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I am trying to prove that the function $\frac{\sin}{\sqrt{x}}$ is not Lebesgue integrable on $[0, \infty]$. The proof I have seen seems to use a comparison test: \begin{align*} \int_0^{\infty} \left \lvert \frac{\sin x}{\sqrt{x}} \right \rvert = \sum\limits_{k = 0}^{\infty} \int_{k\pi}^{(k+1)\pi} \left \lvert \frac{\sin x}{\sqrt{x}} \right \rvert \ dx \geq \sum\limits_{k=0}^{\infty} \frac{2\pi}{\sqrt{k}}. \end{align*} I am having trouble justifying this equalities. The first step seems to involve defining $\sin x$ as a sequence of functions that is equal to $\frac{\sin x}{\sqrt{x}}$ for $x \in [k \pi, (k+1) \pi]$ and $0$ otherwise. I do not know how to justify the second equality. From there, the last series can either be directly compared to a harmonic series or we can draw on the $p$-series test to conclude that it does not diverge, and therefore our integral of $\frac{\sin x}{x}$ must diverge by the comparison test.

Any help on this would be greatly appreciated.

  • Yes, I did. Thank you. –  May 17 '19 at 03:22
  • you also need to put absolute value bars in the integrands. You are comparing above the absolute convergence of $\sin x/\sqrt x$, that is, the integrand is Lebesgue integrable if and only if $\int |\sin x|/|\sqrt x| dx<\infty$ – Masacroso May 17 '19 at 03:43
  • Thanks. I edited the question. I still am not sure how to justify these equalities, however. –  May 17 '19 at 03:48

2 Answers2

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Note that for $x \in [k\pi, (k+1)\pi]$, $\sin x$ has the same sign and by periodicity

$$\int_{k\pi}^{(k+1)\pi} |\sin x| \, dx = \int_0^\pi \sin x \, dx = 2$$

Thus,

$$\int_{k\pi}^{(k+1)\pi} \frac{|\sin x|}{\sqrt{x}} \, dx \geqslant \frac{1}{\sqrt{(k+1)\pi}}\int_{k\pi}^{(k+1)\pi} |\sin x| \, dx = \frac{2}{\sqrt{\pi}}\frac{1}{\sqrt{k+1}} $$

This corrects your inequality and is enough to show divergence of the integral as we have divergence of the series

$$\sum_{k=1}^\infty \frac{1}{\sqrt{k+1}} = + \infty$$

RRL
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  • Thank you for the answer. Could you explain how the inequality came about? It seems you replaced the denominator in the integrand with $\sqrt{(k+1) \pi}$. I am having a hard time understanding the effect this would have on the value of the integral. –  May 17 '19 at 03:59
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    For $k\pi \leqslant x \leqslant (k+1)\pi$ we have $\frac{|\sin x|}{\sqrt{x}} \geqslant \frac{|\sin x|}{\sqrt{(k+1)\pi}}$ which implies $\int \frac{|\sin x|}{\sqrt{x}} , dx \geqslant \frac{1}{\sqrt{(k+1)\pi}}\int |\sin x| , dx$ – RRL May 17 '19 at 04:01
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Simply evaluate $\int_{k\pi}^{(k+1)\pi}|\sin(x)|dx$ and use the fact that $1/\sqrt{x}\geq 1/\sqrt{k+1}$ on each interval

Alex R.
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