Problem 1 is Exercise 7.7.64 from the book. It asks to determine the convergence of

There are several ways of approaching this problem.

Method 1: Begin by noting that

and so

can be evaluated by using the substitution so which transforms the integral into

that can be solved with the trig. substitution so and

Now we return to the problem:

We have found the value of the integral, so in particular, it converges.

Method 2: First,

We use comparison for both and

If Since

we have by comparison that is finite.

If Since

we have by comparison that is also finite.

Hence, converges.

Method 3: As before,

The change of variables transforms into

and

which of course is the same as

So all we need to do is to show that converges, and for this we can use comparison, because

and

Problem 2 is an easier version of Exercise 7.7.63 from the book. It asks to determine the convergence of

The problem here is that we cannot find so we are forced to use comparison. A natural thing to try would be to compare with

Since we have

The problem is that has an asymptote at 0. However, we can split the integral as

The first integral is finite, simply because it is the integral of a continuous function on some finite interval, there is nothing improper here. The second converges by comparison with the -integral with

It follows that the whole integral converges.

A slightly different approach is to use limit comparison rather than direct comparison:

It follows that converges because does; and we have to treat

as above. We cannot just compare with (which happens to diverge), because the limit comparison test requires that the functions that we compare are continuous, and is not continuous at 0.

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Marginalia to a theorem of Silver (see also this link) by Keith I. Devlin and R. B. Jensen, 1975. A humble title and yet, undoubtedly, one of the most important papers of all time in set theory.

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