StackExchange Roulette Ep. 1

[@thederpycat]

I love this!!! ❤❤❤❤

[@carlosgiovanardi8197]

Interesting approach.

[@abc-nd2xt]

I solved the first integral like this. S = Sum (-1)^k /(2k+1)^3 k from 0 to inf and I is the integral. Did the same thing as you to get I = 8S. Then I = 2 Int (x^2)/cosh(x) from 0 to inf, substitute x = arcosh(t) = ln(t+sqrt(t^2-1)), dx=(t+sqrt(t^2-1))/(t^2-1+t*sqrt(t^2-1)) dt. So I = 2 Int (ln(t+sqrt(t^2-1))^2)*(t+sqrt(t^2-1))/(t*(t^2-1+t*sqrt(t^2-1))) dt t from 1 to inf. Now t+sqrt(t^2-1)=u, (u^2-1)/2=t^2-1+t*sqrt(t^2-1), t=(u^2+1)/(2u), dt=(u^2-1)/(2u^2)du. So I = 4 Int (ln(u)^2)/(1+u^2) du u from 1 to inf. Split the integral like int from 0 to inf - int from 0 to 1. The Integral from 0 to 1 is 2S, power series for 1/(1+u^2) and Int x^m (ln(x)^n) formula. The Integral from 0 to inf: f(t)= Int (u^(t-1))/(1+u^2) du from 0 to inf. Write 1/(1+u^2) as Sum (-1)^k/k! * gamma(k+1) * (-1)^k * cos(pi*k/2) u^k, use ramanujans master theorem, f(t)=pi/(2sin(pi*t/2)) calculate f''(t=1)=pi^3/8, so I = pi^3/2-8S and I=8S, we get S=pi^3/32 and I = pi^3/4