Problem 6.9 of Rudin's PMA asks the reader to demonstrate conditions in which indefinite integrals that satisfy the definition $$\int_a^\infty f(x)dx = \lim_{b\to\infty} \int_a^b f(x)dx$$ can sometimes be integrated by parts. Defining $F(x) = \int_a^x f(x)dx$, I have demonstrated that the integral $$\int_a^\infty F(x)g(x) + f(x)G(x) dx = \lim_{b\to\infty}F(b)G(b)-F(a)G(a) = \lim_{b\to\infty}F(b)G(b)$$ converges if $\lim_{b\to\infty} F(b)G(b)$ exists and is finite.

The problem then asks the reader to demonstrate that these conditions are sufficient to show that $$\int_0^\infty \frac{\cos x}{1+x} dx = \int_0^\infty \frac{\sin x}{(1+x)^2}dx$$ and that one of the above integrals converges, while the other converges absolutely.

I have demonstrated the equality of the two integrals, as I have demonstrated that $$\int_0^\infty \frac{\cos x}{1+x} - \frac{\sin x}{(1+x)^2}dx = 0$$ and so converges.

However, I am stuck proving convergence of either integral -- using integration by parts brings me back to where I began. Furthermore, I am at a loss for showing absolute convergence (or the failure thereof) of either integral.

Is it true to say that if $\int_a^\infty H(x)+K(x)dx$ converges, then both $\int_a^\infty H(x)dx$ and $\int_a^\infty K(x)dx$ converge? I can see that if one does, then the other must, but must both converge? Finally, how can I demonstrate absolute convergence other than computing the integral directly, which may not be possible?

  • $\begingroup$ Your second identity does not make any sense. Where does $b$ come from? $\endgroup$ – Mercy King Oct 17 '12 at 13:22
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    $\begingroup$ At some point, you must make estimates. For instance, $$\left| \frac{\sin x}{(1+x)^2}\right| \leq \frac{1}{(1+x)^2}.$$ $\endgroup$ – Siminore Oct 17 '12 at 13:23
  • $\begingroup$ @Mercy The missing limit symbol that I am about to edit in :) $\endgroup$ – Emily Oct 17 '12 at 13:24
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    $\begingroup$ See this question and answer for one of the integrals. $\endgroup$ – Per Manne Oct 17 '12 at 13:25
  • $\begingroup$ @PerManne Thanks :) $\endgroup$ – Emily Oct 17 '12 at 13:38

Another proof to show that $\int_0^{\infty} {\cos(x)\over 1+x}{\rm d}x$ converges:

Let $$A_n =\int_{(n-{1\over 2})\pi}^{(n+{1\over 2})\pi}{\cos(x)\over 1+x}{\rm d}x, \qquad n=1,2,\dots$$ Then $$\int_{\pi \over 2}^{\infty} {\cos(x)\over 1+x}{\rm d}x=\sum_{n=1}^\infty A_n$$ and the latter series converges by the alternating series test.

  • $\begingroup$ And what with $\int_0^{\frac{\pi}{2}} {\cos(x)\over 1+x}{\rm d}x$? $\endgroup$ – Salech Rubenstein Oct 17 '12 at 14:10
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    $\begingroup$ $\int_0^{\pi\over 2} {\cos(x)\over 1+x}{\rm d}x$ is of course a finite number, but without computation it is not completely obvious if it is larger or smaller than $A_1$ in absolute value. To be sure that the terms in $\sum A_n$ were mononically decreasing in absolute value, I omitted this first term. $\endgroup$ – Per Manne Oct 17 '12 at 14:25
  • $\begingroup$ Yes... Thanks. +1. $\endgroup$ – Salech Rubenstein Oct 17 '12 at 14:27
  • $\begingroup$ I absolutely understand this proof of convergence; it makes a lot of sense and is great. However, it's a little too great and I don't want to feel like I am copying it verbatim. I've found a nice happy medium between your approach and @BenjaLim's approach; I've used a clever change of variables to show convergence using the integral test. I'm accepting BenjaLim's answer since it is more of a hint and leaves enough room for me to leave my own mark, but your answer was immensely helpful as well. Thanks! $\endgroup$ – Emily Oct 17 '12 at 18:12
  • $\begingroup$ You're welcome! The proof should really be followed by a drawing; then it would almost not be necessary to write any formulas at all. (I have to learn how to do this...) $\endgroup$ – Per Manne Oct 17 '12 at 19:57

To show that $\int_0^\infty \frac{\cos x}{1+x} dx$ converges you make a change of variables to get the integral

$$\begin{eqnarray*} \int_1^\infty \frac{\cos (u-1)}{u} du &=& M\int_1^\infty \frac{\cos u}{u} du + M'\int_1^\infty \frac{\sin u}{u} du\end{eqnarray*}$$

for constants $M$ and $M'$. Now integrate by parts on these individual integrals to show that they each converge. We can also see that https://in.answers.yahoo.com/question/index?qid=20071231110659AAUZ2hY gives the integral values for sin(x)/x integration.

  • $\begingroup$ Of course, I should have seen this! Thanks! $\endgroup$ – Emily Oct 17 '12 at 13:37

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