I Find Cauchy Principal Value of Improper Integral

[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?

 

[haruspex]

How can I determine whether this improper integral converges in the normal sense

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

 

[mathman]

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

 

[Mr Davis 97]

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

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?

 

[haruspex]

Why does there being a singularity mean that it is not integrable?

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

 

[Mr Davis 97]

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

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

 

[haruspex]

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

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

 

[Mr Davis 97]

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

So when exactly is a singularity non-integrable?

 

[haruspex]

So when exactly is a singularity non-integrable?

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.

 

[Mr Davis 97]

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.

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?

 

[haruspex]

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?

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.