awesome integral with a surprising answer
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- เผยแพร่เมื่อ 10 ต.ค. 2024
- awesome integral with a surprising answer
We calculate the integral of exp(2 cos(x)) from 0 to pi using series and complex analysis, more specifically power series and residue calculus. The result is guaranteed to surprise you and is related to Bessel functions in physics. This is a must-see for any complex analysis lover, enjoy!!
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One could add that all the Besselfunctions have integral representations similar to the one found in this video.
One has I0[z]= 1/π Int Exp[±z cos t] ,0< t< π .These functions arise e. g. in solutions of various differential equations.
awesome integral, awesome shirt, and awesome tattoo! I'm glad I recognized the Bessel function almost instantly because I've been working on something that uses it
My approach to this problem was to write out the series expansion of e^(2cos(x)). Switched the integration-summation order and then found an expression for int(0,pi){cos(x)^n}dx
In the end I had: sum(k, 0, infinity) {2^(2k)/(2k)! * (2k - 1)!!/(2k - 2)!! * pi/2}
(Here I am summing all even integers because the integral of an odd powered cosine is 0)
After simplifying all the double factorials you end up with:
Sum(k, 0, infinity){1/(k!)^2} * pi
Think you forgot to edit out various parts of this :)
I actually did cut things out but the video didn’t finish processing
And what‘s the Limit of this sum?
That's just part of his charm
You look like wearing a whole Galaxy, and which makes your Euler tatoo more magic.
Thank you!!! 😊
Dr Peyam is sparkling today :D
Thank you ✨
"And using this we can transform this integral into a much simpler one." Snip Snip... Dr. Peyam performs a little magic! I love it!
🫰🫰
Dr Peyam, ive taken calc for a month and i love it thanks for all the help
Thats one hell of an amazing integral!!!
Thank you :3
Once again, no closed form, but an awesome approximation using only rational numbers!
Nice. Need to brush up on my complex analysis.
Go for it!
It boils down to 1/2 * the integral from 0 to 2pi of e^(z + 1/z) dz/ iz. By the residue theorem this is pi times the residue of e^(z + 1/z)/z at z = 0. Then use e^(z) = sum n>=0 z^n/n! and e^(-z) = sum n>=0 z^(-n)/n! to realize the constant coefficient of e^(z + 1/z) = e^(z)e^(-z) is the sum over n of 1/(n!)^2. So the overall answer is pi times this. 1800s math was filled with curious identities like this, it's a fascinating lost subject.
Bruh, youre slayin in that long sleeve.
Thank you!!!
Very good outfit indeed
A typical engineering integral from signals and systems
solve the sum to fully solve the problem, it is a convergent sum si it must be solvable
"Laborieux!", but the result is indeed surprising. I was wondering why we have to bother using contour integral since a u substitution could do the trick but there was a reason ...
I’d be interested to see the u sub solution! Contour integrals make this elegant though
@@drpeyam please consider this with caution, this is an answer from an electric engineer!
u = e^ix
du = i e^ix dx= iu dx
dx = du/iu
Then the integration limits go wrong since -pi, pi -> -1, -1
But if we continue we obtain
(1/i) int{ e^u/u^(N+1) }
Then if we develop e^u as 1 + u + u^2/2! + u^3/3! +... and devide it by the denominator u^(N+1), we obtain a sum of 1/u^(N+k) that can be integrated and the result seems to be
Sum(u^(N-k) / (N-k)), k = 0 to N
We found the N! Somehow in the common denominatr.
I have ∫f(x)dx tattooed on my arm.
Nice!!!
Hi Drop. Let me add to the cacophony of super slick starry shirt sycophants. Also, one could consider a Taylor expansion for cos(x) instead.
A beautiful integral
Can we use properties of definite integral. x equal to pi-x
Nice
What is the dominated convergence theorem? Is it a similar condition to uniform convergence of a sequence letting you swap integral and series?
very interesting question
Just curious about a close form for the final summation.
sick
Where have been
2cos(x)=e^ix+e^-ix
I=int[0,pi](e^(e^ix+e^-ix))dx
u=e^ix
du=ie^ix•dx
I=i•int[-1,1](e^(u+1/u)/u)du
u->1/u
du->-du/u^2
I=-i•int[-1,1](e^(u+1/u)/u)du
I=-I
2I=0
I=0
No 1/u is not one to one on [-1,1]
@@drpeyamit is one-to-one, but it's not continuous
regardless, I'm almost certain i did the problem horribly wrong
Awesome video Sir! I could catch on up until the taylor series part but I have no idea what contour integrals and residues are, are there any resources you'd suggest for me to explore on this, preferably something for beginners (I'm in grade 12 in India, that should be calc 2 + whatever's there in vectors, line integrals, partial differentiation (+ gradient/divergence/curl etc.))
Yes of course, check out my complex analysis playlist!
@@drpeyam Sure! Thank you so much!