Відео

Thank you for the explanation.. but i have a question, isn't this integral an improper integral since the function is not defined when x=0 which is the lower bound?

Good topic. Always wanted to learn this.

There is a typo in the sum introduced at 10:52 until 12:09, it should be k instead of n as the summing variable.

The sums are from k equal one to infinity. Be careful with the index used in your sums.

One of my favorite topics. Fun fun fun. There's a good intro to the Mellin transform in "Integral Applications and Their Applications" by Davies , a book that I've used over the years in graduate courses at Georgia Tech. But of course, the really fun part is Ramanujan's Mater Theorem.

I apologise if I made any mistakes or misspoke at any time in this video. The content of this video was comparatively quite involved.

Hi, sum (-1)^(k+1)*k^2/(k^3+1), k=1 to inf

You always have interesting ideas for videos. If an idea for an interesting topic comes to me, I'll try to run it by you.

Use floor functions for those indexes in your binomial expansion at the start.

Your work is way too good. I’ve been studying special functions for a long time. Your derivations are clean, complete and exhaustive. You are a natural teacher explaining hard concepts. Keep working hard, you are going up.

Is a Trigonometric Proof Possible for the Theorem of Pythagoras? Michael de Villiers RUMEUS, University of Stellenbosch CONCLUDING COMMENTS To get back to the original question of whether a trigonometric proof for the theorem of Pythagoras is possible, the answer is unfortunately twofold: yes and no. 1) Yes, if we restrict the domain to positive acute angles, any valid similarity proof can be translated into a corresponding trigonometric one, or alternatively, we could use an approach like that of Zimba (2009) or Luzia (2015). 2) No, if we strictly adhere to the unit circle definitions of the trigonometric ratios as analytic functions, since that would lead to a circularity.

Use the reimann zeta function plug In 2 get the answer

Love this channel.

I love your videos, dude. You deserve more subs and views but I appreciate you not resorting to adopting a 'UA-cam personality' or trying to make click bait thumbnails for your videos (if those like sound like wild approaches to a math channel, well you'd be right, but that doesn't mean it's not ubiquitous). Keep up the great videos. What's next on the upload agenda, or is that a surprise?

Your explanation is very nice. At k=1 what happened for complete elliptic integrals. Whether it has logarithmic singularities at k=1? I

Bro where are you from?

Thank you, sir! 🎉 "Knowing your limits, you are limitless."

Well presented. Articulate ❤

Super

So elegant , so beautiful, just looking like a WOW! I have to say level of rigour in this lecture series is rather impressive and the clarity is top notch, thank you so much for sharing as this is rather a niche topic which is not covered in conventional textbooks!

This video is so great, i wonder why it only got a few visualizations... And also I'm FIRST to comment LESSSSS GOOOOOOOOOOO

Pronunciation: it is Integral, not Inteegral!

i know you found a way around it, but as a beginner in complex analysis i wonder, isn't the ln function defined on the complex plane except the negative real axis? Thanks anyways, well done.

Interesting! I'm sure it's not very significant, but it's amusing that the value of the integral for a, b > 0 is the product of the *logarithm of the Geometric Mean of a and b* and the *Logarithmic Mean of a and b* . Also, if a=b=c>0, then the maximum value for I occurs when c=e, with the integral of log(x) / (x+e)² from 0 to ∞ evaluating to e. If one of {a,b} is equal to 1, then the maximum value of I occurs when the other parameter equals exp(2 + W₀(- 2 e⁻²)) ≈ 4.9215536 where W₀ is the principal branch of the Lambert W aka "product log" function. Thanks for the video. I'm currently fascinated by functions involving products of logs and/or logs raised to powers. Complex analysis is so mysterious and somewhat unintuitive for me, but very powerful!

Nice 🔥

just quite tedious to calculate but it's pretty easy in itself, when you had the integral from zero to infinity of x^(k-1)/(1+x^2) you could have noticed that it's a quite known mellin transform, which instantly evaluates to pi/2*csc(kpi/2) nice problem

Thank you very very nice

impressive, and thanks for showing how the general case translates to the small angle approximation as a first order term

hi sir can you drive this functional equation using the hankel contour

32:53 "A real valued function" Isn't f(x) still complex valued even when you only consider real x? The original integrand for the complex analysis step is at (16:18): f(x) = [(x-i)ln(x+i) - (x+i)ln(x-i)]/[2i(x^2+1)] I don't think that that's real valued on x in [0, infinity). That's the only thing I'm hung up on. Otherwise, this is absolutely awesome!

Leaving this comment so I will remember to come back

Thanks for explaining the graphical part. I had doubt in that part, but now it is clear.

At 1:10 if you make the substitution x = pi/2 - x, then dx = -dx, but you didn't reflect that.

Loved the video. Keep up the good work and congrats on your graduation 🎉

Nice format

I am really unsure whether infinity in the integration limits can be replaced by n . The reason i am unsure is not all infinities are same. I mean i can write as lim m approaches inf , lim n approaches inf integral 0to m (1-t/n)^n * t^(x-1) dt but m can be replaced by n if and only if it is known that m-n or n-m is a finite quantity

nice tutorial, thank you sir

Excellent video!

This is actually really comprehensive. Thanks

If you replace the cosine by a sine in the same integral going from -pi to pi, you will get the same result as with the cosine. We see this using the change of variables x-->pi/2-x and the fact that we are integrating in a full circle (0 to 2pi).

What justifies being able to ignore Cn for n=0 simply because it diverges?

I like the title. The presentation is neat as well.

I am trying to learn the way you show us here... Can I say that point H is 2z - 2y times a rotation of 30 degrees times a magnitude correction so it is 2(z-y) time 3^0.5 / 2 times -1 times (3^0.5 /2 + 0.5 i ) ? or this is "only" CH vector ?

Thank you bro!!! Very excellent solve!