Divergent integrals and the Cauchy principal value.

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[brandongroth]

The double example had me questioning reality for a second.

[sergeipetrov]

Thanks a million, Michael! I haven't known of such a definition of the Cauchy principal value before watching your video.

[TedHopp]

The alternative definition really should have epsilon approaching zero from above.

Also, for 1/x, once one shows that one limit diverges, it would be instructive to mention that the other one does as well, leaving the indeterminate form infinity minus infinity.

[Dan-cwxu]

I've never seen it explained this way. Thanks!

[abrahammekonnen]

6:00 I like how you point out the interplay between trying to create new definitions and realizing the properties of certain integrals.

[CloudSkywalker]

I think that setting one bound as -ε and the other as αε is wrong.
The way we say *ε approaches* <a number> makes some people think that the speed or pace at which it *approaches* matters. And it doesn't.
Taking a limit is not about what happens in our way towards a points but about what happens when we're almost at that point regardless of how fast or slow we get there.

By introducing the α parameter, one of the limits is corrupted. Suppose that a value close enough to our desired boundary exists such that we consider it a good-enough approximation. If we have ε approach that value in a way that we stop when either ε or αε reaches that value (whichever occurs first), one of the two will be a good-enough approximation while the other will not have reached that point. So, basically, you're reaching 0 for one integral and not really reaching it for the other.

Now, knowing that, separately, both integrals diverge (meaning there's infinite area no matter how close you start to the discontinuity); the fact that using the α you get a finite result means: first that together the integral doesn't diverge and second, that you got an arbitrary difference that reflects the arbitrarry offset one limit has with respect to the other. Of course, if you don't include the α offset, the result is specific.

This, together with the fact that we can evaluate this integral using geometry alone to unambiguously determine the result is 0, means that *the Cauchy P.V is also the actual value.*

[leokraushaar]

Really loving the daily uploads. Keep it up! Consistency is the key to growth and success.

[emma]

A nice followup to this would be looking at the logarithmic integral function. For x>1 it's defined as the Cauchy principal value of the integral of 1 / Log[t] dt from 0 to x. It'd be a really interesting video to fully examine it and see how Euler-Mascheroni constant gets involved with that function.

[pierrot]

It remembers my lecture of distribution (those invented by Laurent Schwartz).

[christoskettenis]

Didn't you do the first example twice?

[topologo]

Great video, but there is a minor pen slip around 13:49 where you lose the 3's in front of the radicals. Please, talk more about the meaning of Cauchy's PV in your next videos. Thanks!

[PunmasterSTP]

Cauchy principal value? More like "Completely awesome information that can astound you!" Thanks for continuing to make great videos.

[Pythagoriko]

The determination of the final function in alfa variable, can be taken as the definition of a new transform, like the Laplace Transform or Fourier Transform.

[agentm]

Thank you. This one makes a lot more sense than the one by blackpenredpen on the same concept.

[krisbrandenberger]

@ 9:17 There should be 3's in front of each of the epsilons.

[pianofortexx]

you should convergence tests for integrals

[the_nuwarrior]

CPV is used in electrodynamics to calculate the potencial of charges

[ImKinoNichtSabbeln]

OT: Cauchy's is one of the 72 names engraved on the Eiffel tower.

[sugarfrosted]

I'm wondering if you could have replaced the definition with "for all alpha >0" I suspect that you could break a limit with x and x^2 that worked linearly

[physicsatroeper]

One thing that is kind of important to understand about the P.V., I think, is that zero is the "natural" answer to this integral: it looks like you are integrating an odd function over an even domain, so your answer **SHOULD** be zero. The mathematically correct answer is that the integral doesn't exist, but if it DID exist, you would hope that it should vanish!

Furthermore: the fact that your answer depends on how you take your limit (alpha) makes it even more clear why the integral CANNOT exist: if it did, surely your answer should not depend on how you take your limit. This is even more obvious if you let the lower limit be something even weirder (like epsilon^2 or something). So that is why you cannot just say, "Never mind the official definition, let's just ALWAYS use the P.V. and there is no problem!"

These are issues I faced as a student learning this for the first time, so I hope these comments can help clarify the issues of why (1) the CORRECT answer is that the integral does not exist, and (2) the P.V. is a particularly useful and meaningful regulator of a divergent integral even though it seems rather arbitrary.