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I know that the integral $\int_{-\infty}^{\infty} \frac{\sin(x)}{x}dx = \pi$.

Just for fun, I thought if it would be possible evaluate the same integral but only consider the area below the curve and above the x-axis? Approximately would be fine, if an exact expression is not available.

To make this precise one could define the integral as $$\int_{-\infty}^{\infty} \frac{\sin(x)}{x} \mu_0(\frac{\sin(x)}{x})dx $$

where $\mu_0(x)$ is the unit step function such that

$x<0 \implies \mu_0(x) = 0$

and $x \geq 0 \implies \mu_0(x) = 1$

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  • $\begingroup$ Alternatively, setting $f:=\frac{\sin x}{x}$ you could express it as $$\int_{-\infty}^{\infty}\frac{f(x)+|f(x)|}{2}\ dx,$$ and so the problem is the same as determining $\int_{-\infty}^{\infty}|f(x)|\ dx$. $\endgroup$
    – Servaes
    Commented Apr 17, 2018 at 21:13
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    $\begingroup$ And $$\int_{\mathbb{R}}\left|\frac{\sin x}{x}\right|\,dx $$ is blatantly divergent. $\endgroup$
    – Jack D'Aurizio
    Commented Apr 17, 2018 at 21:15
  • $\begingroup$ So the integral doesn't even converge? $\endgroup$
    – Adam
    Commented Apr 17, 2018 at 21:17
  • $\begingroup$ @Adam: so it seems. By integration by parts $$\int_{0}^{N}\left|\frac{\sin x}{x}\right|\,dx \sim\frac{2}{\pi}\log N.$$ $\endgroup$
    – Jack D'Aurizio
    Commented Apr 17, 2018 at 21:23

1 Answer 1

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The integral will diverge. Without doing the math in detail, each positive section between 2n*pi and (2n+1)*pi will integrate to k/n, for k=2 in the limit as n increases. So the integral on the positive side will be approximately the sum (n=1 to infinity) of 2/n . Which is divergent. (same on the -ve side).

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