$

Question

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...

Answer 1

$$-\int_{0}^{\infty} \frac{\sin\left(\frac{7}{15}x\right)}{x^3}\:dx + \int_{0}^{\infty} \frac{\sin\left(\frac{13}{15}x\right)}{x^3}\:dx + \int_{0}^{\infty} \frac{\sin\left(\frac{17}{15}x\right)}{x^3}\:dx - \int_{0}^{\infty} \frac{\sin\left(\frac{23}{15}x\right)}{x^3}\:dx = \frac{2\pi}{15}$$

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.

Answer 2

\begin{multline} \int_0^\infty \frac{\sin(x)}{x^3}dx = \int_0^1 \frac{\sin(x)}{x^3}dx +\int_1^\infty \frac{\sin(x)}{x^3}dx \\> \int_0^1 \frac{x/2}{x^3}dx +\int_1^\infty \frac{\sin(x)}{x^3}dx = \frac{1}{2}\int_0^1 \frac{1}{x^2}dx +\int_1^\infty \frac{\sin(x)}{x^3}dx = \infty \end{multline}

The integral diverges.

Answer 3

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.