Advanced Integration

By integration by parts we may obtain

$\displaystyle \begin{aligned} \int \frac{\sin^4 x}{x^4}\, \mathrm{d}x &= -\frac{1}{3} \frac{\sin^4 x}{x^3} + \frac{4}{3} \int \frac{\sin^3 x \cos x}{x^3}\, \mathrm{d}x \\ &= -\frac{1}{3} \frac{\sin^4 x}{x^3} - \frac{2}{3} \frac{\sin^3 x \cos x}{x^2} + \frac{2}{3} \int \frac{3 \sin^2 x \cos^2 x - \sin^4 x}{x^2}\, \mathrm{d}x\\ \displaystyle&= -\frac{1}{3} \frac{\sin^4 x}{x^3} - \frac{2}{3} \frac{\sin^3 x \cos x}{x^2} + \frac{2}{3} \int \left(\frac{\sin^2 2x}{x^2} - \frac{\sin^2 x}{x^2}\right)\, \mathrm{d}x \end{aligned}$

Hence,

$\displaystyle \begin{aligned} \int^\infty_0 \frac{\sin^4 x}{x^4}\, \mathrm{d}x &= \frac{2}{3} \int^\infty_0 \frac{\sin^2 2x}{x^2}\, \mathrm{d}x - \frac{2}{3} \int^\infty_0 \frac{\sin^2 x}{x^2}\, \mathrm{d}x\\ &= \frac{2}{3} \int^\infty_0 \frac{\sin^2 x}{x^2}\, \mathrm{d}x\\&= \frac{\pi}{3} \end{aligned}$

Then, the required answer is $\displaystyle \frac{\pi}{3} \approx \boxed{1.047}$

In addition, this is the generalization of $\displaystyle \int^\infty_0 \left(\frac{\sin x}{x} \right)^n\, \mathrm{d}x$

$\displaystyle \begin{aligned} \int^\infty_0 \left(\frac{\sin x}{x} \right)^n\, \mathrm{d}x &= \lim_{\epsilon \to 0^+} \frac{1}{2} \int^{+\infty}_{-\infty} \left(\frac{\sin x}{x - i\epsilon}\right)^n\, \mathrm{d}x\\ &= \lim_{\epsilon \to 0^+} \frac{1}{2} \int^{+\infty}_{-\infty} \frac{1}{(x - i\epsilon)^n} \left(\frac{\mathrm{e}^{ix} - \mathrm{e}^{-ix}}{2i}\right)^n\, \mathrm{d}x\\ &= \lim_{\epsilon \to 0^+} \frac{1}{2} \frac{1}{(2i)^n} \int^{+\infty}_{-\infty} \frac{1}{(x - i\epsilon)^n} \sum_{k=0}^n (-1)^k {n \choose k} \mathrm{e}^{ix(n-2k)}\, \mathrm{d}x\\ &= \lim_{\epsilon \to 0^+} \frac{1}{2} \frac{1}{(2i)^n} \sum_{k=0}^n (-1)^k {n \choose k} \int^{+\infty}_{-\infty} \frac{ \mathrm{e}^{ix(n-2k)}}{(x - i\epsilon)^n} \, \mathrm{d}x \end{aligned}$

Then, by using Cauchy we obtain

$\displaystyle \begin{aligned} \int^\infty_0 \left(\frac{\sin x}{x} \right)^n\, \mathrm{d}x &= \frac{1}{2} \frac{1}{(2i)^n} \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}(-1)^k {n \choose k} \frac{2 \pi i}{(n-1)!} \frac{\mathrm{d}^{n-1}} {\mathrm{d}x^{n-1}} \mathrm{e}^{ix(n-2k)} \left. \right|_{x=0}\\ &= \frac{\pi}{2^n(n-1)!} \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k {n \choose k} (n -2k)^{n-1}\end{aligned}$

Putting $\displaystyle n=4$ we have $\displaystyle \frac{\pi}{3} \approx \boxed{1.047}$ as the answer.