One of the best integrals I've solved

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[@aidenmcdonald5605]

9:40 should have been 4x^4 on the bottom but it's still correct because x=1

[@PRABALBAISHYA-xi1fd]

We could also Feynman's technique here. Consider a new integral as sin(ax)/(e^x-1). Notice that our result integral is the derivative of the new integral at a=1. The new integral can now be solved which yields π/2(coth(πa)-1/πa). ( Here the pole expansion of cothz comes in handy).

[@AB-nu5we]

Factors of 2 and minus signs. Sounds like a physics thing. Cool integral, cool result.

[@nicobianchi1477]

WoW, a really cool one, keep it up man, loving the uploads, could you do something involving matrices?

[@bartekabuz855]

When I saw the miniature I was thinking how to solve it. And you did exaclty how I thought, nice!

[@maths_505]

Awesome

[@zunaidparker]

Not a big fan of the new thumbnail style vs the OG. Also, I still think you should indore the graph of the function in the thumbnail, would be cool to visualize what we're actually tackling.

[@natepolidoro4565]

Agree

[@giuseppemalaguti435]

Σ[(2n-1)^2-1]/[(2n-1)^2+1]^2+Σ(4n^2-1)/(4n^2+1)^2...n=1,2...ho usato semplicemente la serie geometrica e integrazione per parti,ma non riesco a raggrupparle... comunque sembra corretto

[@Qrudi234]

I thought it had something to do with the integration rep. Of zeta(1). Their integrands are very similar in structure

[@kazagucci]

I had a slightly different solution to this one. At 4:15, you can rewrite the sum as Re(ζ(2,1-i)) which equals Re(ψ(1,1-i)). Then apply the identity Re(z) = (z + z*)/2 to get (ψ(1,1-i) + ψ(1,1+i))/2 and rewrite ψ(1,1+i) as -1/i² + ψ(1,i). Then we can apply the digamma reflection formula (differentiating first) to get (1 + π²csc²(πi))/2 = (1 - π²csch²(π))/2.

[@random22453]

I thought of using the Zeta function but I'm not into complex analysis yet so don't really know how Zeta function and digamma function are related

[@manstuckinabox3679]

we had the same method until evaluating the sum wilst you use a precise and interesting method, my method was to give up and see what you did; this is truely a really awesome integral, and now you hyped me up to watch that sum video!

[@nicolascamargo8339]

Genial

[@random22453]

Pls make a marathon video on contour integrals complex analysis

[@danielc.martin]

I expected the lemniscate constant, but great, though xd

[@MrWael1970]

Thanks for your effort.

[@AlexanderMontes-hr6xv]

what program/app are you using?

[@CM63_France]

Hi, Awesome! Is there a reduced form for the sum of the inverse of the squares of the primes numbers? That is : sum{p=2,p prime}^infty { 1 / p^2 } ? That is : 1/4 + 1/9 + 1/25 + 1/49 + 1/121 + ...

[@maths_505]

I'm afraid not bro

[@CM63_France]

@@maths_505 Ok thanks so I just stop studying this subject 😆

[@PI...--_.----_....-]

​​@@maths_505Hey, I noticed something this sum appears to approach 1/π^ln(2). What do you think?

[@maths_505]

@@PI...--_.----_....- do you have a rigorous proof?

[@maths_505]

If so then that would be amazing

[@MGoebel-c8e]

I fail to see the point in your newer videos. The core idea is delivered as a fact out of nothingness (or in a link to another video after watching 2/3 of the current one) and the rest is arithmetics albeit growing in complexity. Few ideas, few ingenuity. That’s not how you started your channel. Few of your watchers are viewing your videos in sequence, so referrals to your past oeuvre makes less sense than it feels for you.

[@crendler9912]

The references are needed to prevent the video from being repetitive. Also integrals are often about transforming the function into something that can be worked with. The part about too few ideas does not make sense either. Many integrals are similar and use the same techniques. I agree that the concept to solve is delivered out of nothingness, but he does show how it makes sense. Maybe he could do a slightly better job, but I think he does it fine. Finally if these videos are not for you there are countless math videos that you can find.

[@spiderjerusalem4009]

Calm down, andreescu. You're a genius, we get it

[@altissimo4158]

It must have been 4x⁴ downstairs, but does not matter for 1.

[@GeoPeron]

My dumbahh got to the sum and called it a day 😭🫠