A RIDICULOUSLY AWESOME INTEGRAL

A superb integral featuring some of my favorite tools leading to a beautiful result and an equally beautiful related result.
My complex analysis lectures:
• Complex Analysis Lectures
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[@Tosi31415]

luckily you use a black background since im watching this at midnight before sleep

[@satyam-isical]

Now Kamaal will take an immediate action to this very dangerous problem😂😂

[@pyrite2060]

Same

[@vladimir10]

Awesome development!!
Another little thing I've learned is that ζ(1/2) is negative!
Thanks for that!!!

[@txikitofandango]

You brought out the exquisite succulence of this ravishing integral via a luscious solution development

[@sohaib_mer-]

oh to have such vocabulary

[@plasmusss056]

8:50 my man, for a mathematician your handwriting is impeccable and very clear, dont worry about it

[@namansanghi1940]

Was finding some interesting problem to do in your channel and u just spawned with a brand new one 😂

[@subhrayanbarman1654]

You could also do it using Integration by parts ,differentiating x and integrating x/(e^x²+e^-x²)

[@ralfbodemann1542]

Awesome instrumentalization of the geometric series, really love your solution!
Since the integrated function is a square, the integral should be non-negative.
Is Zeta(1/2) a negative number?

[@tapasmazumdar3831]

Yes Zeta(1/2) < 0 and also notice he wrote sqrt(pi)-sqrt(2*pi) in the numerator which is also negative. So the answer is positive.

[@ericthegreat7805]

How come when you differentiate the series 1/(1+x) wrt x and make the transformation x > e^-2x^2, you dont use the chain rule?

[@maths_505]

I differentiated the series and then just plugged in the exponential function. If you plug in the exponential and then differentiate, you need the chain rule and you'll get the same result.

[@leroyzack265]

At the stage we had x²exp(-kx²) I will have gone for integration by parts but substituting till recovering the gamma function was much more smart.

[@giobrach]

Looks very statistical-mechanics-y

[@isaacmalik3714]

oooook cool

[@lalomedina6495]

I love watching your videos about crazy series and integrals. Do you have a course or a list of books to improve our skills in maths, specially in calculus?

[@gianpaolosoligo646]

Can be solved with contour integration?

[@giuseppemalaguti435]

Diventa un integrale gaussiano dopo avere fatto l'integrazione per parti...I=(√π/8√2)(1-1/√2+1/√3-1/√4+1/√5.....si può sintetizzare con la zeta function a valori alterni..la serie è (1-√2)ζ(1/2)

[@nerddosnumeros]

When you changed from x to e^-2x², where did the k that was in the sum with the x go. I've watched it 5 times and still can't understand where it went lol

[@bzzz179]

Was asking myself that as well.
Edit: he notices it later and changed it.

[@bandishrupnath3721]

sir why did u too x times root 2k as u ?

[@emanuellandeholm5657]

Honestly, geometric series transformations are just so powerful. They need to nerf that!

[@stevenkrantz6580]

does the chain truly not apply when differentiating the geometric series expansion?

[@takemyhand1988]

It applies

[@qetzing]

Why does at 6:59 du = 1/2sqrt(t) dt? If I differentiate t=u^2 by u, shouldn’t I get 2u / du = 1/2u dt?

[@madsheller5242]

I think you have it all right, despite a few missing parentheses. Just notice that t=u^2 implies that u=sqrt(t) and plug that in in the end.

[@qetzing]

@@madsheller5242 Yes, I see, thank you. I tipped this on my tablet and it didn't come out quite as I wanted it to. I didn't know that you could also replace the u with the t in a substitution like this, thank you!

[@vit1leman14]

I was thinking couldn’t we start by an integration by parts where we differentiate x and integrate x e^-2x^2/ (1+e^-2x^2)^2 then we would have to integrate from 0 to infinity 1/4(1+e^-2x^2) and then use the series ? Or did i miss something ?

[@vit1leman14]

I stopped the video so I didn’t see the end

[@nicolascamargo8339]

Genial

[@hashirama868]

does zeta one half diverges??

[@maths_505]

Nope

[@waarschijn]

The series definition of ζ(s) diverges for s ≤ 1, but ζ(½) itself is finite. Use one of the other definitions of ζ if you want to compute its value at ½.
It's like the function 1/(1-x), which has a finite value if x ≠ 1 but the geometric series diverges for |x| > 1.

[@hashirama868]

@@waarschijn Thanks for your explanation, I appreciate it

[@CM63_France]

Hi,
"terribly sorry about that" : 0:57 , 3:55 , 8:26 ,
"ok, cool" : 1:23 , 1:54 , 7:48 .

[@MrWael1970]

Very smart solution. Thank you.