# Nonlinear Euler sums using Nielsen formula

According to Nielsen we have the following :

If $$f(x)= \sum_{n= 0}^\infty a_n x^n$$

Then we have the following

$$\tag{1}\int^1_0 f(xt)\, \mathrm{Li}_2(t)\, dx=\frac{\pi^2}{6x}\int^x_0 f(t)\, dt -\frac{1}{x}\sum_{n=1}^\infty \frac{a_{n-1} H_{n}}{n^2}x^n$$

Now let $a_n = H_n$ then we have the following

$$f(x)=\sum_{n=1}^\infty H_n x^n=-\frac{\log(1-x)}{1-x}$$

$$-\int^1_0 \frac{\log(1-xt)}{1-xt} \mathrm{Li}_2(t)\, dx=-\frac{\pi^2}{6x}\int^x_0 \frac{\log(1-t)}{1-t} dt-\sum_{n=1}^\infty \frac{H_{n-1} H_{n}}{n^2}x^{n-1}$$

Hence we have the following by gathering the integrals and $x\to 1$

$$\sum_{n=1}^\infty \frac{H_{n-1} H_{n}}{n^2}=\int^1_0\frac{\log(1-x)\left(\mathrm{Li}_2(x)-\zeta(2)\right)}{1-x} dx$$

Integrating by parts we have

$$\sum_{n=1}^ \infty \frac{H_{n-1} H_{n}}{n^2}=-\frac{1}{2}\int^1_0\frac{\log(1-x)^3}{x} dx$$

Hence we have

$$\sum_{n=1}^\infty\frac{ H^2_{n}}{n^2}=\sum_{n=1}^\infty \frac{ H_{n}}{n^3}-\frac{1}{2}\int^1_0\frac{\log(1-x)^3}{x} dx=\frac{17 \pi^4}{360}$$

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