# Showing that a limit exists and showing $f$ is not integrable.

Show that $\lim\limits_{n \to \infty} \int_{1}^{n} f$ exists while $f$ is not integrable over $[1,\infty)$.

Define $f(x)=\frac{\sin(x)}{x}$ for $1 \leq x <\infty$.

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:See my answer and do not worry about the downvote. –  Mhenni Benghorbal Nov 19 '12 at 7:26

$\def\limn{\lim_{n\to\infty}}$We have by partial integration \begin{align*} \limn\int_1^n \frac{\sin x}x \,dx &= \limn \left(\left.-\frac{\cos x}x\right|_1^n - \int_1^n \frac{\cos x}{x^2}\, dx\right)\\ &= \cos 1 - \int_1^\infty \frac{\cos x}{x^2}\, dx \end{align*} where the latter integral converges as one can see by comparision with $\int_1^\infty x^{-2}\, dx$.

For the non-integrability, note that one has $$\int_{n\pi}^{(n+1)\pi} \frac{|\sin x|}x \, dx\ge \frac 1{(n+1)\pi}\int_0^{\pi}|\sin x|\, dx = \frac 2{(n+1)\pi}$$ and hence $$\int_1^\infty \frac{|\sin x|}x\, dx \ge \sum_{n=1}^\infty \frac 2{(n+1)\pi} = \infty.$$

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Recalling the Sine integral

$$Si(x)=\int_{0}^{x} \frac{\sin(x)}{x}dx ,$$

then we can write our integral as

$$\lim_{n \to \infty} Si(n) = \frac{\pi}{2} .$$

See the link below to see that the integral converges. Off course, there are several techniques to evaluate the above integral. On the other hand, it is a well known fact that the integral

$$\int_{0}^{\infty} \left|\frac{\sin(x)}{x}\right| dx$$ does not converge. Then by the theorem "A function $f$ is Lebesgue integrable if and only if $|f|$ is.", one can show directly that $\frac{\sin(x)}{x}$ is not Lebesgue integrable.

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Presumably emka is asking about how to prove both of these "well known" results. –  Antonio Vargas Nov 19 '12 at 6:47
@AntonioVargas: There is no reason to downvote this answer. –  Mhenni Benghorbal Nov 19 '12 at 6:52
I disagree. When you originally posted the answer you simply transformed emka's question into a statement, replacing "show that" with "there are several techniques to evaluate..." and "it is a well known fact that...". However, the answer now links to a source with some actual content, so I will remove my downvote. –  Antonio Vargas Nov 19 '12 at 12:22