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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?

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    Your second identity does not make any sense. Where does $b$ come from?2012-10-17
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    At some point, you must make estimates. For instance, $$\left| \frac{\sin x}{(1+x)^2}\right| \leq \frac{1}{(1+x)^2}.$$2012-10-17
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    @Mercy The missing limit symbol that I am about to edit in :)2012-10-17
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    See [this question and answer](http://math.stackexchange.com/questions/87960/is-int-0-infty-frac-cosxx1-dx-divergent?rq=1) for one of the integrals.2012-10-17
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    @PerManne Thanks :)2012-10-17

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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.

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    And what with $\int_0^{\frac{\pi}{2}} {\cos(x)\over 1+x}{\rm d}x$?2012-10-17
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    $\int_0^{\pi\over 2} {\cos(x)\over 1+x}{\rm d}x$ 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.2012-10-17
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    Yes... Thanks. +1.2012-10-17
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    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!2012-10-17
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    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...)2012-10-17
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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.

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    Of course, I should have seen this! Thanks!2012-10-17