Integration Bee Training for Advanced #12.6 - Constructive Summation

Memorable Timestamp:
00:27 - An Intro Demonstration
7:13 - Why we construct.
18:30 - Another Log Example
23:09 - An Oxford MAT Integral Example
27:03 - A Tricky Example

[@bra1nwave172]

34:26 x->1/x and then realize that we only account for intervals where the argument of the sine function goes from an odd integer multiple of pi to an even multiple of pi (floor equals -1). After that you have to recognize the form of the series.

[@bra1nwave172]

The last one is the hardest to do without constructed summation. First, we let x->1/x and u=2x. Thus the integrand takes the form 2⌊cos(π/2*u)⌋/u^2 with bounds from 2 to infinity. Next, we look for integer intervals that contribute to the sum. Unfortunately we have two sequences of sets: (4n+2, 4n+3) & (4k+1,4k+2) with n>=0 and k>=1. To make the sum evaluation less tedious I combine them noting that I have to remove the integral from 1 to 2 of 2⌊cos(π/2*u)⌋/u^2.
Now we have 1 plus the integral from 1 to infinity of 2⌊cos(π/2*u)⌋/u^2. Make the integral into the sum where n∈N_0 of the integral from 4n+1 to 4n+3 of -2/u^2 => 1+2*sum[1/(4n+3)-1/(4n+1)]. Here I expanded and rearranged to acquire: 1-2(1-1/3+1/5-1/7+1/9-1/11+...)=1-2(π/4)=1-π/2.

[@Silver-cu5up]

@@bra1nwave172 woah 0o0
Nice! Thats the correct answer!

[@evankniffen]

bro is cranking these out; best week ever

[@bra1nwave172]

18:04 I got there with the forced u-sub by letting the index n range from 0 to infinity and inverting the bounds of the sequence of integrals. That way the outside was comfortable and on the inside of the integral we just choose the lower bound for the floor function which was -(n+1) the rest was the same.

[@Silver-cu5up]

Ohhhhh, you did -(n+1). I couldnt figure out an easy way to do this.

[@joshpradhan3292]

Yeah this is definitely the hardest part of section 12.
Last q: I got -1/3 - 2(1/5 - 1/7 + 1/9 - 1/11 + ... )
= 1 - π/2 using arctanx Taylor series
Is this cooked or fire?

[@Silver-cu5up]

You got it!! Nice job!!! 🤌🔥

[@phat5340]

Don't wanna be that guy but at the end were you break up the integral don't you presuppose were it converges ie the reimann summation theorem

[@Silver-cu5up]

Im not tooooo familiar with the riemann summation theorem. Idk how this can be applied.