APOLOGIES I made a mistake with the Kummer series at the 10:10 mark. The index k starts at 2 and not 1 That's why the first odd number you'll see in the series after plugging in x=3/4 is 3 and not 1. And that's why we have a (-1)^(k+1) term because the first sine term you'll have in the series is sin(9pi/2) which is +1.
Hi, I think I found the problem finally. Your expression for the Kummer series is in fact right as the k=1 term is nulled by the ln(k) term. U simply state that f(k) = sin(3*pi*k/2) = (-1)^(k+1) for k odd, which is wrong as (-1)^(k+1) == 1 for k odd; considering only the odd k, we have: f(3) = 1 , f(5) = -1 , f(7) = 1 etc… Hence we get this alternating pattern among all odd k, so the expression is slightly more complex, it is: f(k) = (-1)^[(k-1)/2 + 1] precisely. Going to the next step, u corrected this own mistake in the summand. Essentially by making a k = 2l+1 substitution (k=3 -> l=1 first non zero term), f(k) does indeed transform to (-1)^(l+1), and the sum term does become: 1/pi * sum(l>=1) (-1)^(l+1) ln(2l+1)/(2l+1) as u wrote down, with the implicit dummy variable change back to k. I hope that clears any confusion!
Awesome! This has been my all-time favorite integral result. It's - dare I say - as satisfying as having a warm brownie on a cold, snowy day! I gotta say, good job with keeping up with the variable chaos, you've hardly made any mistakes!
If possible could you also show us what your friend Myers did, how he arrived at the solution by differentiating the zeta function?. I think that would make for a pretty awesome video
here's where I reaached: let I = integral from pi/4 to pi/2 : log(tan(x))^s dx, by letting u = tan(x) we get log(u)^s/x^2+1 now we let f(z) = log(z)^s/z^2+1, if we integrate f(z) round a keyhole contour, the sum of the residues = pi^s*i^(s-1)*(1-(-1)^s)/2^s. now the big and small circle vanish if we let s 0
Hey, excellent video and I had never heard of the series for log of the gamma function until now! Just wondering, should sin(3pi k/2) have a different answer?we see that sin(3 pi/2) = -1 for k = 1, sin(9 pi/2) = 1 for k = 3, sin(15 pi /2) = -1 for k = 5 etc.. , so rather the formula should be: sin(3k pi/2) = (-1)^k for k == 1 mod 4, and sin(3k pi/2) = (-1)^(k+1) for k == 3 mod 4? Thanks again for the crystal clear explanations!
Hey! I came up with a pretty cool integral but I’m struggling to solve it using integration techniques. It is the integral from 0 to infinity of e^(-x)ln(x)dx which by use of the laplace transform property that says: the integral from 0 to infinity of f*g = L(f)InverseL(g) where L and inverse L are laplace transforms. Using this and saying f=ln(x) and g=e^(-x) the integral miraculously resolved to the negative of the Euler mascheroni constant 0.57721… do you think it’s possible to evaluate this without using this laplace transform method?
Can you solve this multiple integral 4 definite integrals with limits from 0 to 1 and the 4-variable function is ((1-2x) (1-y) (1-z) (1-w)) /(1-(1-(1-xy) z) w) . I mean this is a quadruple integral.
I transformed k into 2k+1 So I'm plugging in k=1,2,3.... And I'm getting 3,5,7.... Think of it like writing k=2n+1 but the dummy variable (index) written back as k
4:10 What does "there are no problems" and "we can in fact perform the switch up using Fubini's theorem" means? How does this Measure Theory theorem translate into this particular problem? Is this the absolute convergence (just a guess here) that allows that or something else? Why not to explain this properly? 😢
I prefer your videos solving integrals based on standard techniques. Pulling Kummer out of your hat like a magician does with a rabbit is not much more insightful than just giving the result of the integral in the first place. This type of videos makes you look smart (legitimately so!) but is of little instructive value to your viewers since you fill most of the 15 min with standard algebra while handwaiving over the actual integration problem …
