Let's summarize what we've found with these p integrals.

If we integrate along a tail going to infinity,

then the p integral converges when p is strictly bigger than one.

It diverges when p is less than or equal to one.

For a blow up singularity at zero, these are reversed.

And we get a convergent integral when p is less than one, and

a divergent integral when p is bigger than or equal to one.

Note, that at the particular value of one, it's always divergent.

No matter whether you're going from zero to one or from one to infinity.

Now you don't need to remember the actual values of the convergent p integrals.

But you do need to remember this chart, this listing

of when the integral converges and when is diverges.

And the reason you need to remember this is because it will help

you determine convergence or divergence of other integrals.

Integrals whose anti-derivatives who may not be so easy to compute.

Let's look at an example.

Consider the integral of dx/(square root of x

squared + x) as x goes from zero to one.

This is a finite domain, however there is a singularity, or

a blow up, at x equals zero.

So how shall we proceed?

I don't know the anti-derivative to this.

It doesn't look like it's going to be terribly easy.

So, let us consider what the leading order behavior for this integrand is.

We're going to think in terms of Taylor series.

Now, this integrand doesn't have a well-defined Taylor series at x equals

zero, since the function is not even defined, and it blows up.

But notice that if we factor out a square root

of x from the denominator, then we're left with

1/(square root of x+1) as a factor.

Now let's rewrite that,

thinking that we are going to be looking at what happens as x is near zero.

I can write this as (x to the -1/2) times quantity (1+x) to the -1/2.

And this then is helpful, why?

Because the binomial series says that whenever you have (1+x) to the alpha,

as x is small, then this is of

the form one plus something in big O of x.

Now if we apply that to the (1+x) to the -1/2 term,

then we get the integral as x goes zero to one of x to the -1/2

times quantity 1 + something in O(x).

And we see that the leading order term

in this integrand is x to the -1/2.

So I'm going to split this up into two integrals.

The first is a p integral with p equals 1/2.

The second is an integral whose precise form I haven't written down,

but it's something that is in O(square root of x), and indeed, is bounded.

And I'm integrating it over the domain from zero to one.

The integral on the right definitely converges.

What about the integral on the left?

Well remember, I said you had to memorize some of these.

Let's go back and recall that when p equals one half,

for a blow up p integral, it converges.

Therefore, we know that this entire integral converges.

We may not know the value, but we know it converges.