Showing $\sum_{n=1}^\infty\frac1{n^2}\left(1+\frac1{2^2}+\cdots+\frac1{n^2}\right)=\frac{7\pi^4}{360}$ Help me calculate the amount in a simpler way:
\begin{aligned}
\mathcal{S}&=\sum\limits_{n=1}^{\infty }\frac{1}{n^2}\left ( 1+\frac{1}{2^2}+\ldots+\frac{1}{n^2}  \right )=\sum\limits_{n=1}^{\infty }\frac{H^{\left ( 2 \right )}_n}{n^2}=-\sum\limits_{n=1}^{\infty }H^{\left ( 2 \right )}_n\int\limits_{0}^{1}x^{n-1}\ln xdx=&&\\
        &=-\int\limits_{0}^{1}\frac{\ln x}{x}\sum\limits_{n=1}^{\infty }H^{\left ( 2 \right )}_nx^ndx=\int\limits_{0}^{1}\frac{\ln x}{x}\frac{\operatorname{Li}_2x}{1-x}dx=-\int\limits_{0}^{1}\left ( \frac{1}{x}+\frac{1}{1-x} \right )\ln x\operatorname{Li}_2xdx=&&\\
        &=-\int\limits_{0}^{1}\frac{\ln x\operatorname{Li}_2x}{x}dx-\int\limits_{0}^{1}\frac{\ln x\operatorname{Li}_2x}{1-x}dx=&&\\
        &=-\frac{\ln^2 \operatorname{Li}_2x}{2}\Bigg|_0^1-\frac{1}{2}\int\limits_{0}^{1}\ln^2 x\cdot \frac{\ln \left ( 1-x \right )}{x}dx+\ln x\operatorname{Li}_2x\ln \left ( 1-x \right )\Bigg|_0^1-&&\\
&-\int\limits_{0}^{1}\ln \left ( 1-x \right )\left ( \frac{\operatorname{Li}_2x}{x}-\frac{\ln x\ln \left ( 1-x \right )}{x} \right )dx=&&\\
        &=-\frac{1}{2}\int\limits_{0}^{1}\frac{\ln^2 x\ln \left ( 1-x \right )}{x}dx+\int\limits_{0}^{1}\frac{\ln^2 \left ( 1-x \right )\ln x}{x}dx+\int\limits_{0}^{1}\operatorname{Li}_2xd\left ( \operatorname{Li}_2x \right )=&&\\
        &=\frac{1}{2}\sum\limits_{n=1}^{\infty }\frac{1}{n}\int\limits_{0}^{1}x^{n-1}\ln^2 xdx+2\sum\limits_{n=1}^{\infty }\left ( \frac{H_n}{n}-\frac{1}{n^2} \right )\int\limits_{0}^{1}x^{n-1}\ln x+\frac{\operatorname{Li}^2_2x}{2}\Bigg|_0^1=&&\\
        &=\frac{1}{2}\sum\limits_{n=1}^{\infty }\frac{1}{n}\cdot \frac{2}{n^3}+2\sum\limits_{n=1}^{\infty }\left ( \frac{H_n}{n}-\frac{1}{n^2} \right )\left ( -\frac{1}{n^2} \right )+\frac{\pi^4}{72}=&&\\
        &=\zeta \left ( 4 \right )-2\sum\limits_{n=1}^{\infty }\left ( \frac{H_n}{n^3}-\frac{1}{n^4} \right )+\frac{\pi^4}{72}=\frac{\pi^4}{90}-2\left ( \frac{\pi^4}{72}-\frac{\pi^4}{90} \right )+\frac{\pi^4}{72}=\frac{\pi^4}{30}-\frac{\pi^4}{72}=\frac{7\pi^4}{360}
\end{aligned}

You don’t need any integrals.
$$S=\sum_i \sum_{j<=i} 1/(i^2 j^2)= \sum_i \sum_j 1/(2i^2 j^2) + \sum_i 1/(2i^4) =(\zeta(2)^2 +\zeta(4))/2= (\pi^4/36+\pi^4/90)/2=7\pi^4/360$$
