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 ∫∞0cosx1+x dx and ∫∞0sinx(1+x)2 dx converges absolutely, and which one does not?
My Attempt:
For any b>0, and for all x∈[0,b], the following inequality holds:
|sinx(1+x)2|≤1(1+x)2,
which implies (by virtue of Theorem 6.12 (b) in Baby Rudin) that
∫b0|sinx(1+x)2| dx≤∫b01(1+x)2 dx=−11+b−(−11+0)=1−11+b;
moreover,
lim
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?
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