m-1-n-1-1-n-1-m-2-n-mn-2-

Question Number 195952 by mnjuly1970 last updated on 13/Aug/23

Ω = \underset{m = 1}{\sum^{\infty}} \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1}}{m^{2} n + {m n}^{2}} = ?

- - - - -

Answered by witcher3 last updated on 14/Aug/23

\frac{1}{mn (n + m)} = \frac{1}{m^{2}} (\frac{1}{n} - \frac{1}{n + m})

A = \sum_{m ⩾ 1} \frac{1}{m^{2}} \sum_{n ⩾ 1} \frac{{(- 1)}^{n - 1}}{n} = \frac{π^{2} \ln (2)}{6}

B = \sum_{m ⩾ 1} \frac{1}{m^{2}} \sum_{n ⩾ 1} \frac{{(- 1)}^{n - 1}}{n + m}

= \sum_{m ⩾ 1} \frac{1}{m^{2}} \sum_{n ⩾ 1} {(- 1)}^{n - 1} \int_{0}^{1} x^{n + m - 1} dx

= \int_{0}^{1} Σ \frac{x^{m}}{m^{2}} \sum_{n ⩾ 1} {(- x)}^{n - 1} dx

= \int_{0}^{1} {Li}_{2} (x) . \frac{dx}{1 + x}

= \int_{0}^{1} \frac{{Li}_{2} (x)}{1 + x} = \ln (2) ζ (2) + \int_{0}^{1} \frac{\ln (1 - x) \ln (1 + x)}{x} dx

= \ln (2) ζ (2) + \frac{1}{4} \int_{0}^{1} \frac{{(\ln (1 - x^{2}))}^{2} - \ln^{2} (\frac{1 - x}{1 + x})}{x} {dx}_{= I}

I = \frac{1}{8} \int_{0}^{1} \frac{\ln^{2} (1 - t)}{t} dt - \frac{1}{4} \int_{0}^{1} \frac{{2 \ln}^{2} (t)}{(1 - t^{2})} + ζ (2) \ln (2)

= \frac{1}{8} \int_{0}^{1} \frac{\ln^{2} (t)}{1 - t} dt - \frac{1}{2} \int_{0}^{1} \frac{\ln^{2} (t)}{1 - t^{2}} dt + ζ (2) \ln (2)

\int_{0}^{1} t^{n} \ln^{2} (t) dt = \frac{Γ (3)}{{(n + 1)}^{3}}

= \frac{2}{8 {(1 + n)}^{3}} - \sum_{n ⩾ 0} \frac{1}{{(1 + 2 n)}^{3}} + \frac{π^{2} \ln (2)}{6}

= \frac{1}{4} ζ (3) - (\frac{7}{8} ζ (3)) = - \frac{5}{8} ζ (3)

Ω = A - B = \frac{π^{2}}{6} \ln (2) - (\frac{π^{2}}{6} \ln (2) - \frac{5}{8} ζ (3))

\sum_{n, m ⩾ 1} \frac{{(- 1)}^{n - 1}}{{nm}^{2} + {mn}^{2}} = \frac{5}{8} ζ (3)

Answered by qaz last updated on 14/Aug/23

Ω = \underset{m, n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1}}{n m (n + m)} = \underset{m, n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1}}{n m} \int_{0}^{1} x^{n + m - 1} d x

= - \int_{0}^{1} \frac{l n (1 + x) l n (1 - x)}{x} d x = \frac{5}{8} ζ (3)

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