Prove-that-p-1-q-1-1-pq-p-q-1-pi-2-3-

Question Number 89936 by ~blr237~ last updated on 20/Apr/20

P r o v e t h a t \sum_{p ⩾ 1, q ⩾ 1} \frac{1}{p q (p + q - 1)} = \frac{π^{2}}{3}

Answered by maths mind last updated on 20/Apr/20

\sum_{p, q ⩾ 2} \int_{0}^{1} \frac{x^{p + q - 2}}{p q} d x

= \int_{0}^{1} (\sum_{q ⩾ 1} \frac{x^{q - 1}}{q}) (\sum_{p ⩾ 1} \frac{x^{p - 1}}{p}) d x

= \int_{0}^{1} (\frac{{l n}^{2} (1 - x)}{x^{2}}) d x

l n (1 - x) = - u

x = 1 - e^{- u}

= \int_{0}^{+ \infty} \frac{u^{2} e^{- u}}{{(1 - e^{- u})}^{2}}

\frac{1}{1 - e^{- u}} = \sum_{k ⩾ 0} e^{- k u}

\Rightarrow \frac{e^{- u}}{{(1 - e^{- u})}^{2}} = \sum_{k ⩾ 1} {k e}^{- k u}

= \sum_{k ⩾ 1} \int_{0}^{+ \infty} {k u}^{2} e^{- k u} d u

= \sum_{k ⩾ 1} \int_{0}^{+ \infty} \frac{w^{2} e^{- w}}{k^{2}} d w

= \sum_{k ⩾ 1} \frac{1}{k^{2}} . \int_{0}^{\infty} w^{2} e^{- w} d w = Γ (3) . \sum_{k ⩾ 1} \frac{1}{k^{2}} = Γ (3) ζ (2) = 2 . \frac{π^{2}}{6} = \frac{π^{2}}{3}