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- เผยแพร่เมื่อ 5 ก.ค. 2024
- Another one bites the dust....just a sprinkle of complex analysis and a nice result needed.
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Super interesting! I love it when the result remains elementary without invoking infinite series expansions or exotic constants. Just one special cot formula and the rest is simple trig. Nice!
With a sin, the integral is trivial : a simple transformation from the x world to the 1/x world shows it is equal to 0
I tried it, and somehow the integral changes into its negative. Therefore it's 0 :)
@ around 9:06 The second term in the numerator should be -4i*sinh pi/3*cosh pi/3, not -2i*sqrt(3)*sinh pi/3*cosh pi/3.
Hi,
5:00 : May be better to write cot z = i (e^iz + e^-iz) / (e^iz - e^-iz)
5:00 To remember the values of sin for 0° 30° 45° 60° 90°
sqrt(0)/2 sqrt(1)/2 sqrt(2)/2 sqrt(3)/2 sqrt(4)/2
To show that int_0^infty { (sin(ln x) dx)/(1+x+x^2) } is null, just calculate int_0^1 + int_1^infty and do u=1/x in the first integral.
"terribly sorry about that" : 2:57 , 5:09 , 8:27 ,
"ok, cool" : 3:14 , 5:31 , 5:43 , 7:07 , 9:16 , 9:58 .
im checking regularly these videos now just to check if the french guy will tell us exactly when he says 'ok cool' ... thanks! merci
Thank you for this innovative integral and solution.
I did this from scratch and it took me 3 hours. I started the same but instead of using the formula for the integrals, I calculated them manually. I split the integral into 0 to 1 and 1 to infinity, then expanded the denominator as a geometric series. After integrating term by term with power rule, I was left with 2 infinite series which I combined into 1. It was a sine fourier series where b_n = (-1)^n*n/(n^2+1) evaluated at x=pi/3. Based on the 1/(n^2+1) and only odd terms, I correctly guessed f(x)=c*sinh(x). The most simplified version of the answer is tau/sqrt(3)/(1+2cosh(tau/3)).
I whacked this in Mathematica, and it took a little bit, but I got your result which simplifies to \frac{2 \pi}{\sqrt{3}} (1 + 2 \cosh(2\pi/3))^{-1}. Nice work dude.
and as always using your vid to fall asleep
Bro really said Kamal's videos are boring
I see a Mellin transform in the middle of your calculation. Mathematica actually gets the original integral (and the Mellin transform), although it takes a long time. Yes, I have the "similarity" super power.
I would divide interval of integration to [0,1] and [1, infinity]
Use substitution t = 1/x on integral on interval [1,infinity]
Expand numerator and denominator by (1-t)
Use geometric series expansion with common ratio t^3
Exchange order of summation and integration
and I would need to evaluate two integrals
\int\limits_{0}^{1}t^{3n}\cos{\ln{t}}du
\int\limits_{0}^{1}t^{3n+1}\cos{\ln{t}}du
and it can be done by parts
with du = t^{3n} dt , v = \cos{\ln{t}}
The slightly condescending way in which you give partial credit to WolframAlpha for getting the other integral is already on par with how my professors do it.
Fantastic!!!
Awesome integral and solution development, Kamaal!🔥 I believe the final answer can be simplified to [(π sqrt(3)) / 3] x [1 / (cosh^2(π/3) + sinh^2(π/3))]. This is achieved by adding and subtracting another sinh^2(π/3) term in the denominator.
That is nicer indeed
@@maths_505 Whoops, I made a mistake… the correct “simplified” answer should be [(π sqrt(3)) / 3] x [1 / (cosh^2(π/3) + sinh^2(π/3) + 1/2)]. I forgot the *+* *1/2* in the denominator. Perhaps this isn’t any more simplified than your final answer after all.
@@johnanderson290 yours is still better than Wolfram alpha 😂
@@maths_505 Thanks! 🤣 I love, love, love your channel Kamaal, and your incredible content!
So does wolfram alpha return any answer?
Very nice. Could you start out sketching the function in a graph?
U-sub using exp(-u), change bounds to -inf, inf and it will work..... Cheers
Life received
You know it's good when even our lord and savour WA is stumped
Hi could you link a resource to the principal value claim you made? Thanks
Sir I think here is only one mistake you have made is in numerator the term will be "-- i 4 sinh(pi/3)cosh(pi/3)= -- i 2sinh(2*pi/3)= -- i 8 " May be I am wrong just check it.
Similarly:
Do you like integrals ?
I have following integral for you
Let P(x) be orthogonal polynomial
Let f(x) be given function (We can assume that f(x) is polynomial of degree n)
[t_{1};t_{2}] is interval (fe for Chebyshev and Legendre it is interval [-1;1])
w(x) is weight function
Calculate \frac{1}{p}\int\limits_{t_{1}}^{t_{2}}f(x)w(x)P_{k}(x)dx
where p = \int\limits_{t_{1}}^{t_{2}}P_{k}^2(x)w(x)dx
k runs from 0 to n when f(x) is polynomial of degree n
I can evaluate this integral but I am interested in evaluation as quick as possible
My attempts end on time complexity O(n^3) and space complexity O(n)
and possibly could end in time complexity O(n^2) and space complexity O(n^2)
but can it be calculated quicker and cheaper ?
Mind adding the link in the description to how you derived the result of the cot? I'm sure it's not trivial and I'd like to see the work
Bisogna conoscere la formula a 2.27..dopo sono solo passaggi matematici...io ho provato con beta function ma 1-x^3 Is bad
I tried solving this with contour integration but got stuck. I don't know how you do it man.
i've solved it using keyhole contour
I can attest that keyhole integral do work on this integral
x^i is very bad idea.
And why would that be
@@azmath2059 Its definition depends on which branch of the logarithm you use, and it won't be continuous everywhere in the complex plane. It's fine to use it if you're careful though.