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

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?

• Hint: it's comparable to $\sum_{n=1}^\infty\frac{(-1)^n}{n}$. – Arthur Aug 4 '17 at 16:33

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 $1/x$. – Bungo 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 5:43
• @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 $c > 3$. – Bungo Dec 17 '18 at 5:47
• Edited the answer to clarify. – Bungo Dec 17 '18 at 5:55

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.