# I Cauchy Principal Value

1. Mar 23, 2017

### Mr Davis 97

Say I am given the integral $\displaystyle \int_0^{\infty} \frac{x \sin (ax)}{x^2 - b^2} dx$. How can I determine whether this improper integral converges in the normal sense, or whether I should just look for the Cauchy Principal Value?

2. Mar 24, 2017

### haruspex

There are three regions for x where there could be a problem. Consider each separately.

3. Mar 24, 2017

### mathman

Singularity (not integrable) at |x-b|=0. Need Principal value.

4. Mar 24, 2017

### Mr Davis 97

Why does there being a singularity mean that it is not integrable? Couldn't we try to evaluate it as any other improper integral, using limits on the integral as x approaches b and seeing if it converges?

5. Mar 24, 2017

### haruspex

I believe mathman is saying it is a non-integrable singularity. It is equivalent to integrating 1/x through 0.

6. Mar 24, 2017

### Mr Davis 97

Would it only be an integrable singularity of it were a removable or jump discontinuity?

7. Mar 24, 2017

### haruspex

No, you can integrate x-0.5 through 0. The limit as a tends to zero of ∫abx-0.5.dx exists.

8. Mar 24, 2017

### Mr Davis 97

So when exactly is a singularity non-integrable?

9. Mar 24, 2017

### haruspex

When the limit, as a bound approaches the singularity, does not exist. That's when, in order to integrate through the singularity, you have to play questionable games letting the negative infinities on one side cancel the positive infinities on the other.

10. Mar 24, 2017

### Mr Davis 97

Can I tell just from looking at the integral whether the limit, as a bound approaches a singularity, does or does not exist? Or do I explicit need evaluate the integral to the point where I can show that the limit does not exist?

11. Mar 24, 2017

### haruspex

You can usually spot the asymptotic behaviour, rather than integrating exactly. For (x2-b2)-1, you can factor out the x+b and observe that it will be roughly 2b in the neighbourhood of x=b. Therefore it converges if and only the integral of (x-b)-1 does.