How would you tackle this series by real analysis?∑n=1∞∑m=1∞1m4(m2+n2)

Since:∑n≥11m2+n2=−1+mπcoth⁡(mπ)2m2our sum equals:−ζ(6)2+π2∑m≥1coth⁡(mπ)m5=−πζ(5)−ζ(6)2+π∑m≥11m5(e2mπ−1) (1)and maybe the last series has a nice closed form.Then:∑m≥11m5(e2mπ−1)=∑m≥1Li5(e−2πm)=12πi∫c−i∞c+i∞Γ(z)(2π)zζ(z+5)ζ(z)dzIf now we set f(z)=Γ(z)(2π)zζ(z+5)ζ(z), we have:Res(f(z), z=0)=−ζ(5)2Res(f(z), z=−1)=π5540Res(f(z), z=−3)=−π5540Res(f(z), z=−4)=2π43ζ′(−4)Res(f(z), z=−5)=−π51890hence a closed form for (1) just depends on a closed form for ζ′(−4), or, by the reflection formula, on a closed form for:ζ′(5)=−∑n≥1log⁡nn5

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Zaynah Dunn

Answered question

2022-02-22

How would you tackle this series by real analysis?

\sum_{n = 1}^{\infty} \sum_{m = 1}^{\infty} \frac{1}{m^{4} (m^{2} + n^{2})}

Answer & Explanation

Jett Brooks

Beginner2022-02-23Added 7 answers

Since:

\sum_{n \geq 1} \frac{1}{m^{2} + n^{2}} = \frac{- 1 + m π \coth (m π)}{2 m^{2}}

our sum equals:

- \frac{ζ (6)}{2} + \frac{π}{2} \sum_{m \geq 1} \frac{\coth (m π)}{m^{5}} = - \frac{π ζ (5) - ζ (6)}{2} + π \sum_{m \geq 1} \frac{1}{m^{5} (e^{2 m π} - 1)}

(1)
and maybe the last series has a nice closed form.
Then:

\sum_{m \geq 1} \frac{1}{m^{5} (e^{2 m π} - 1)} = \sum_{m \geq 1} L i_{5} (e^{- 2 π m}) = \frac{1}{2 π i} \int_{c - i \infty}^{c + i \infty} \frac{Γ (z)}{{(2 π)}^{z}} ζ (z + 5) ζ (z) d z

If now we set

f (z) = \frac{Γ (z)}{{(2 π)}^{z}} ζ (z + 5) ζ (z)

, we have:

R e s (f (z), z = 0) = - \frac{ζ (5)}{2}

R e s (f (z), z = - 1) = \frac{π^{5}}{540}

R e s (f (z), z = - 3) = - \frac{π^{5}}{540}

R e s (f (z), z = - 4) = \frac{2 π^{4}}{3} ζ^{'} (- 4)

R e s (f (z), z = - 5) = - \frac{π^{5}}{1890}

hence a closed form for (1) just depends on a closed form for

ζ^{'} (- 4)

, or, by the reflection formula, on a closed form for:

ζ^{'} (5) = - \sum_{n \geq 1} \frac{\log n}{n^{5}}

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