Double integral of absolute value of $\frac{\sin(x)}{x}$

Question

Can someone help me with the following question: $$\iint_{R} \left| \frac{\sin(x)}{x} \right| dA$$ where R is the rectangle formed when $x$ goes from 0 to $\infty$ and $y$ goes from 0 to 1?

I don't actually know to begin this one, because the absolute value makes things slightly strange. Wolfram Alpha can't seem to compute it either.

Answer 1

A slight variant on @YvesDaoust's answer, bounding the $\frac1x$ instead of the sine, is to note that$$\int_0^\infty\left|\frac{\sin x}{x}\right|dx=\sum_{k\ge1}\int_{(k-1)\pi}^{k\pi}\left|\frac{\sin x}{x}\right|dx\ge\sum_{k\ge1}\frac{1}{k\pi}\int_{(k-1)\pi}^{k\pi}\left|\sin x\right|dx=\sum_{k\ge1}\frac{2}{k\pi},$$so$$\int_{-\infty}^\infty\left|\frac{\sin x}{x}\right|dx=2\int_0^\infty\left|\frac{\sin x}{x}\right|dx\ge\frac{4}{\pi}\sum_{k\ge1}\frac1k,$$a lower bound proportional to the famously divergent harmonic series.

Answer 2

The integral of $$\left|\frac{\sin x}x\right|$$ diverges because you can find an interval of finite length in every half-period of the sine such that the sine exceeds, say $0.5$. Then the integral can be bounded below by a multiple of the harmonic series.