Gaussian Integral 12 Ramanujan Way

[@leonardromano1491]

Isn't the Ramanujan Formula kinda redundand as you LITERALLY get the definition of the Gamma function from this substitution... However I saw the formula for the first time and already like it!

[@Debg91]

Great series, but I'm still not convinced the Gaussian integral is √π… Just kidding! This is what I like the most about math: you can arrive at the same result from many different ways, which shows how beautifully consistent it is! I'd say my favourite way is 11: the complex way. But this one is also awesome. Ramanujan is my favourite mathematician in terms of originality. I'm hopping there will be more series like this. My suggestion: different regularization methods for divergent series!

[@magnetonerd4553]

This is probably the most beautiful method to show this identity that I have seen. Thank you for the demonstration.

[@irishkqings]

After watching this full series I realized that your Full Lecture presents, Math discussion 90% + 10% Comedy act😅= 100% That's why I never get bored while watching. I wish you make more videos again. ❤❤❤

[@bouteilledargile]

~series ends~ top 10 saddest anime moments

[@abdulalhazred5924]

definitely needs a second season

[@eliasarguello9961]

So cool! I absolutely loved those series! My favorite method had to be the complex analysis one (because complex analysis is truly the best) but this one also has to be one of my favorites! What a great job ❤️

[@AKSHATAHUJA]

Proud to be indian And Salute to Sir Ramanujan

[@jesusalej1]

India and its neighbours... the best education in the world.

[@jesusalej1]

@@aashsyed1277 some people doesnt believe in god but arabic i give you that my friend.

[@БогданЛисянський]

Thank you Dr. Peyam for all your great videos. I'm going to write math olympiad tomorrow and I know that things, I learnt from your videos will really help me.

[@soutriksarangi5580]

Богдан Лисянский which olympiad are u gonna write?....just curious cuz I am also gonna write Team Selection Test for IMO in a few days

[@БогданЛисянський]

This is olympiad for getting a little bit more points on my entrance exams to college. P.S. I decided to study math in college because of you and bprb

[@rome8726]

I love Ramanujan.

[@sharathgowtham2157]

Ramazhunuzhan

[@frozenmoon998]

Finally! The Ramanujan way

[@Czeckie]

nice cliffhanger for the second season, where we gonna explore Mellin transform and other integral transforms

[@neilgerace355]

Congratulations on getting to the end yourself :) this was very informative and even fun along the way

[@SmileyHuN]

Mellin transform is so badass

[@thomasfritz8174]

Well, \phi might be one for integer arguments, but how do we know that it is also one for all other values as $s$ and 1/2?

[@yrcmurthy8323]

He is the next big mathematician, his favourite is Gaussian Integral. His future name "Dr. Pegauss"

[@yrcmurthy8323]

Thanks for heart sir

[@jesusalej1]

Full genius. Greetings from Argentina.

[@jacoboribilik3253]

At school we approximate the gaussian integral with a triangle. Our predictions are terrible but at least probabilities lie within 0 and 1.

[@MilanStojanovic9]

i dont know why but i never imagined ramanujan doing calculus. it seems to mechaniclak for his imagination and creativity

[@jayamitra4656]

Damn... No more Gaussian integrals.... What are we supposed to do with our lives now? 😂🤣😂🤣

[@neilgerace355]

Linear algebra!

[@Gamma_Digamma]

Topology

[@jesusalej1]

Musician? Perhaps...

[@PeterBarnes2]

I didn't know about the Mellin transform, before. I tried Ramanujan's Master theorem to get an alternative definition of the 'Taylor Transform' I made up before, but that didn't seem to work when I applied it to e^(x^p) . On slightly different work, I think that you could define a function as a series of exponentials of monomials, though I don't know if it's completely correct: f(x) = {sum n=0 to inf. of} F[ Γ(z+1) * T[f](z) ](n) * e^x^n where F[g](z) is the Fourier transform of 'g,' and T[g](z) = (D^z)[g](0) / Γ(z+1) (D^z) is the complex fractional-derivative operator for the zth derivative. (In this case, defining T with 1/Γ(z+1) is redundant, but I use the same definition for other work where it occasionally isn't redundant.) It might be the case that the correct inverse transform here is the improper integral from -inf. to inf. instead, but that can't be known without finding the Taylor Transform of e^x^p for all p and z. I do know that something like this is possible, because you can trivially define the Taylor Transform of e^x^n for z on the integers, and it's periodic to n.

