Question

Real Analysis:
When is \( \int_a^\infty \) said to converge absolutely? Show that \( \int_1^\infty {\sin (x) \over x^2} dx \) converges absolutely.

Assume that:
(i) \( f \) is integrated over \( [a,t] \) for all \( t \geq a\) and there exists \( M > 0 \) such that for all \( t \geq a \). Show that
\[ \large \left | \int_a^t f(x) dx\right | \leq M \]
ii) \( g(x) \) is monotonically decreasing to 0 as \( x \rightarrow \infty \). Prove that \( \int_a^\infty f(x) g(x) dx \) exists.

Answer 1

For the first part, absolute convergence of an integral implies the integral of the absolute value of the function you're integrating on is finite. That's probably a poor explanation. Wikipedia does a better job:

http://en.wikipedia.org/wiki/Absolute_convergence

I would assume you can do the integral yourself. It follows straight from the definition.

For (i), I'm not entirely sure how to approach this as there's a multitude of proofs for it depending on how you defined integration. Riemann integration is going to force you to first bound f, which you can do since it is integrable. Then, using theorems you've probably been given, you can show that the absolute value of the function is bounded and is less than or equal to the integral of the absolute value. There's a full proof in Steven Lay's "Analysis (with an Introduction to Proof)" in Chapter 30.

If you've defined it by Lebesgue integration, then this is a consequence of Lebesgue Dominated Convergence Theorem. I would assume this isn't what you're doing though.

(ii) I'm fairly sure you have to assume f(x) is integrable. From there, use the Cauchy-Schwartz inequality. After that, you can use the fact that f is bounded (from part (i)) and g is integrable (a theorem you should have), which gives you that the product is integrable.

Anyway, sorry if this is a bit vague. I don't have any of your theorems and there's multiple ways to do these.

Answer 2

it's definitely Riemann integration. I don't get the second and the third problem.