Which class of question is this?

ᴡᴏᴡ😊

You are wrong infinity minus infinity will be a bit tricky

Thank you Sir !

Thank you so much 🙏

Brilliant👍😎

keep it up

hi do ypu give classes or advice? how can i connect you? i have WhatsApp

Thanks man

thank you

nice

Liked and subscribed, hoping to see more videos on this topic.

Brother... with all respect I just want to say that I think you made some mistake... you took the limit function as f(t)=|t| and have a conclusion that it is not continuous because of the mod. But brother f(t)=|t|=t because 't' is in [0,1]. So in [0,1] f(t)=|t| is continuous. Hope you understand. Keep learning and Keep Growing.❤

Thanks for nice videos!

Nice video. Thanks!

দাদা সারা ইন্টারনেট খোঁজার পর আপনার ভিডিও তেই ans pelam 😃😃

Thanks for all. Flawless

😮wow

Nice sir thank u

👍 nice

Answer is [0 1)

Good presentation.

THANKS IT WAS VERY HELPFUL👍

thanks!

Please explain basic properties of zeta function based on the text Complex analysis by Elias M stein andRami shakarchi

awesome!

🔥🔥👍🏻

Thanks for the video

Elliptic integral bro pls

Bhai, that Euler's theorem not Fermat's Little theorem.

great explanation

Yrr hindi me bna le video

but at zeta at 1s the euler maschorini constant = lim n tends to infinity ( harmonic number - ln(n) )

Well explained. I come from Sweden and I am an complete amateur so this was good for me!

Nice explanation. 🔥

Happy to see that you r making vedio keep it up 👍🏻👍🏻👍🏻

Der Beweis durch Widerspruch führt bestimmt zum Ziel .

Where is 4?

Please take one example to the non trival zero sum.

(3n+1)--problem ={{{{{{{{ 3n+1]=[{n+3}= n^4 }}}}}} |||| integral of n+3= n+3 po versijos n+3=o n-7 +3=o-7 ](n-4)^0=0 =[ n=4] tada 1 kada n kada n=1 tada 4

Vai zeta function solve korchis ajkal? Bdw, nice video.

After 3 there is 4, not 5

There will be an improvised version on this topic :) very soon

Million dollar hypothesis Riemann Zeta function ζ(s) = 0 = 1

You should know that s there is a complex no😂😂😂

it's not easy to be a Ramanujan !

Anna please give more exercise on order of U(n). Like that, no.of.elements of order 20 in U(100)

Nice explanation Anna...