AN ABSOLUTE BEAST!!! And we're solving it using Feynman's technique

Another example showing exactly why Feynman's technique is the coolest way to integrate.
Proof video for the integration result used in the video:
• A breathtaking integra...

[@pnintetr]

I didn't notice parental advisory explicit content in thumbnail until now 🤣🤣😂😂 Indeed, that technique is way too beautiful for most children to comprehend! 🤣

[@skakistisA]

For the engineers among us, that's around 0.51 and the integrated function looks somewhat flat in (0,1)

[@danielc.martin]

Γ(1/4) in terms of lemniscate constant deduction in the channel?

[@meteor3033]

13:01 "I'm not gonna overthink it right now, I'll overthink it later." Such a mood.

[@insouciantFox]

Γ(¼) seems to me just scream duplication formula but it doean't get you very far. Since ¼+½=1-¼,the reflection formula leads you in circles. Best I got was I = ln[sqrt(2π)/Γ²(¾)] or I = ln[Γ(¼)/(Γ(¾)Γ(½)]

[@maths_505]

The result is more compact in terms of the lemniscate or gauss constants.

[@nikko2505]

In other words I = ln(2*G), were G is Gauss constant?

[@trelosyiaellinika]

I think we can simplify the answer further =ln(2ω/π)=ln(2G) [G the Gauss constant]

[@iqtrainer]

Similar to the one solved by PK

[@salaheddineog5965]

Hello i am so thankful for you for these videos 🙏 . I have a good integral for is from -∞ to +∞ of e^(x^2 + 1) /(x^2 + 1) And thank you so much 🙏😊❤️

[@yoav613]

Noice! I could solve it with no problems,because i am subscribed to mathematics mi channel ,who also solves nice integrals like this one.

[@mtl.wraith9981]

would it be possible to integrate this function from 0 to infinity? the function converges, but i’m not sure how far that would get you during integration

[@maths_505]

That's alot easier in fact. After differentiation you get the integral representation for Eular's reflection formula.

[@aleksandervadla9881]

Does the integral from 1 to inf of cos(logx)/x converge?

[@maths_505]

The integral you're talking about is equivalent to int cos x from 0 to infinity which is divergent. Dirichlet's theorem doesn't work here because you just took a divergent integral and performed a u sub. It'll stay divergent.

[@jochemvandolder5046]

Isn't the reciprocal of ln(x) unbounded at the interval (0,1)? ln(1)=0 thus 1/ln(1) must tend to either infinities depending on the direction we approach it right? Please correct me if i misunderstand

[@giuseppemalaguti435]

S(k=0 to inf)(-1)^k ln(2k+2/2k+1)=ln2-ln4/3+ln6/5-ln8/7+ln10/9...circa. 0,51(is correct)...

[@kgopotsomasela4627]

Can you give recommendation on texts on the differentiating under the integral

[@maths_505]

Well you won't exactly find much on textbooks. Just try solving integrals using the technique. You'll get the hang of it (sorta) eventually.

[@GGBOYZ583]

Google “Inside Interesting Integrals”, u will find a free online pdf of the whole book. Chapter 3 of this book gives a pretty good rundown of the technique, and some nice questions to solve. It also has answers at the back of the book. Pdf: galoisian.files.wordpress.com/2018/11/undergraduate-lecture-notes-in-physics-paul-j-nahin-inside-interesting-integrals-2015-springer-1.pdf

[@MrWael1970]

Very smart solution. Thanks.

[@mab9316]

beautiful solution

[@Rory626]

Can you do this with a contour integral?

[@maths_505]

Mate that's way too outrageous even for me😂

[@軒軒-n9l]

I have a question at 2:14. You said that the integral is convergence since it's the product of bounded function and the decreasing function. How do you get this result?

[@maths_505]

That's just Dirichlet's convergence theorem

[@軒軒-n9l]

@@maths_505 Nice! Thanks for helping!

[@okeizh]

Except that to apply Dirichlet, the decreasing function needs to be non-negative and 1/ln x isn't non-negative, so there's a problem here with the proof.

[@okeizh]

The fix is to show that (x^a - 1)/ln x is continuous and therefore bounded on (0,1). The zero of the numerator cancels the pole from one over the denominator in the limit as we approach 1.

[@okeizh]

Just to add, l'Hôpital will give the value of the limit at 1.