Another Absolute BEAST!!! Introducing the Vardi Integral

A legendary battle with a legendary integral!

[@maths_505]

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.

[@asparkdeity8717]

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!

[@orionspur]

Wow. Shocking that any log-log-trig integral has any hope of resolving to a closed form.

[@martiribapons]

Shouldn't the 4th root of pi be inside the ln at the end? Anyway amazing video as always, such a beautiful result!

[@maxvangulik1988]

yeah

[@judecarter6095]

i was so sad when the euler macaronis cancelled 😢

[@shreyanshmehta5810]

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

[@meteor3033]

Yesss! @Maths 505

[@maths_505]

Sure

[@daddy_myers]

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!

[@asparkdeity8717]

Thank you Daddy Myers for making this video possible!

[@manstuckinabox3679]

10:03 This really feels like plugging in a cheat code, thank you papa Kummer!

[@maths_505]

But sir I know this means alot to you but you have to understand the budget constrain- WE'RE CONTOUR INTEGRATING THIS B*TCH INTO OBLIVION!!!!

[@manstuckinabox3679]

@@maths_505 ITS BEEN 9 DAYS THAT I'M TRYING TO!

[@manstuckinabox3679]

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

[@manstuckinabox3679]

I just realized how goofy this intire journey was. update: I changed contours, I'll be using the indented semi-circular one

[@jyotsanabenpanchal7271]

Awesome! 💯💯

[@MrWael1970]

In the last line, the fourth root is the denominator of gamma 3/4 not for natural logarithm. Thanks for your nice effort.

[@maths_505]

Yes you're right....I missed that in haste Thank you so much for such kind and positive feedback as always....it really means alot

[@anjanbiswas302]

Truly beautiful !!!

[@muzamilnazir3983]

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.

[@asparkdeity8717]

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!

[@maths_505]

See the pinned comment

[@Mephisto707]

11:34 plugging k = 1, we have sin 3pi/2 which is -1, so the answer can’t be (-1)^(k+1) for odd k.

[@maths_505]

I made a mistake writing out the kummer series. See the pinned comment

[@Mephisto707]

But then plugging k=5, we have sine 15pi/2, which is -1, while (-1)^(5+1) is 1.

[@maths_505]

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

[@hectorjosedelarosagutierre8998]

Legendary

[@scarletevans4474]

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? 😢

[@Sty5A467]

What about Integral of 1/(sin(x)+x) ?💀

[@Nathan_Drake707]

Yo sir whats good remember me? Its me hannan if you remember…Also keep up the good work i hope you have a mil subs!

[@maths_505]

Of course I remember bro Thanks mate Really means alot

[@bobingstern4448]

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?

[@maths_505]

Well this integral is in fact the integral representation of the eular masceroni constant so there's actually no need to evaluate it

[@asparkdeity8717]

Yes, watch Dr Peyam’s video on it

[@suvosengupta4657]

niceeeeeeee

[@maths_505]

SUIIIIIIIIIIIIIIII

[@Unidentifying]

can you do more vids with the gamma fn ?

[@maths_505]

I got a whole playlist and there are more coming up

[@Unidentifying]

@@maths_505 awesome bro

[@mahdielzein85]

What level calculus is this? Calc 4, analysis,?

[@maths_505]

You're gonna need some exposure to complex analysis for this....and by exposure I mean self teaching and researching.

[@rohitashwaKundu91]

Hey, can you recommend some good books for a freshman year Maths major?

[@maths_505]

Oh yes ofcourse Calculus by Thomas Advanced engineering mathematics by Erwin Kreyzig Linear algebra by Anton Mathematical methods by Boas

[@rohitashwaKundu91]

@@maths_505 Thanks man!😊

[@natepolidoro4565]

"structures"

[@maths_505]

Yeah I always call em that....I have a love for form....which explains why I'm in love with Gigi Hadid 😂

[@Walczyk]

11:44 limits are wrong here. k+1 is wrong

[@MGoebel-c8e]

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 …

[@erhanturker9325]

WHAT