# How to evaluate the integral $\int_0^{+\infty} \frac{\sin^4{x}}{x^4}dx$? [duplicate]

As we know that $\int_0^{+\infty} \frac{\sin{x}}{x}dx=\pi/2$,but how to evaluate the integral $\int_0^{+\infty} \frac{\sin^4{x}}{x^4}dx$?

## marked as duplicate by Robert Z, user228113, Did, user99914, ShaileshNov 25 '17 at 5:16

• It should be $\frac{\pi}{3}.$ – Michael Rozenberg Nov 24 '17 at 12:57
• @MichaelRozenberg ah I figure it out. By a lot of integral by parts. It’s $\pi/3$ exactly. – 闫嘉琦 Nov 24 '17 at 13:08
Surely you know that$$\int\limits_0^{\infty}dx\,\frac {\sin^2x}{x^2}=\frac {\pi}2$$ which can be proven using integration by parts. Therefore, we take your integral by using integration by parts twice and a trigonometric identity to deduce\begin{align*}I & =-\frac {\sin^4x}{3x^3}\,\Biggr\rvert_{0}^{\infty}+\frac 43\int\limits_0^{\infty}dx\,\frac {\sin^3x\cos x}{x^3}\\ & =-\frac {2\cos x\sin^3x}{3x^2}\,\Biggr\rvert_0^{\infty}+\frac 23\int\limits_0^{\infty}dx\,\left(\frac {\sin^2 2x}{x^2}-\frac {\sin^2x}{x^2}\right)\end{align*}Where the second equation is used with the identity$$3\cos^2x\sin^2x-\sin^4=\sin^22x-\sin^2x$$Hence\begin{align*}I & =\frac 23\int\limits_0^{\infty}dx\,\frac {\sin^2 2x}{x^2}-\frac 23\int\limits_0^{\infty}dx\,\frac {\sin^2x}{x^2}\\ & =\frac 43\int\limits_0^{\infty}dx\,\frac {\sin^2x}{x^2}-\frac 23\int\limits_{0}^{\infty}dx\,\frac {\sin^2x}{x^2}\\ & =\color{blue}{\frac {\pi}3}\end{align*}