- So, our methods are pretty much the same, but I used a generalized form of the Mellin transform of 1/1+x^b (proved using contour integration, review Gamelin's exercise on it) at 1 and at 3 to get 2*(pi/5sin(pi/5) + pi/5sin(3*pi/5)) ... - hey, I thought of beta too! it's amazing how contour integration and beta function are so similar when it comes to solutions, I think Churchill had a nice exercise exploring the beta function in light of complex analysis. Man, I'm loving the content my man! keep it up! and don't fear experimenting with the content as well!
Using the substitution x = arctan(t), which turns the integral domain to the entire positive real axis, and simplifying the rational function gets: (t^2+1)/(t^4-t^3+t^2-t+1) Which is an even function. Applying the residue Theorem solves the problem, but all four poles (only two are enclosed by the contour) are irrational.
11:32 asumme y is the golden ratio Sqrt5=2y-1 ysqrt5=2y²-y ysqrt5=2y+2-y ysqrt5=y+2 the denominator can be simplyfy to 5sqrt(y+2) where y is the golden ratio.
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i see 5, so the answer must be something with golden ratio
- So, our methods are pretty much the same, but I used a generalized form of the Mellin transform of 1/1+x^b (proved using contour integration, review Gamelin's exercise on it) at 1 and at 3 to get 2*(pi/5sin(pi/5) + pi/5sin(3*pi/5)) ...
- hey, I thought of beta too! it's amazing how contour integration and beta function are so similar when it comes to solutions, I think Churchill had a nice exercise exploring the beta function in light of complex analysis.
Man, I'm loving the content my man! keep it up! and don't fear experimenting with the content as well!
Using the substitution x = arctan(t), which turns the integral domain to the entire positive real axis, and simplifying the rational function gets:
(t^2+1)/(t^4-t^3+t^2-t+1)
Which is an even function. Applying the residue Theorem solves the problem, but all four poles (only two are enclosed by the contour) are irrational.
11:32 asumme y is the golden ratio
Sqrt5=2y-1
ysqrt5=2y²-y
ysqrt5=2y+2-y
ysqrt5=y+2
the denominator can be simplyfy to 5sqrt(y+2) where y is the golden ratio.
Interesting. I probably would have tried factoring the denominator, and simplifying.
Very interesting integral, and very smart solution plan. Thank you.
I=(π/√5)(1/√(5-2√5)+1/√(5+2√5))+(2/√5√(5-2√5))arctg1/(√5-2√5)-2/(√5√(5+2√5))artg(1/√(5+2√5))=3,46...
How about this form,
(2pi/5 sec pi/10)*phi^2 😅
*radical sign and five*
Coorporate needs you to find the difference between these two images.
Maths 505: they're the same picture
Just got your gamma function hoodie, will be wearing it to school on Monday
WOAH 🔥
The golden ratio satisfies fi*fi = 1-fi and not fi*fi = 1+fi. With this, the result is slightly different but similarily compact 😊
Nicee
Just not hold on ß function yet still the same procedure I apply earlier
Nicee