The discussion revolves around the convergence of the improper integral ##\displaystyle \int_0^{\infty} \frac{x \sin (ax)}{x^2 - b^2} dx##, specifically whether it converges in the normal sense or if the Cauchy Principal Value should be considered. Participants explore the implications of singularities in the integrand and the conditions under which they affect integrability.
Participants express differing views on the implications of singularities for integrability, with no consensus reached on whether singularities always indicate non-integrability or if limits can be evaluated to determine convergence.
Participants discuss various types of singularities and their effects on integrability, but the definitions and conditions under which singularities are considered integrable or non-integrable remain unresolved.
There are three regions for x where there could be a problem. Consider each separately.Mr Davis 97 said:How can I determine whether this improper integral converges in the normal sense
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?mathman said:Singularity (not integrable) at |x-b|=0. Need Principal value.
I believe mathman is saying it is a non-integrable singularity. It is equivalent to integrating 1/x through 0.Mr Davis 97 said:Why does there being a singularity mean that it is not integrable?
Would it only be an integrable singularity of it were a removable or jump discontinuity?haruspex said:I believe mathman is saying it is a non-integrable singularity. It is equivalent to integrating 1/x through 0.
No, you can integrate x-0.5 through 0. The limit as a tends to zero of ∫abx-0.5.dx exists.Mr Davis 97 said:Would it only be an integrable singularity of it were a removable or jump discontinuity?
So when exactly is a singularity non-integrable?haruspex said:No, you can integrate x-0.5 through 0. The limit as a tends to zero of ∫abx-0.5.dx exists.
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 said:So when exactly is a singularity non-integrable?
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 said: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.
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.Mr Davis 97 said: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?