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It’s really nice to hear comments like this from my viewers. Thank you so much.

It’s definitely an interesting topic. I will try to get to it in the future. Thanks for your interest in the channel.

@FireSermon83 thanksss

@@joeeeee8738 Yes! Please see the video on my channel titled “Euler MacLaurin Summation Formula” where I derive them. Also see my video “Bernoulli Exponential Generating Function” where I give another definition. Very important subject and not covered in general math curriculum for some reason! Thanks for watching!

@@ronitmehta1464 Thanks! I’m glad you enjoyed the video!

@@randomacc246 Awesome! Glad you enjoyed it! Thanks.

Thanks so much for the positive feedback. If you enjoy these videos please tell your friends!

@@boody6380 I’m glad you found it helpful!

@@howdadogdoin729 Thanks for your positive feedback!

@@Ourhealingchannel I am happy to be proven wrong here, but my understanding is that the infinite product definition of the Gamma function converges for all complex z except z = 0, -1, -2, -3 etc.

@@jeromejean-charles6163 Thanks for the positive feedback. I always try to motivate my videos with some question- I do think that the great mathematicians who made all these discoveries were trying to answer specific questions. It helps us understand how these great mathematical discoveries came about. Thanks again.

@@knivesoutcatchdamouse2137 Thanks for the positive feedback. I’m glad you are enjoying them!

@@souldrip2000 Thanks. I take Khan Academy videos as my inspiration. I’m glad you find the format helpful!

@@boriskogan666 Thanks for your comments. I do worry that some of the voice-over is repetitive. I’ll take that into account going forward. Glad you enjoyed the videos!

@@Calcprof Thanks. I’ll be doing a video about that soon!

Thanks! I’m glad you enjoyed it!

@@nin10dorox I’m glad you enjoyed the video. Here are a couple sources that you might find helpful if you want to see Euler’s original argument: arxiv.org/pdf/0806.4096 And: www.ms.uky.edu/~sohum/ma330/files/euler_zeta_ayoub.pdf

@@eiseks3410 Thanks for your question. You are absolutely correct in observing that the Bernoulli numbers are very closely related to values of the zeta function. I have a video coming up about this. Stay tuned!

Thank you for the positive feedback! I’m glad you enjoyed the video. This is some pretty advanced subject matter and sometimes the sources leave out important steps or ideas. I am an amateur trying to master this material myself, so in the process of learning this math and turning it into presentations, I try to make everything in my videos as clear as possible. Thanks again!

Thanks for your feedback. I think you are correct that in the voice narration for the video I am stating the bounds of integration opposite to the way they are usually read out loud. I do not mean to confuse anyone here and may change this when I have time. Hopefully the written portion of the video is very clear (always my goal), as there are moments in this video series where I show that reversing the bounds of integration changes the sign. My habit of reading the integral out loud opposite to convention probably stems from my thinking about the integral in terms of its evaluation by the Fundamental Theorem of Calculus- we take the antiderivative of the integrand first at the upper bound then subtract the value of the antiderivative at the lower bound- hence “from b to a”. But again, I do not mean to assert that this is how anybody should say it. I hope you are aware that I am a math amateur. Studying and teaching math is not my profession. I am not part of the professional mathematical community, and have not been in a math classroom for many years, so there are probably several things I say and write in these videos that reflect my own idiosyncrasies, or worse, misunderstandings. I hope viewers like yourself will continue to take the time to comment and offer corrections or improvements to the presentations. Thanks again for the helpful feedback. I’m glad you are still enjoying the videos!

Thanks for your feedback. I agree this topic doesn’t seem to be addressed much outside of specialized sources. But it’s an interesting and important result! I see your point about the plus/minus signs. In future videos I will try to make this more clear, especially if it is a discussion of infinite series or products. Thanks again!

Thanks! I’m glad you found it helpful.

Thanks for your supportive feedback! I’m glad you enjoyed the video.

Thanks so much!

