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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.2012-11-19

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$\def\limn{\lim_{n\to\infty}}$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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    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.2012-11-19