# Calculate the integral $\int_{0}^{\infty}|\frac{\sin{x}}{x} |dx$

I'm trying to solve this problem from Real Analysis of Folland but can't find any solution for it. Can anyone help me ?. Thanks so much.

$$\mbox{Show that}\quad \int_{0}^{\infty}\left\vert\,{\sin\left(x\right) \over x}\,\right\vert\,{\rm d}x =\infty$$

And also, can we calculate the similar integral $\int_{0}^{\infty}{\sin\left(x\right) \over x}\,{\rm d}x$ ?. Please help me clarify this. I really appreciate.

• I don't know about the main question, but the other integral $\int\limits_0^\infty \dfrac{\sin x}{x}dx$ that you mentioned can be solved using various methods and the value of that is $\dfrac{\pi}{2}$. See Sine Integral. – taninamdar Mar 8 '14 at 12:09
• compute it from $0$ to $\pi$. – OBDA Mar 8 '14 at 12:19

$$\int\limits_0^\infty \left|\frac{\sin x}{x} \right| \mathrm{d}x \\ =\sum\limits_{n = 0}^\infty \int\limits_{n\pi}^{(n+1)\pi} \left|\frac{\sin x}{x} \right| \mathrm{d}x \\ \geq \sum\limits_{n = 0}^\infty \int\limits_{n\pi}^{(n+1)\pi} \left|\frac{\sin x}{(n+1)\pi} \right| \mathrm{d}x \\ = \sum\limits_{n = 0}^\infty \frac{1}{(n+1)\pi}\int\limits_{n\pi}^{(n+1)\pi} \left|\sin x \right| \mathrm{d}x \\ = \sum\limits_{n = 0}^\infty \frac{2}{(n+1)\pi}\\ = \frac{2}{\pi}\sum\limits_{n = 0}^\infty \frac{1}{n+1}\\ = \frac{2}{\pi}\left(1+\frac{1}{2}+\frac{1}{3}+ \dots\right) = \infty$$
• Because $\frac{1}{(n+1)\pi}$ is independent of $x$. – Priyatham Mar 8 '14 at 12:29