Vrftjkry

In my attempt to solve the this improper integral, I employed a well known improper integral (part of the Borwein family of integrals):

$$ int_{0}^{infty} frac{sinleft(frac{x}{1}right)sinleft(frac{x}{3}right)sinleft(frac{x}{5}right)}{left(frac{x}{1}right)left(frac{x}{3}right)left(frac{x}{5}right)} : dx = frac{pi}{2}$$

To begin with, I made a simple rearrangement

$$ int_{0}^{infty} frac{sinleft(frac{x}{1}right)sinleft(frac{x}{3}right)sinleft(frac{x}{5}right)}{x^3} : dx = frac{pi}{30}$$

From here I used the Sine/Cosine Identities

$$ int_{0}^{infty} frac{frac{1}{4}left(-sinleft(frac{7}{15}xright)+ sinleft(frac{13}{15}xright) + sinleft(frac{17}{15}xright) -sinleft(frac{23}{15}xright) right)}{x^3} : dx = frac{pi}{30}$$

Which when expanded becomes

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

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

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

Which simplifies to

$$ -frac{120}{15^2}int_{0}^{infty} frac{sin(x)}{x^3}:dx = frac{2pi}{15}$$

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...

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

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.

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.

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
$$int_epsilon^inftyfrac{sin x}{x^3},dx=epsilon^{-1}-frac{pi}{4}+O(epsilon).$$
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
$$int_0^inftyfrac{sin x}{x^3},dx=frac12int_{-infty}^inftyfrac{sin x}{x^3},dx.$$
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.