Finding Euler

n=1Hnnq=ζ(q)log2q2ζ(q+1)+k=1qη(k)η(qk+1) \sum_{n=1}^{\infty} \dfrac{\overline{H_{n}}}{n^q} = \zeta(q) \log 2 - \dfrac{q}{2} \zeta(q+1) + \sum_{k=1}^{q} \eta(k) \eta(q-k+1)

Prove the equation above.

Notation : Hn=k=1n(1)k1k\displaystyle \overline{H_{n}} = \sum_{k=1}^n \frac{(-1)^{k-1}}{k} denotes the Alternating Harmonic Number.

ζ(s)=k=11ks\displaystyle \zeta(s) = \sum_{k=1}^{\infty} \dfrac{1}{k^s} denotes the Riemann Zeta Function.

η(s)=k=1(1)k1ks\displaystyle \eta(s) = \sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k^s} denotes the Dirchlet Eta Function.

Note by Ishan Singh
3 years ago

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Since Hn=k=1n(1)k1k=011(x)n1+xdx\displaystyle \overline{H_n} = \sum_{k=1}^{n}\frac{(-1)^{k-1}}{k}=\int_{0}^{1} \frac{1-(-x)^n}{1+x}dx ,

Summing both sides we obtain n=1Hnnq=01ζ(q)1+xdx01Liq(x)1+xdx=ζ(q)ln201Liq(x)1+xdxI\displaystyle \sum_{n=1}^{\infty} \frac{\overline{H_n}}{n^q} = \int_{0}^{1}\frac{\zeta(q)}{1+x}dx - \int_{0}^{1}\frac{Li_q(-x)}{1+x}dx=\zeta(q)\ln 2 -\underbrace{\int_{0}^{1}\frac{Li_q(-x)}{1+x}dx}_{\color{#D61F06}{I}}

Now applying IBP we have, I=01Liq(x)1+xdx=[Liq(x)ln(1+x)]01+01Liq1(x)ln(1+x)xdx\displaystyle \color{#D61F06}{I} = \int_{0}^{1}\frac{Li_q(-x)}{1+x}dx = [Li_q(-x)\ln(1+x)]_{0}^{1}+\int_{0}^{1} \frac{Li_{q-1}(-x)\ln(1+x)}{x}dx

Since Liq(1)=η(q),η(1)=ln2\displaystyle Li_q(-1)=-\eta(q),\eta(1)=\ln 2 and applying IBP again,

I=η(q)η(1)[Liq1(x)Li2(x)]0101Liq2(x)Li2(x)xdx=η(q)η(1)η(q1)η(2)+01Liq2(x)Li2(x)xdx\displaystyle \color{#D61F06}{I} = -\eta(q)\eta(1)-[Li_{q-1}(-x)Li_2(-x)]_{0}^{1} - \int_{0}^{1} \frac{Li_{q-2}(-x)Li_2(-x)}{x}dx = -\eta(q)\eta(1)-\eta(q-1)\eta(2)+ \int_{0}^{1} \frac{Li_{q-2}(-x)Li_2(-x)}{x}dx

Similarly applying IBP pp times , we will get

I=r=1qη(qr+1)η(r)η(q+1)\displaystyle \color{#D61F06}{I}=-\sum_{r=1}^{q}\eta(q-r+1)\eta(r) -\eta(q+1)

Therefore , n=1Hnnq=ζ(q)ln2+k=1qη(qk+1)η(k)+η(q+1)\displaystyle \sum_{n=1}^{\infty} \frac{\overline{H_n}}{n^q}=\zeta(q)\ln2+\sum_{k=1}^{q}\eta(q-k+1)\eta(k)+\eta(q+1) for odd qq

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Note that when you apply IBP, every time the sign of the integral changes. So your formula will regenerate the integral only if qq is odd. In other words, your formula is correct, when qq is odd. For instance, I have solved the regular Euler Sum here, in a similar way before.

Ishan Singh - 3 years ago

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So I may assume that the result as quoted by you of this summation is for the even qq ?

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@Aditya Narayan Sharma No, it is for all qq, even or odd.

Ishan Singh - 3 years ago

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@Aditya Narayan Sharma There is another mistake in your formula, from what you have done, there should be a (1)k(-1)^k inside the summation

Ishan Singh - 3 years ago

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@Ishan Singh The formula for the alternate harmonic sum is fine .it's written (-x) and not x

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@Aditya Narayan Sharma As I said, there is a sign error in the second red line, it should be +01Liq2(x)Li2(x)xdx+ \int_{0}^{1} \dfrac{\operatorname{Li}_{q-2}(-x) \operatorname{Li}_2(-x)}{x} \mathrm{d}x and not 01Liq2(x)Li2(x)xdx- \int_{0}^{1} \dfrac{\operatorname{Li}_{q-2}(-x) \operatorname{Li}_2(-x)}{x} \mathrm{d}x (even if we take (-x)).

Ishan Singh - 3 years ago

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@Ishan Singh Thanks for that , tou said there should be a (1)k(-1)^k in your summation. Summation reffers only to the integral representation of the alternate harmonic number. But that's fine. I said there is a (x)(-x) in that formulae

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@Aditya Narayan Sharma I meant in the eta summation, there should be a (1)k(-1)^k in the sum. Compare with original Euler Sum in my answer here.

Ishan Singh - 3 years ago

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Also, there seems to be a sign error in the second red line, it should be + integral 0 to 1 ((Li)(q-2) (x) Li(2) (x))/x dx

Ishan Singh - 3 years ago

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