Thanks for your question. The orientation you take around the branch cut affects the sign of the integral, but does not determine whether the branch point is included or excluded. For the branch point to be included in the sense I think you mean it, the contour would have to literally go through it, or be closed around it. The reason on the limit contour we have to make z>0 is just that we are integrating a function whose leading term is r^(z-1) over a circular contour, and importantly, taking r to zero. The function diverges if z<0 because 1/r^z diverges as r goes to 0. That would be true on a circular contour anywhere in the complex plane, not just one circling a branch point. We avoid this issue with the Hankel contour however because there r is fixed. The whole beauty of the Hankel contour is that it is open- so it has no “inside” or “outside.” It skirts around the branch point and branch cut without ever touching these points, and remains open so that it does not actually enclose them. Later in the video, when we actually do close the contour (adding the pieces C1 and C2) to show that the integral along Hankel Contour and limit contour are equal, we close the contour in such a way that the branch point and branch cut lie outside the contour. Our closed contour is shaped a bit like pac-man- with the branch point and cut sitting in Pac-man’s mouth. I hope this explanation helps.

@@FireSermon83 Yes thank you very much. I think the point I was missing is that for the branch point to be INcluded, the contour around it would need to be closed, which of course it is not. One follow-up question: I hope you do not mind: Is it then irrelevant what the direction of travel along the open Hankel contour is? I have always wondered why the contour is specified as +inf to +inf with a CCW open circle round the branch point. Could we also do +inf to +inf with a clockwise open circle?

Yes. We could perform the integration in the opposite orientation- starting below the real line at + infinity, traveling toward the origin, circling the origin clockwise, then traveling out towards + infinity again above the real line. That would just change the sign of the integral without changing its absolute value.

@@FireSermon83 Perfect. Many thanks for your help.

Thanks so much! I’m glad you enjoyed the video!

You’re welcome! I’m glad you continue to find these videos worthwhile.

@@FireSermon83 They are more than worthwhile. I really appreciate the depth and care you put into them.

That’s a good question. I believe that the fact that the absolute value of the integral converges for Real(z) > 0 is sufficient to prove that the complex Gamma function also converges in this region. But I am not a math professor- so someone with a more detailed knowledge of complex analysis might want to weigh in here.

Thanks for your helpful feedback. I will try to answer in a general way that I hope will be helpful. L’Hopital’s rule tells how to evaluate limits when we have an expression that equals either “0/0” or “infinity/infinity.” The rule says that to find the limit, we just have to differentiate the numerator and denominator of the expression repeatedly until we can differentiate no further, and then take the limit. So here we have (1-x^0)/0. As long as x itself does not equal zero, which we will have to assume here, x^0 always equals 1. So (1-x^0)/0 = 0/0. That expression is undefined unless we treat it as a limit to be evaluated! The way to make sense of this expression then is to rewrite it as (1-x^z)/z and take the limit as z goes to zero and use L’Hopital’s rule to evaluate.

Sorry- was there anything in particular in the video you found unclear? While I am trying to make these videos at a level accessible to as many people as possible, I am generally assuming some experience with calculus. I have videos on my channel about complex numbers and Euler’s identity, which are helpful background for this video.

​@@FireSermon83 I think the video was perfectly clear and explained the steps well

Hi- Thanks for your comment. I am glad you enjoyed the video. I agree that it is a fascinating topic, which is why I put so much energy into making the videos clear and accessible to as many viewers as possible. Also as I mention in the heading to my channel page, I am just an amateur, not a professional mathematician, so I am also learning. I welcome any feedback or corrections my viewers have to offer. To your question: You are absolutely right- every time we perform the integration by parts, we reach a new formula for the Gamma function that is convergent for a slightly larger domain, and we find that this is so because we have performed mathematical operations on the original function (taking integrals, derivatives and evaluating limits) that allow us to discover each time a new, more general formula for the Gamma Function. To take for example the continuation of Gamma from the reals > 0 to reals > -1, yes, the new formula has to converge for reals greater than zero, but in this new form, it will also converge for reals greater than -1! Moreover, the new formula computes all the same values for Gamma as it was originally defined on reals greater than 0. So each step is essentially a “continuation” of the gamma function from its original domain to the new, larger domain. It is not contradicting the original definition, just supplying new values where the function can be computed. Also - and I don’t go into this in my video- it turns out that analytic continuation is unique! We are not in any way “making up” a new definition for the Gamma function. No other continuation of the function from its original domain to this new domain is valid! It’s a fascinating topic. Thanks again for your comment and I hope this response is helpful!

Thanks for your comments. I’m glad you enjoyed the video. I am working on a video for the Hankel contour right now. It should be ready later this month. Thanks for watching and stay tuned!

@@FireSermon83 Thank you for your response! It will be interesting to see your thoughts on Riemann’s ingenious derivation. Looking forward to it!

Thank you for the positive comments! I'm glad you enjoyed this video.