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