If the integral ∫ 0 2 1 5 x tanh − 1 x d x = a b c 5 ( ln d − e c 2 F 1 ( q p , 1 , q r , s 1 ) ) where all the unknowns are positive integers with a , b , d are primes, then find the sum of all positive integers.
The original problem belongs to Sir Srinivasa Raghava where he proposed to evaluate the integral in closed form, ie; ∫ 0 2 1 5 x tanh − 1 x d x
Notation: 2 F 1 ( a , b ; c ; z ) is Gaussian hypergeometric function.
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Matlab brought 0.166555580713847 as the required answer.
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Lazy to do latex work :D so I'm just sharing the closed form that I found. ∫ 0 2 1 5 x tanh − 1 x d x = 5 2 1 9 5 ( ln 3 − 9 5 2 F 1 ( 1 0 9 , 1 ; 1 0 1 9 , 4 1 ) ) = 0 . 1 6 6 5 5 5 5 8 0 ⋯ Note that the integral can be represented in the form two separate sums that are convergent one of those is trival to evaluate the next involves the hypergeometric expression.