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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...

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

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.

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    $\begingroup$ Thank you! That's exactly what I was looking for! Is there a theorem for that property? and if so, do you know the name? $\endgroup$ – user150203 Nov 16 '18 at 4:59
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    $\begingroup$ This is a basic property of integration. I'm sorry I don't know the name. $\endgroup$ – Kemono Chen Nov 16 '18 at 5:00
  • $\begingroup$ No worries. Thanks for your post. Much appreciated. $\endgroup$ – user150203 Nov 16 '18 at 5:00
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\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.

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  • $\begingroup$ I thought that was the case. What can we say of the result above then? What mistake(s) have I made? $\endgroup$ – user150203 Nov 16 '18 at 4:57
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    $\begingroup$ @DavidG The error was the separation of the four integrals. Each one individually diverges--the integrand only converges in the full integral because the divergent parts of the sum cancel out. $\endgroup$ – eyeballfrog Nov 16 '18 at 4:59
  • $\begingroup$ Yes Kemono Chen just commented that. Do you know the name of the Theorem that speaks to the invalidity of the expansion in this case? $\endgroup$ – user150203 Nov 16 '18 at 5:00
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    $\begingroup$ @DavidG I don't know that there is a name for it. Just a general note of caution to always check for convergence before doing things in calculus. $\endgroup$ – eyeballfrog Nov 16 '18 at 5:04
  • $\begingroup$ I will indeed. Thanks again for your post. $\endgroup$ – user150203 Nov 16 '18 at 5:05
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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:

enter image description here

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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  • $\begingroup$ Thanks so much for that post - fascinating stuff. I'm sure I'm wrong in comparing the two, but it reminds me of the $\sum_{i}^{\infty} i = -\frac{1}{12}$ where if we 'ignore' a certain rule in the Field of Reals we get an answer that has value in String Theory. I often wonder if that's because we are yet to discover the link that allows this incorrect result to be correct... in the same way that Complex Numbers allowed us to understand aspects of Real Numbers. But hey, maybe I'm just a dreamer. $\endgroup$ – user150203 Nov 16 '18 at 7:37

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