Problem#3.

In the "An interesting integral with the floor function."-video the youtuber claims that the value of the integral is $I=1-2 \ln (2)$, which is a negative value: $$I=\int_0^1 (-1)^{\left\lfloor \frac{1}{x}\right\rfloor } \ dx=\ldots =1-2 \ln (2)\approx -0.386294$$ Imho the Wolfram L code for this expression should be:

\[CapitalIota] = Integrate[(-1)^Floor[1/x], {x, 0, 1}] (* //N  out=0.4375 *)

However, Mathematica outputs the integral uncomputed and the NIntegrate[] function returns the positive value $0.4375$. WolframAlpha outputs $0.4375$ for the value of the integral, too. Did I make a user error or is our software indeed wrong?

Raspi:

On the 2nd problem.

In my day job writing and optimizing SQL, I run into interesting situations like this all the time. The direct and logical approach just doesn't work, or doesn't work in a reasonable amount of time.

In those situations, I have found that I can sometimes get what I need by first getting more than I need and then filtering, What amazes me most is that sometimes I can get what I need sub-second using the round-about approach when the direct and logical approach can run for hours.

Using this indirect approach, I think this solves this problem:

 Select[
 FindInstance[{(*OddQ[y]==True,*)0 <= x <= 100, 0 <= y <= 100, 
   1/x + 4/y == 1/12}, {x, y}, Integers, 100], 
 OddQ[#[[2]][[2]]] &
 ]

Of course, I would agree with anyone who said that this approach is cheating. But, it has helped me overcome challenges that others gave up on.

Hi Raspi,

Seems like FindInstance requires numerical equations, booleans do not work. This works

FindInstance[{0 <= x <= 100, 0 <= y <= 100, 1/x + 4/y == 1/12, Mod[y, 2] == 1}, 
  {x, y}, Integers, 100]
(* {{x -> 76, y -> 57}} *)

Table[NProduct[((1 + x^(n + 1))/(1 + x^n))^(x^n), {n, 0, Infinity}, 
  WorkingPrecision -> 15], {x, .99999999, 1, .000000001}]

looks like it goes to 2/e.