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Here are the links to three earlier posts of mine on Prob. 9, Chap. 6, in Baby Rudin here on Math SE.

Prob. 9, Chap. 6, in Baby Rudin: Integration by parts for improper integrals

Prob. 9, Chap. 6, in Baby Rudin: Integration by parts for improper integrals with one infinite limit

Prob. 9, Chap. 6 in Baby Rudin: Integration by parts for an improper integral

Now my question is the following.

Which one of the integrals $\int_0^\infty \frac{ \cos x }{ 1+x } \ \mathrm{d} x$ and $\int_0^\infty \frac{\sin x}{ (1+x)^2 } \ \mathrm{d} x$ converges absolutely, and which one does not?

My Attempt:

For any $b > 0$, and for all $x \in [0, b]$, the following inequality holds:
$$ \left\lvert \frac{ \sin x }{ (1+x)^2 } \right\rvert \leq \frac{1}{(1+x)^2}, $$ which implies (by virtue of Theorem 6.12 (b) in Baby Rudin) that $$ \int_0^b \left\lvert \frac{ \sin x }{ (1+x)^2 } \right\rvert \ \mathrm{d} x \leq \int_0^b \frac{1}{(1+x)^2} \ \mathrm{d} x = - \frac{1}{1+b} - \left( - \frac{1}{1+0} \right) = 1 - \frac{1}{1+b}; $$ moreover, $$ \lim_{b \to \infty} \left( 1 - \frac{1}{1+b} \right) = 1. $$ So we can conclude that $$ \int_0^\infty \left\lvert \frac{ \sin x }{ (1+x)^2 } \right\rvert \ \mathrm{d} x = \lim_{ b \to \infty} \int_0^b \left\lvert \frac{ \sin x }{ (1+x)^2 } \right\rvert \ \mathrm{d} x \leq 1. $$ That is, the improper integral $\int_0^\infty \left\lvert \frac{ \sin x }{ (1+x)^2 } \right\rvert \ \mathrm{d} x $ converges, which is the same as saying that the integral $\int_0^\infty \frac{ \sin x }{ (1+x)^2 } \ \mathrm{d} x $ converges absolutely.

Am I right?

If so, then, as suggested by Rudin, the other integral does not converge absolutely.

But how to show this directly?

Saaqib Mahmood
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2 Answers2

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Observe that on any interval of the form $[(k-1/3)\pi, (k+1/3)\pi]$ (where $k \in \mathbb Z$), we have $|\cos(x)| \geq 1/2$. Therefore the integral

$$\int_0^{\infty}\left|\frac{\cos(x)}{1+x}\right|\ dx$$ is at least as large as $$\sum_{k=1}^{\infty}\int_{(k-1/3)\pi}^{(k+1/3)\pi} \frac{1}{2(1+x)}\ dx$$ As the integrand is monotonically decreasing for positive $x$, it follows that on the interval $[(k-1/3)\pi, (k+1/3)\pi]$ (with $k > 0$) we have $$\frac{1}{2(1+x)} \geq \frac{1}{2(1 + (k + 1/3)\pi)}$$ and therefore $$\sum_{k=1}^{\infty}\int_{(k-1/3)\pi}^{(k+1/3)\pi} \frac{1}{2(1+x)}\ dx \geq \sum_{k=1}^{\infty} \frac{\pi}{3(1+(k+1/3)\pi)}$$ which diverges by limit comparison with $\sum 1/(3k)$.

  • I don't think that comparison test says that the last summation diverges, see this. Its the integral test, which is applicable. That is, $\frac{\pi}{3\left(1+\left(x+\frac{1}{3}\right)\pi\right)}$ never greater than $1/x$ for positive values. Thanks for this lovely answer by the way. +1 – Silent Dec 16 '18 at 13:30
  • @Silent Compare with $1/(3x)$, not $1/x$. –  Dec 16 '18 at 23:08
  • Sorry to bother you again, but even with this modification, inequality holds only for finitely many negative integers only. See this (scroll down). – Silent Dec 17 '18 at 05:43
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    @Silent Right, but asymptotically the difference is negligible: $\displaystyle \lim_{k \to \infty}\frac{\left(\frac{\pi}{3(1 + (k+1/3)\pi)}\right)}{\left(\frac{1}{3k}\right)} = 1$. So by the limit comparison test, either both series diverge or both series converge. Since we know $\sum \frac{1}{3k}$ diverges, so must the original series. You can compare with $1/(4k)$ if you want term-by-term inequality for all sufficiently large $k$. Similarly, $1/(ck)$ will work for any $c > 3$. –  Dec 17 '18 at 05:47
  • Edited the answer to clarify. –  Dec 17 '18 at 05:55
  • Thank you very much! This was very helpful. – Silent Dec 17 '18 at 06:10
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Your proof of convergence is correct.

To show that (a) does not converge, consider that $$ \int_0^\infty \left| \frac{\cos x}{1+x}\right| \geq \sum_{n=0}^\infty \int_{2n\pi}^{2n\pi+\pi/6} \left| \frac{\cos x}{1+x}\right| \geq \sum_{n=0}^\infty \int_{2n\pi}^{2n\pi+\pi/6} \left| \frac{1/2 }{1+x}\right| \\ \geq \frac{\pi}{12} \sum_{n=0}\frac{1}{1+x} $$ and the last sum diverges.

Mark Fischler
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