I need to show that
$$
\int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3}
$$
I have already derived the result $\int_0^\infty \frac{\sin^2x}{x^2} = \frac{\pi}{2}$ using complex analysis, a result which I am supposed to start from. Using a change of variable $ x \mapsto 2x $ :
$$
\int_0^\infty \frac{\sin^2(2x)}{x^2}dx = \pi
$$
Now using the identity $\sin^2(2x) = 4\sin^2x – 4\sin^4x $, we obtain
$$
\int_0^\infty \frac{\sin^2x – \sin^4x}{x^2}dx = \frac{\pi}{4}
$$
$$
\frac{\pi}{2} – \int_0^\infty \frac{\sin^4x}{x^2}dx = \frac{\pi}{4}
$$
$$
\int_0^\infty \frac{\sin^4x}{x^2}dx = \frac{\pi}{4}
$$
But I am now at a loss as to how to make $x^4$ appear at the denominator. Any ideas appreciated.
Important: I must start from $ \int_0^\infty \frac{\sin^2x}{x^2}dx $, and use the change of variable and identity mentioned above
Best Answer
You are likely expected to integrate by parts (twice) $$ \begin{eqnarray} \int \frac{\sin^4(x)}{x^4} \mathrm{d}x &=& -\frac{1}{3} \frac{\sin^4(x)}{x^3} + \frac{4}{3} \int \frac{\cos(x) \sin^3(x) }{x^3} \mathrm{d} x \\ &=& -\frac{1}{3} \frac{\sin^4(x)}{x^3} -\frac{2 \cos(x) \sin^3(x)}{3 x^2} + \frac{2}{3} \int \frac{3 \cos^2(x) \sin^2(x) - \sin^4(x)}{x^2} \mathrm{d} x \\ &=& -\frac{1}{3} \frac{\sin^4(x)}{x^3} -\frac{2 \cos(x) \sin^3(x)}{3 x^2} + \frac{2}{3} \int \left(\frac{\sin^2(2x)}{x^2} - \frac{\sin^2(x)}{x^2} \right) \mathrm{d}x \end{eqnarray} $$ where the last equality used $$\begin{eqnarray} 3 \cos^2(x) \sin^2(x) - \sin^4(x) &=& 3 \cos^2(x) \sin^2(x) - \sin^2(x) (1-\cos^2(x)) \\ &=& \left(2 \sin(x) \cos(x) \right)^2 - \sin^2(x) = \sin^2(2x) - \sin^2(x) \end{eqnarray} $$ Now $$\begin{eqnarray} \int_0^\infty \frac{\sin^4(x)}{x^4} \mathrm{d}x &=& \frac{2}{3} \int_0^\infty \frac{\sin^2(2x)}{x^2} \mathrm{d} x - \frac{2}{3} \int_0^\infty \frac{\sin^2(x)}{x^2} \mathrm{d}x \\ &=& \frac{4}{3} \int_0^\infty \frac{\sin^2(y)}{y^2} \mathrm{d} y - \frac{2}{3} \int_0^\infty \frac{\sin^2(x)}{x^2} \mathrm{d}x \\ &=& \frac{2}{3} \int_0^\infty \frac{\sin^2(x)}{x^2} \mathrm{d}x = \frac{\pi}{3} \end{eqnarray} $$