A wonderful floor integral connected to the Basel problem

[@BadlyOrganisedGenius]

Great video! If you want to try some more integrals similar to this one, you could try: Integral from 0 to infinity of {e^x} -1/2 That's the fractional part of e^x, subtract 1/2 or alternatively: integral from 0 to infinity of ({x}/x)^2 Both have very interesting solution developments!

[@maths_505]

Woah Awesome! I'll give em a shot

[@MathematicFanatic]

I did not do the substitution and looked at it as a sum of trapezoids of decreasing width from 0 to 1. This was much more elegant with the substitution because it is easy then to turn the floor into a sum. Very nice!

[@MrWael1970]

Really clever solution. Thanks for your fascinating video.

[@andy_lamax]

Beautiful one this one was. Sooo elegant

[@samning4141]

Pretty cool integral!

[@GreenMeansGOF]

Here’s how I would solve it: Write it as the sum from n=1 to infinity of integral from 1/(n+1) to 1/n of the function. Then the floor function part is just n. So each integral equals n(x^2/2) from 1/(n+1) to 1/n. This simplifies to (2n+1)/(2n(n+1)^2). After doing partial fractions I also get the telescoping series and zeta(2) term. EDIT: Rather than combining using common denominator, we can split the n/(n+1)^2 term similar to how you did in the video.

[@MathOrient]

Wow. Never had heard of Basel problem :)

[@jonsmith8579]

Math 505 I love you

[@suryamgangwal8315]

So the answer is eta(2)

[@Unidentifying]

bro check this, there might be some very mysterious monsters right there or in other forms en.wikipedia.org/wiki/Common_integrals_in_quantum_field_theory