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I am trying to compute the following limit, $$ \lim_{k \to \infty} \int_0^1 \left| \frac{\sin(kx)}{x}\right| dx $$ After some numerical calculations, it loos like the limit is $\infty$. To prove it, I tried to use Riemann sums but such approach is not working.

Any hints on how one should prove this ?

Thanks in advance.

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    $\begingroup$ Note that this integral approximates $\int_0^1{\frac{1}{x}} dx$ $\endgroup$
    – John Lou
    May 2 '17 at 4:06
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Changing variables ($y=kx$) this is $$\lim_{k\to\infty}\int_0^k\frac{|\sin y|}{y}\,dy =\int_0^\infty\frac{|\sin y|}{y}\,dy.$$ This improper integral is well-known to diverge. For instance $$\int_{(n-1)\pi}^{n\pi}\frac{|\sin y|}{y}\,dy \ge\frac{1}{n\pi}\int_0^\pi\sin y\,dy=\frac{2}{n\pi}.$$ Adding these up, and considering the behaviour of the harmonic series shows the integral diverges.

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  • $\begingroup$ How do you get the inequality ? $\endgroup$
    – Leo Sera
    May 2 '17 at 5:15
  • $\begingroup$ $1/y\ge1/(n\pi)$ on that interval. $\endgroup$ May 2 '17 at 5:18
  • $\begingroup$ Right ! Thanks again ! $\endgroup$
    – Leo Sera
    May 2 '17 at 5:23

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