Multiple integral related to Euler sums

Posted by Cornel Ioan Valean on Facebook

Show that
$$\small \int^1_0 \cdots \int^1_0 \frac{\mathrm{d}\varphi_1\cdots \mathrm{d}\varphi_5}{(\varphi_1+\varphi_2+\varphi_3+\varphi_4+\varphi_5+1)(\varphi_1+\varphi_2+\varphi_3+\varphi_4+\varphi_5+1)} = \frac{59}{32}\zeta(5)-\frac{1}{2}\zeta(2)\zeta(3)$$

$$\textit{proof}$$

Consider the integral
$$I = \int^1_0 \cdots \int^1_0 \frac{\mathrm{d}\varphi_1\cdots \mathrm{d}\varphi_5}{(\varphi_1+\varphi_2+\varphi_3+\varphi_4+\varphi_5+1)(\varphi_1+\varphi_2+\varphi_3+\varphi_4+\varphi_5+1)}$$

Use the series expansion
$$I = \sum_{n=0}^\infty \sum_{k=0}^\infty \frac{(-1)^{n+k}}{n+1} \int^1_0\cdots \int^1_0 (\varphi_1\cdots \varphi_4)^{n+k}\,\mathrm{d}\varphi_1\cdots \mathrm{d}\varphi_5$$

Which reduces to
$$I = \sum_{n=0}^\infty \sum_{k=0}^\infty \frac{(-1)^{n+k}}{n+1} \left( \int^1_0 \varphi^{n+k}\,\mathrm{d}\varphi\right)^4 $$

Hence we have the sum
$$\begin{align}I= \sum_{n=0}^\infty \frac{(-1)^n}{n+1} \sum_{i=n}^\infty \frac{(-1)^{i-n}}{(i+1)^4} &=\sum_{n=0}^\infty \frac{1}{(n+1)} \sum_{i=n}^\infty \frac{(-1)^{i}}{(i+1)^4} \\ &= \sum_{i=0}^\infty \frac{(-1)^i}{(i+1)^4}+\sum_{n=1}^\infty \frac{1}{(n+1)} \sum_{i=n}^\infty \frac{(-1)^{i}}{(i+1)^4} \\ &= \sum_{i=0}^\infty \frac{(-1)^{i}}{(i+1)^4} \sum_{n=0}^i\frac{1}{(n+1)}\\ &= \sum_{i=0}^\infty \frac{(-1)^{i}H_{i+1}}{(i+1)^4}\\&= \sum_{k=1}^\infty\frac{(-1)^{k+1} H_k}{k^4} \\&=\frac{59}{32}\zeta(5)-\frac{1}{2}\zeta(2)\zeta(3) \end{align}$$

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