[VA] SRC #008 - 2021 is here !

[Albert Chan]

(01-07-2021 09:37 PM)Valentin Albillo Wrote:  

The funny thing about this integral is that its value remains the same if you replace 2.021 by any positive real value !!

Can you explain why ?

With the help of HP Prime emulator, here is the proof for k = 2:

I = ∫(ln(sin(x)^2/2 + (cos(x)+sin(x)/sqrt(2))^2) / sin(2x), x = 0. .. pi)
  = ∫(ln(1+sin(2x)/sqrt(2)) / sin(2x), x = 0. .. pi)

I = ∫(ln(1+sin(2y)/sqrt(2)) / sin(2y), y = -pi/2. .. pi/2)       -- y = pi/2-x, dy=-dx
  = ∫(2*atanh(sin(2y)/sqrt(2)) / sin(2y), y = 0. .. pi/2)       -- ln((1+x)/(1-x)) = 2*atanh(x)

I = ∫(atanh(sin(z)/sqrt(2)) / sin(z), z = 0. .. pi)                  -- z = 2*y, dz=2*dy
  = ∫(2*atanh(sin(z)/sqrt(2)) / sin(z), z = 0. .. pi/2)            -- sin(z) symmetry around pi/2

I = ∫(2*atanh(sqrt(2)*t/(1+t^2))/t, t = 0. .. 1)          -- t = tan(y/2), dt=(1+t^2)/2*dx

CAS> series(atanh(√2*t/(1+t^2))/(√2),t,0,18,polynom)

t -1/3*t^3-1/5*t^5+1/7*t^7+1/9*t^9 -1/11*t^11-1/13*t^13+1/15*t^15+1/17*t^17

2*atanh(√(2)*t/(1+t*t))/t
= √(8) * (1 - t^2/3 - t^4/5 + t^6/7 + t^8/9 - t^10/11 - t^12/13 + t^14/15 + t^16/17 - ...)

Integrate term by term, t from 0 to 1:

I = √(8) * (1 - 1/3² - 1/5² + 1/7² + 1/9² - 1/11² - 1/13² + 1/15² + 1/17² - ...)

Cas> s1 := simplify(sum(1/(2k-1)^2, k=1..inf))                       // pi^2/8
Cas> s2 := simplify(sum(1/(8k-1)^2 + 1/(8k+1)^2, k=1..inf)) // (√(2)*pi^2+2*pi^2-32)/32
Cas> simplify(sqrt(8) * (2 - s1 + 2*s2))                                  // I = pi^2/4

I can proof s1 by hand, but not s2 (any help appreciated)