Prob. 9, Chap. 6, in Baby Rudin: Which one of these two improper integrals converges absolutely and which one does not?

Question

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?

Answer 1

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 $\sum 1/(3k)$.

Answer 2

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.