I have to decice if the following function is Lebesgue-integrable on $[0,1]$:
$$g(x)=\frac{1}x\cos\left(\frac{1}x\right) $$ where $x\in[0,1]$.
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Use the substitution $t = 1/x$ and follow the same steps as in this answer. – Ayman Hourieh Oct 17 '13 at 16:02
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But I can't use Riemann as in the example – Blanca Oct 17 '13 at 19:27
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You can use the monotone convergence theorem to show that the Lebesgue integral is the limit of the Riemann integral in the example. – Ayman Hourieh Oct 19 '13 at 07:51
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@AymanHourieh: The Riemann integral exists, because the improper Riemann integral is defined as $$ \lim_{\epsilon\to0+}\int_\epsilon^1\frac1x\cos\left(\frac1x\right),\mathrm{d}x $$ However, the Lebesgue integral requires that the integral of the absolute value exist. – robjohn Oct 20 '13 at 07:28
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@robjohn Yes. I'm aware of this. The answer I linked to considers the integral of the absolute value. – Ayman Hourieh Oct 20 '13 at 10:16
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@AymanHourieh: I was referring to the comment that "the Lebesgue integral is the limit of the Riemann integral in this example" which seems to say that the Lebesgue integral exists. I must be misunderstanding that comment. – robjohn Oct 20 '13 at 13:44
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@robjohn Perhaps my comment was a bit vague. The Lebesgue integral I was referring to was the integral of the absolute value of the function. Since the limit of the Riemann integral of the absolute value goes to $\infty$, the Lebesgue integral doesn't exist indeed. – Ayman Hourieh Oct 20 '13 at 14:16
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@AymanHourieh: okay, that makes sense. – robjohn Oct 20 '13 at 14:22
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$$ \int_{\large\frac1{n\pi}}^1\left|\frac1x\cos\left(\frac1x\right)\right|\,\mathrm{d}x =\int_1^{n\pi}\left|\frac1x\cos(x)\right|\,\mathrm{d}x $$ However, $\left[\frac\pi2,n\pi-\frac\pi2\right]\subset[1,n\pi]$ and $$ \begin{align} \int_{\pi/2}^{n\pi-\pi/2}\left|\frac1x\cos(x)\right|\,\mathrm{d}x &=\sum_{k=1}^{n-1}\int_{k\pi-\pi/2}^{k\pi+\pi/2}\left|\frac1x\cos(x)\right|\,\mathrm{d}x\\ &\ge\sum_{k=1}^{n-1}\frac2{(2k+1)\pi}\left|\int_{k\pi-\pi/2}^{k\pi+\pi/2}\cos(x)\,\mathrm{d}x\right|\\ &=\sum_{k=1}^{n-1}\frac4{(2k+1)\pi} \end{align} $$ and since the Harmonic Series diverges, $\frac1x\cos\left(\frac1x\right)$ is not Lebesgue integrable on $[0,1]$.
robjohn
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I don't understand why do you chose that values for the interval of the integral. Could you explain it please? – Blanca Oct 21 '13 at 20:36
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@Blanca: I have added some to my answer that may help to explain. If not, let me know. – robjohn Oct 21 '13 at 21:24