How to $\int_{0}^\infty {\sin^3(x)\over x}dx$

Question

How to evaluate : $$\int_{0}^\infty {\sin^3(x)\over x}dx$$ I don't know how to do it. I tried to finish it using integration by parts, but it doesn't work? Can someone tell me how to evaluate the integral?

Answer 1

We know that $\displaystyle\int_{0}^{\infty}\dfrac{\sin x}{x}\,dx = \dfrac{\pi}{2}$. Also, $\sin^3 x = \dfrac{3}{4}\sin x - \dfrac{1}{4}\sin 3x$.

Therefore, $\displaystyle\int_{0}^{\infty}\dfrac{\sin^3 x}{x}\,dx = \dfrac{3}{4}\int_{0}^{\infty}\dfrac{\sin x}{x}\,dx - \dfrac{1}{4}\int_{0}^{\infty}\dfrac{\sin 3x}{x}\,dx$.

Can you finish from here?

Answer 2

Starting with $$\sin(3x)=3\sin(x)-4 \sin^3(x)$$ you will easily obtain that the antiderivative is $$\int{sin^3(x)\over x}dx=\frac{1}{4} (3 \text{Si}(x)-\text{Si}(3 x))$$ Now, taking into account the bounds and the properties an the sine integral $$\int_{0}^\infty {\sin^3(x)\over x}dx=\frac{1}{4}(3\frac{\pi}{2}-\frac{\pi}{2})=\frac{\pi}{4}$$

Answer 3

I would recommend following the simplification steps in the other answers and reaching

$$\begin{align}I &=\dfrac{1}{4}\int_0^\infty \frac{3\sin(x) -\sin(3x)}{x} \mathrm{d}x\\&= \dfrac{1}{4}\int_0^\infty \frac{2\sin(x)}{x} \mathrm{d}x+\dfrac{1}{4}\int_0^\infty \frac{\sin(x) -\sin(3x)}{x} \mathrm{d}x\end{align}$$

The first integral is a well known value and the second is of the Frullani type and can immediately be seen to be zero.

Therefore,$$I = \dfrac{1}{4}\int_0^\infty \frac{2\sin(x)}{x} \mathrm{d}x = \frac{\pi}{4}$$

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For those interested, Frullanian integration arises with integrals of the form $$\int_0^\infty \frac{f(ax)-f(bx)}{x} \mathrm{d}x$$

For convergent integrals involving a function $f(x)$ with continuous derivative, we can write

$$\int_0^\infty \frac{f(ax)-f(bx)}{x} \mathrm{d}x = \lim_{x\to\infty} (f(x) - f(0))\log(a/b)$$

In an instance where the limit does not exist it is also possible to write

$$\int_0^\infty \frac{f(ax)-f(bx)}{x} \mathrm{d}x = \left(\lim_{x\to\infty} \frac{1}{x}\int_0^x f(x)\, \mathrm{d}x - \lim_{x\to 0}x\int_0^{x}f(x)\, \mathrm{d}x\right)\log(a/b)$$