A deceivingly difficult integral

When Forest Grump said "life is like a box of chocolates, you never know what you're going to get", I'm sure he wasn't talking about integral calculus. However this integral fits the bill quite well.
Proof for ln(cos(x)) series:
• My take on this on won...
My complex analysis lectures:
• Complex Analysis Lectures
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[@Tosi31415]

the result happens to be extremely close to the euler mascheroni constant aswell

[@ericthegreat7805]

Oily-Macaroni

[@CM63_France]

Hi, I am going to make a catalog of all these constants or functions defined by series, because I do not know them. "Terribly sorry about that" : 0:07 , 3:58 , 4:03 , 4:11 , 5:39 , 12:49 , 14:41 , 14:45 , 16:37 , 17:51 , 22:27 , "ok, cool" : 4:11 , 7:11 , 9:12 , 14:55 , 15:36 , 17:04 , 18:53 , 20:23 .

[@maths_505]

Wow there were alot in this video

[@CM63_France]

@@maths_505 Yes, and you can notice that at 4:11 I spotted them both, the reason for this is that, at this time, you said something about my counting, unfortunately I don't understand what, but it ends with "thank you very much", or some thing like that, so I'm not that worried 😃. It's a pleasure for me. By the way could you make a video about the alternate ways(1) of prooving that zeta(2)=pi^2 / 6 . There is one that starts from int_0^infty { int_0^infty { dx dy / (1-xy) }} that is not that easy, because it involves a couple of variable substitutions a bit tricky. (1) otherwise than the famous Euler proof.

[@xloppyschannel4881]

23 minute video has cost me my entire 8 hours of sleep, thanks.

[@johnanderson290]

Wow Kamaal, this was a beautiful beast! Most impressive! 👏

[@thomasrichard7054]

Very good! The result is remarkably close to sqrt(3)/3, so numerical integration may fool us. 😁

[@juniorcyans2988]

This is a cool problem worth trying out. Thanks a lot!

[@txikitofandango]

I understand how you did the integration by parts, but I can't imagine myself ever escaping from uv - integral of v du paradigm

[@Mathematician6124]

It's very easy using fourier series. I got it 😊. Dear Friend

[@mcalkis5771]

Look at that, the gang's all here. I guess e and γ didn't get the invite though. And I suppose Ω isn't really friends with anyone these days. Although at this point I fully trust that you would be able to find an integral whose value contains everything. π, e, γ, G, ζ(3), ln2

[@hewhomustnotbenamed5912]

8:08 We can take comfort in the fact that you're speaking from experience.

[@MrWael1970]

There are some simple mistakes, but overall is very smart solution. Thank you indeed.

[@joanofArc6]

8:38 the famous bansal sir of india who used to teacher for jee gave it the name kings property.... Back here in india, we also have queen and jack rules❤❤

[@keyaanmatin4804]

4:37 is so relatable

[@txikitofandango]

Didn't think a math video could be so salty :-p Result is very close to (error less than 0.1% of) the Euler-Mascheroni constant as well as the others mentioned

[@romanvolotov]

the amount of missing du's in this one is insane lmao

[@christophercalvaire2014]

from ln（1+tan u），we could expand to series with bernoulli number at once

[@christophercalvaire2014]

or I =int x ln （1+ tan au ） dx，using feymann trick, but it seems that I got an much simpler solution through this ode, I am not sure if I am right.dI/da=C-I*2/a

[@sarahakkak408]

Great solution, but you was can use king rule in the first step

[@Anonymous-Indian..2003]

I guess i solved that question some weeks ago.🤔 But i didn't remember the book in which I've found that. Btw, your solution is also amazing. Love❤ you bro !(No HOMO)

[@Anonymous-Indian..2003]

I remembered now, I solved by simplifying. Integral Term is: Int(0 - π/2) (x²/2) { tan(x/2) + tan(π/4 - x/2) }dx = Int(0 - π/2) (1/2) { x² + (π/2 - x)² } tan(x/2) dx Now it'll be easy, Substitute, (1/2)tan(x/2) = sinx - sin2x + sin3x - sin4x + ......... And booooom.

[@mihaipuiu6231]

Sir math 505,...I try to understand a very nice integral from "a deceivingly difficult integral" and I met a formula, which I couldn't find proof. If you are kind to help me find this demonstration, because I can't continue to watch your integral.the formula is...ln(cosx) in terms of cos(2kx). Thank you very much for your help.If you agree to help the beginners like me to locate on the internet formula that you use without proof. This will be a huge help for people like me.I'm engineer and for me math is a HOBBY. Again thanks.

[@maths_505]

kzbin.info/www/bejne/o6Kzhal7ga-Dg5I

[@malino0527]

Could you make a video about the integral from 0 to infinity of 1/(x^ln(x))? The result is the fourth root of e times the square root of pi, which I think is really beautiful. It seems that it uses the error function, if you plug it into wolframalpha you'll see.

[@maxvangulik1988]

now factor the result in terms of pi

[@kingzenoiii]

F, definitely watching cuz i have no friends 😭

[@maths_505]

(with musical effects) you've got a friend in me

[@kingzenoiii]

@@maths_505 🤩

[@Aditya_196]

😂 I understand the nomenclature being kings rule like it's something great but in reality and significance it's just simple , idk myself why they say it maybe something like chess related where u swap King and took just moving around, it helps a lot in problem solving despite being so simple

[@yoav613]

Very nice!

[@PritamDas-zu9im]

sir i think there is a mistake in 9:17 .you forgot to write cosine of log. its ln(cos{pi/4-u})

[@uggupuggu]

dw he still wrote cos in the next line

[@Shindashi]

Your pronunciation of Catalan's constant 18:44 threw me off, almost like Caplan.

[@DestroManiak]

How is it deceivingly difficult lol, it definitely looks difficult.

[@maths_505]

It honestly looks kinda innocent....throw in an x² and a couple trig functions....and then you question all your life decisions leading up to that point 💀

[@Shindashi]

7:47 lmao thanks for that. F.

[@aravindakannank.s.]

F

[@insouciantFox]

Delicious

[@xdShaty]

i got roasted but it was true.. F

[@mohammedfarhaan9410]

F