As a wizard, i feel happy that people like you still acknowledge our existence!! Feel really down and ignored because people keep forgetting us but thanks to you, my hope in humanity is back!!(Retiming our invansion, send your location so that we dont invade yours) Also, still trying to understand where the hyperbolic functions came from And no gamma function this time?
As I have said before I only really try these integrals if I can us contour integration. This one looked like it could be done that way. I used the same approach as you writing cos(Lnx) as Real Part x**i and using a key hole contour with a branch cut along the real axis. I rather wish I hadn’t. The integral was easy enough but then I was left with a long expression involving exponential functions which took me for ever to bring it down to the same form as your answer. I was beginning to doubt the whole thing but then I saw the trick which put our answers in agreement. This came as a big relief and I am modestly pleased with myself which is the reason for writing.
I think it could be fun if you got a soundboard. Maybe play a funny meme noise whenever the gamma function pops up or you make a writing error etc. Just a wandering thought from one of your coshiners
Great video bro. But I have a question, do these integrals even converge in the first place because when x tends towards 0 the cos(ln(x)) and sin(ln(x)) both do not exist. While the rational function part tends to 1.
Indeed they don't so the best thing we can do is be thankful that they're bounded and evaluate the integrals in the limit as b (the lower limit of the integral) approaches zero from the right.
@@moeberry8226 a substitution x=e^u gives us a cos(u) term instead and the function \frac{e^u(1+e^2u)}{1+e^4u} doesn't cause any problems either. The integrand in this form (limits being +♾️ and -♾️) makes convergence of the integral a bit more clear along with looking at a graph of the integrand.
@@maths_505 thanks bro I appreciate it. But the closer this integrand gets to 0 the more it skips between -1 and 1 I still don’t see how it can converge to the value you obtained.
The function not converging is not the same as the integral not converging. Here's the convergence test: Take the absolute value: |sin(ln x) * (1+x^2)/(1+x^4)|
Great! Split the sin function into 0-1 and 1-infin then substitute t=1/x in one and the two integrals cancel out, confirming your algebra 🤗
COSHINE LEAGUE, ASSEMBLE!!
me and bro shining shoes together
bro has seen lebron too much
As a wizard, i feel happy that people like you still acknowledge our existence!!
Feel really down and ignored because people keep forgetting us but thanks to you, my hope in humanity is back!!(Retiming our invansion, send your location so that we dont invade yours)
Also, still trying to understand where the hyperbolic functions came from
And no gamma function this time?
Nah I'm cool with it. Wizards would do a much better job than modern politicians.
As I have said before I only really try these integrals if I can us contour integration. This one looked like it could be done that way. I used the same approach as you writing cos(Lnx) as Real Part x**i and using a key hole contour with a branch cut along the real axis. I rather wish I hadn’t. The integral was easy enough but then I was left with a long expression involving exponential functions which took me for ever to bring it down to the same form as your answer. I was beginning to doubt the whole thing but then I saw the trick which put our answers in agreement. This came as a big relief and I am modestly pleased with myself which is the reason for writing.
This somehow feels familiar... I can't say what but i think its the 1+x^2/1+x^4 but i remember a rather stunning MSE integral with some log form
Speaking of log....differentiating the function will yield ln(x) times the other function.
Aaand i also think I have an integral for you but how do I suggest it? Do I mail it or something?
Yep, but I'm still searching for the MSE one(although I got sidetracked looking at the absolutely gorgeous cleo integral which is also another ln boi)
@@shardulkakade9365 Instagram or email
@@maths_505I don't have Instagram so I'll mail you. And thank you for the help and the ethereal integrals
Hi,
"ok, cool" : 1:07 , 2:25 ,
"terribly sorry about that" : 3:06 , 3:32 , 5:06 , 5:21 .
I think it could be fun if you got a soundboard. Maybe play a funny meme noise whenever the gamma function pops up or you make a writing error etc. Just a wandering thought from one of your coshiners
I've used the residue theorem on z^i*(1+z^2)/(1*z^4) and found the same result ^.^
Thank you for your amazing effort.
Bro you can directly put x=1/t and then after simplifying you can easily get zero😅
Btw this method was also good 😊
You did something involving cos(lnx) and sin(lnx) derived from x^i in a previous video I think.
This would be the best intro in your video 😅😅
Without that formula, wpuld ypu ise partial fraction decomposition?
But can you do the same but with tan instead of cos or sin?
jokes apart i used to call it coshine for real
You forgot ‘Mathamagician” :)
Basta utilizzare la Re(e^ix) e Im(e^ix),la funzione beta...risulta I=(√2π/2)ch(π/4)/ch(π/2)..e I=0
Great video bro. But I have a question, do these integrals even converge in the first place because when x tends towards 0 the cos(ln(x)) and sin(ln(x)) both do not exist. While the rational function part tends to 1.
Indeed they don't so the best thing we can do is be thankful that they're bounded and evaluate the integrals in the limit as b (the lower limit of the integral) approaches zero from the right.
@@maths_505 if they don’t converge how did we get a answer? Wouldn’t that imply that they converged to a real value.
@@moeberry8226 a substitution x=e^u gives us a cos(u) term instead and the function \frac{e^u(1+e^2u)}{1+e^4u} doesn't cause any problems either. The integrand in this form (limits being +♾️ and -♾️) makes convergence of the integral a bit more clear along with looking at a graph of the integrand.
@@maths_505 thanks bro I appreciate it. But the closer this integrand gets to 0 the more it skips between -1 and 1 I still don’t see how it can converge to the value you obtained.
The function not converging is not the same as the integral not converging. Here's the convergence test:
Take the absolute value: |sin(ln x) * (1+x^2)/(1+x^4)|
I will forever say coshine now
Awesome
Okay cool 🔥🔥🔥
First