100%. I'm sure it's frustrating for math wizards, but this is one of the few channels tackling these problems that I don't need to incessantly pause (except for his poor writing sometimes)
@@Charky32 im in high school but im preparing for the jee exam. i didnt use the feynmen technique however. i used the series expansion of e^(ilnx) to get terms. then i evaluated all the lnx terms by getting a generalized formula for int (lnx)^n (which turns out to be (-1)^n *n!), then it becomes an expansion for arctan.
@@snow5064 you are very smart, I am only in year 4 primary school, but this type of calculus is still very far away from my knowledge level, I certainly applaud you my friend.
My approach is t = -ln(x) -t = ln(x) x = exp(-t) dx = -exp(-t)dt Int(-sin(-t)/(-t)exp(-t),t=infinity..0) Int(sin(t)/t*exp(-t),t=0..infinity) Calculate Laplace transform of sin(t)/t and plug in s = 1 To calculate L(sin(t)/t) we probably need double integral unless we know formula for L(f(t)/t) I know that purpose of this video is presentation of Leibniz rule but I have read something about differential equations lately and I can see that some integrals can be calculated the same way as Laplace transform which is defined as integral (definite and improper)
I used another approach: substitute y = ln x. Then one gets in the next step the integral I of sin y/y *Exp[y]. Define I(a) = ∫ sin y/y *Exp[-a y] ( limits 0 and inf). Then one gets after simple steps I[a]= arctan a. Hence I = π/4
I first use feynman technique to cancel out the ln(x), so that I made a substitution (y=lnx), after that I used euler's formula, I also got pi/4 By the way, thank you for the job you have done in your chanel, I haven't finished school yet, but I'm learning calculus by my own and videos like yours or bprp's are really helpful Sorry if my english were bad, I'm still learning
My aproach : I use the series expension of sinx so that the two ln(x) cancel out then we notice that we can switch the integral and sum sign and we have : Int [0, 1] (ln(x)^2k) dx Using Feymann technique we prove that this integral is equal to (n)! if n is even since 2k is even we now have : Sum [0, infinity] ( [(-1)^k * (2k)!]/(2k+1)! the (2k)! and (2k+1)! cancel out to give us 2k+1 This sum we have is the Dirichlet beta function evaluated at 1 which is pi/4 So this integral is equal to pi/4
Here’s my solution - Perform a substitution : u = - ln x Then the integrand becomes, (sin u)/(u) * e^(-u) And the bounds go from 0 to infinity. This can be done using simple Feynman’s technique, just take the parametrisation as e^(-αu) and then the answer is pretty obvious. It comes out to π/4 (I am wroting this before watching the video so I hope I am right, otherwise, have a good laugh at how I embarrassed myself :) )
Thank you for going extra slow; it makes me feel less of a dumby.
My friend you're not dumb by any means
You display a passion and discipline for learning and that is the core of intelligence.
100%. I'm sure it's frustrating for math wizards, but this is one of the few channels tackling these problems that I don't need to incessantly pause (except for his poor writing sometimes)
What a clean integral! Definitely a holiday special :)
Yeah I especially loved how Eular's formula and Feynman combined to give a power rule integration problem
dude what a coincidence, i just solved this yesterday, our teacher gave this as a challenge, im super happy that i was able to do it
what year are you?
@@Charky32 im in high school but im preparing for the jee exam. i didnt use the feynmen technique however. i used the series expansion of e^(ilnx) to get terms. then i evaluated all the lnx terms by getting a generalized formula for int (lnx)^n (which turns out to be (-1)^n *n!), then it becomes an expansion for arctan.
@@snow5064 you are very smart, I am only in year 4 primary school, but this type of calculus is still very far away from my knowledge level, I certainly applaud you my friend.
@@snow5064 good luck for your JEE preparation fellow aspirant
@@Charky32 Damn that you know what this stuff even is while being in primary school is impressive!
My approach is
t = -ln(x)
-t = ln(x)
x = exp(-t)
dx = -exp(-t)dt
Int(-sin(-t)/(-t)exp(-t),t=infinity..0)
Int(sin(t)/t*exp(-t),t=0..infinity)
Calculate Laplace transform of sin(t)/t and plug in s = 1
To calculate L(sin(t)/t) we probably need double integral unless we know formula for L(f(t)/t)
I know that purpose of this video is presentation of Leibniz rule but I have read something about differential equations lately
and I can see that some integrals can be calculated the same way as Laplace transform which is defined as integral (definite and improper)
I used another approach: substitute y = ln x. Then one gets in the next step the integral I of sin y/y *Exp[y]. Define I(a) = ∫ sin y/y *Exp[-a y] ( limits 0 and inf). Then one gets after simple steps I[a]= arctan a. Hence I = π/4
Already made a video on this integral using this approach.
I SMELLED COMPLEX ANALYSIS THE MOMENT YOU SAID "DIFFERENT APPROACH"!
Yeah I decided we should try being complex men instead of real men for a change😂😂😂
@@maths_505 It's the time of the year to apply a morbius transformation on our manlyhoods .
@@manstuckinabox3679 you mean mobius right cuz I don't think there's a morbin time in math😂
@@maths_505 lol yeah it was a "my dad is a math professor and trying to relate to the young people" kind of Dad joke.
Very cool integral. Thanks.
I first use feynman technique to cancel out the ln(x), so that I made a substitution (y=lnx), after that I used euler's formula, I also got pi/4
By the way, thank you for the job you have done in your chanel, I haven't finished school yet, but I'm learning calculus by my own and videos like yours or bprp's are really helpful
Sorry if my english were bad, I'm still learning
My aproach :
I use the series expension of sinx so that the two ln(x) cancel out
then we notice that we can switch the integral and sum sign and we have : Int [0, 1] (ln(x)^2k) dx
Using Feymann technique we prove that this integral is equal to (n)! if n is even
since 2k is even we now have : Sum [0, infinity] ( [(-1)^k * (2k)!]/(2k+1)!
the (2k)! and (2k+1)! cancel out to give us 2k+1
This sum we have is the Dirichlet beta function evaluated at 1 which is pi/4
So this integral is equal to pi/4
Nice Integral Maths 505
Thanks bro
Happy holidays
The Bob Ross of calculus!
8:48 the Immaginary part of a real number is 0
very nice question
Thanks Nico
Happy holidays
Happy hollydays !!!!!
Here’s my solution -
Perform a substitution : u = - ln x
Then the integrand becomes, (sin u)/(u) * e^(-u)
And the bounds go from 0 to infinity.
This can be done using simple Feynman’s technique, just take the parametrisation as e^(-αu) and then the answer is pretty obvious. It comes out to π/4
(I am wroting this before watching the video so I hope I am right, otherwise, have a good laugh at how I embarrassed myself :) )
You're correct.....this is actually the 2nd video I made on this integral....the first video involved exactly the technique you've described.
We can even use Laplace Transform.
Please solve Integral of Cos(lnx)dx
That's the real part of the integral of x^I
You can evaluate that using the power rule
Sorry,I haven't seen that video.
🤓