[@fgdgjgjhc]

I honestly don't get this. It all seems to depend heavily on your choice of Phi, and I cannot see, why this is the only valid choice. If I instead choose Phi(n)=cos(2 pi n), then I still get the series expansion for e^(-x), because this Phi is 1 if n is a whole number. However, then for the Mellin transform, I get Gamma(s)cos(-2 pi s). If I plug in 1/2 into this, I get zero. So why does your choice of Phi give the correct result and mine doesn't? Is this similar trickery to the sum of all numbers being equal to -1/12?

[@CousinoMacul]

Thank you for this series.

[@yaaryany]

👍👍👍 Keep up the good work!

[@aliali.m4841]

great!!! Dr peyam

[@was2zur-holle]

He peyam ,Plz give me the link which you solved this integral I want it integration of ln(ln(1/x))/(1-x+x²)

[@mounirhayani3059]

It's amazing that you spent 6 minutes to show that Gamma (1/2) is the integral of x^(-1/2)exp(-x) (which is the definition of Gamma) 😂😂

[@cbbuntz]

Oh... wait. That's what the Mellin transform does?!?! Okay, that is useful. I always suspected there was some kind of relationship between the gaussian integral and the fact that gamma(1/2) is sqrt(pi).

[@felicsmoses1771]

I am proud that I live in the district (City) adjacent to Ramanujan's native district (City) just 32 miles (50 km) away

[@yeahyeah54]

Where is the dx on the integral???

[@shanmugasundaram9688]

While applying Ramanujan's formula you have left unexplained the term pi(-s) in the end.Is it Euler's pi function?

[@Pterry23real]

So now the general form ;) It looks like that the integral from 0 to infinity of a^(-x²) is 1/2 sqrt(pi/ln(a)). Furthermore it also looks like the integral from 0 to inf of a^(-x^b) dx = Γ(b+1/b)/ln(a)^(1/b). It looks like a and b has to have some constraints, but what do I know xD I think wolfram alpha uses the ramanujan way to calculate these kind of general gaussian integrals.

[@aadfg0]

I thought the Ramanujan way was to just write down the answer without any explanation.

[@drpeyam]

Hahaha

[@darkseid856]

Lol. No but seriously , he did proved some of his theorems / formulas. But sadly he died at a young age . :(

[@rishinandha_vanchi]

what about trial and error being the 13th method?

[@mohamedabdelkareem9443]

this Fi(n) in the number theory ?

[@shandyverdyo7688]

Huahahahaha! Finally. The last of Gaussian Integral.

[@chandankar5032]

If someone mail you another elegant way to compute gaussian integral,then will you make video on that?

[@drpeyam]

Sure, I can look into that, although it’s not guaranteed :)

[@sofianeafra7023]

Dr peyam does ln(0) exist in complexe analysis ?

[@drpeyam]

No, not even in complex analysis

[@sofianeafra7023]

Dr Peyam but when we calcul the integral of tan(x) from 0 to π/2 we are obliged to calcul -ln(|cos(x)|) and x equals to π/2 and cos(π/2) is 0 and ln(0) is undifened ?

[@drpeyam]

It’s an improper integral, you’re doing a limit as t goes to 0 of ln(t)

[@sofianeafra7023]

Dr Peyam i suggest that -ln(|Cos(x)|) equals to -ln(0) and we know that e^ix = cos(x) + i sin(x) so we have e^i(π/2) = cos(π/2) + i sin(π/2) and by calculing we will find e^i(π/2) - i = 0 which means that is equal to cos(π/2) and we substitute that formula to get e^i(π/2)=i and entering -ln function to back to our equation so we get -ln(e^i(π/2)=-ln(i) and finally we calcul to get -i(π/2) = -ln(i) by multiplying both sides by -1 we get i(π/2)=ln(i) and as a conclusion we get the integral of tan(x) from 0 to π/2 equals to ln(i) 🔥 how cool is that but is that true ? and thank you 🙏

[@gellogellogello2915]

Eulerian integrals!

[@juandiegoparales9379]

Dr Peyam, after watching the whole series, I can confidently say that you missed my method 😎

[@drpeyam]

Great 👍

[@Rundas69420]

If we make a meme out of this, I bet it would be something like this: Who would win: An integral which got at least 12 videos on this channel, or 1 indian boi?

[@drpeyam]

Hahaha, I’m not Indian though! 😂

[@Rundas69420]

@@drpeyam I meant Ramanujan, but an interesting thought indeed xD

[@drpeyam]

Oh yeah, just realized that 😂😂😂

[@balajidodda7701]

I did it using double integrals.

[@drpeyam]

Check out the playlist

[@PraneshPyaraShrestha]

Wooah!!!

[@ritesharora6032]

Nice accent. Where are you from

[@drpeyam]

Originally from Iran but grew up in Austria, went to a French school, and lived 12 years in California

[@PaoloACostantino]

ouroboro

[@hajsaifi3842]

Bêta mieux