Ramanujan would be proud of this integral

Ramanujan would be proud!!

We calculate the integral from 0 to infinity of 1/x^n + 1 using Ramanujan's master theorem. This involves by writing the function out as a Taylor or Maclaurin series and then converting it into the Gamma function, which is a generalization of the factorial function. We also use Euler's formula which relates the gamma function with sin and cos. This is a must see for anyone who appreciates Srinivasa Ramanujan's work and math and calculus in general

Note: Big thanks to Hari Kishan for recommending me the problem and its solution

[harikishan]

Such a lovely theorem, Thank you so much for featuring this !🎉🎉

[weiderchang]

I am not a math major but it seems there is a problem about the power expansion of 1/(1+u). Since 0<u<inf, the power expansion does not converge. The power expansion converges only if abs(u) < 1. Maybe a different substitution or additional substitution is required to make the power expansion work.

[battlecoder]

What a beautiful journey that was. Thanks for showing us this.

[afmikasenpai]

I like how pi just appears out of nowhere.

[CDChester]

the RMT will always be my favorite integral technique u ever showed me

[benniepieters]

The very first video I saw of you was you tackling this integral using complex analysis

[JayOnDaCob]

This actually has a really cool solution to me, but mainly cause I’ve been taking a linear systems analysis course. I believe the solution can be rewritten as 1/sinc(1/N)

[ajitandyokothakur]

Simply beautiful! Dr. Ajit Thakur (USA).

[romainmorleghem]

I notice something cool when N is large :) 1/1+x^n tends to a dirac delta and its integral is equal to 1 according to the formula ^^

[CynthiasDepravities]

And, here I am wondering how Ramanujan arrived at his master theorem.

[KobiFlax]

The expansion of 1/(1+x) is valid only for 0<=x<1....

[gosufd]

This integral is found in Euler's "Institutionum calculi integralis" vol1 part 1 section 1 chapter 8, 9.

[Ju_et_LG]

As a French guy, your french is very good. Thank you for this video

[mertaliyigit]

Ramanujan never fails to disappoint, edit: impress, i mean never fails to impress

[ChaineYTXF]

As a Frenchman, I have to say I'm cery impressed by your French!

[HAMIDSUHAIL-ri]

The series expansion for 1/(1+x) works only when
|x|<1 hiw can we use that on R^+

[benjaminbrat]

Amazing! Could you explain the logic behind this theorem? Is there an easily understandable proof?

[sciencelover-cj]

Can you solve this integral by using regular methods of integration, or another way more simple

[mauricebre]

I am not so confident with your calculation: You used the series expansion of 1/(1+x) for x in R but you do must more careful by this step. I found a much more acurate way by substituting: x^N/(x^N+1)=u. With this you find the Integral is equal with 1/N \int_0^1 (1-u)^{1-1/N-1}*u^{1/N-1} du. The Integral is precisly the Beta function with z1=1-1/N and z2=1/N if N>1. Then you get the

[FrankHarwald]

I wonder: can this approach be changed or generalized to compute certain special cases of elliptic integrals?