Finally, an interal worthy of flexing over my new math friend. imma tell em to sub obviously. I'm so going to use you for next sem's complex analysis to also rizz on my professor, THANKS MATHS 505 KAMAL!
@rishabhhappy9 күн бұрын
The Result is Cool as always
@cdkw29 күн бұрын
I have a physics exam today, its 2 AM why am I watching this...
@yurfwendforju7 күн бұрын
you know when someone mastered math when he's watching your video and all he does is the skip five second button the likes and clicks off. i was impressed when I saw him in the library today
@mikecaetano8 күн бұрын
The new audio track feature translating your remarks into other languages is neat. I haven't seen that before. I checked out Spanish and Japanese. The voices sound natural rather than artificially generated. Cool. Expand your audience. Now back to the original from the start.
@kro_me9 күн бұрын
i really enjoy these vids when i get them recommended lol
@Leonhardeuler22199 күн бұрын
Congrats for the job champs 🎉🎉
@slavinojunepri76487 күн бұрын
Excellent
@sundaresanabishek51279 күн бұрын
Hey broo fan from Sri Lanka ❤🎉
@maths_5059 күн бұрын
@@sundaresanabishek5127 hey broo friend from Pakistan ❤️🎉
@rishabhshah87549 күн бұрын
hi, nice video, i have a few questions where did you study advanced integrals, especially involving the gamma and beta functions? and, where did you do you high school schooling from?
@maths_5059 күн бұрын
@@rishabhshah8754 I learned alot of math simply in the process of solving integrals 😂 just look for integrals to solve and you should be covered. As far as highschool is concerned, I'm Pakistani and I just did the local high school thing we have here.
@АндрейПергаев-з4н8 күн бұрын
Выражение под логарифмом в числителе представляет собой сумму геометрической прогрессии.
@BottlePack9 күн бұрын
If the alternating sum is replaced by (1+x+x^2...x^2n), wolframalpha says it is not convergent. Is it right?
@BottlePack9 күн бұрын
split interval to (0,1) and (1,infinity), we have same result still
@DanielPascuasTijero8 күн бұрын
The result is correct, but the method of solution is wrong!! You have to avoid the non convergent integral \int_0^\infty 1/(1+x). In order to do that, split the integral \int_0^\infty (x dx)/(1+x^3) as the sum of the integral between 0 and 1 plus the integral between 1 and \infty. Perform the substitution t=1/x in the second integral to show it is equal to \int_0^1 dx/(1+x^3). Thus our integral equals to the arctangent integral (2/3)\int_0^1 ((1+x)dx)/(1+x^3)= (2/3)| \int_0^1dx/(x^2-x+1) which can be easily computed.
@الْمَذْهَبُالْحَنْبَلِيُّ-ت9ذ2 күн бұрын
Where do you see the integral of 1/(1+x) in the video?
@tiagobeaulieu17458 күн бұрын
How did no one in the comments realize you missed a factor of x squared in the numerator? Pleeeeease it pains me to see a solution development for an integral that is not the one at the start of the video 😭😭😭