Thank you for solving such integrals. So, for non specialists and undergraduate students, we shall proof each integral or theorem before introducing solutions. This channel is the best, and i want to be the best all the time.
@stefankraft54158 ай бұрын
At 3:15 I can see a difference of two integrals, where both integrals do not exist for instance for s=2 (but Re(s=2)
@giuseppemalaguti4358 ай бұрын
Ecco il passaggio decisivo...moltiplicare (1+x) sopra e sotto...io non c'ero arrivato ...complimenti
@shivamdahake4528 ай бұрын
Bro is starting a math cult I am all here for it Also I got JEE advanced exam on may 26, hoping it goes well.
@helloworld43098 ай бұрын
Can you tell how to integrate root sinx from 0 to π/2 is this integral can be done using calculus 1
@jackkalver46448 ай бұрын
cosh^2 n+3sinh^2 n=1+4sinh^2 n
@tsa_gamer0078 ай бұрын
0:16 "You have not been scammed" Reality: Here's a starter integral before we go to the main course lunch
@orionspur8 ай бұрын
According to Wolfram these last two simplify to... √⅓ / (4 cosh(π/3) - 2), and... -½ π sinh(π/6) sech(π/2)
@whippyPoo8 ай бұрын
What software are you using for writing this math on?
Who else used contour integration and residue theorem. ?
@Cs-vs7qk8 ай бұрын
Woah
@holyshit9228 ай бұрын
I played with u=1/x substitution and partial fraction decomposition of the factor (1+x^2)/(x^4+x^2+1)I have got \int_{0}^{1}\frac{\cos{\ln{x}}}{x^2+x+1}dx + \int_{0}^{1}\frac{\cos{\ln{x}}}{x^2-x+1}dx Now i would expand numerator and denominator to get \int_{0}^{1}\frac{(1-x)\cos{\ln{x}}}{1-x^3}dx + \int_{0}^{1}\frac{(1+x)\cos{\ln{x}}}{1+x^3}dx and now i would try to use geometric series expansion
@christophercalvaire20148 ай бұрын
quoted theorem is actually really hard, I tried MMA to understand the details 😢but finally get this in youtube int x^(m-1)/(x^n+1) kzbin.info/www/bejne/hZ-3n2aIbsSgZ7Msi=sqvK0IGvkC9ED9Uh
@christophercalvaire20148 ай бұрын
things become much much easier in the complex number field❤