$\sum\limits_{m\geq1}\sum\limits_{n\geq1}\frac{(-1)^n}{n^3}\sin\left(\frac{n}{m^2}\right)=\frac{\pi^6}{11340}-\frac{\pi^4}{72}$ Numerical evidence

Question

I am looking for numerical evidence that $$\sum_{m\geq1}\sum_{n\geq1}\frac{(-1)^n}{n^3}\sin\left(\frac{n}{m^2}\right)=\frac{\pi^6}{11340}-\frac{\pi^4}{72}$$ I have proven it, but I just want to be extra sure. Desmos only gives me accuracy to the third decimal place, but I know that some of you (@Claude Leibovici) are able to give me extremely high decimal accuracy.

Proof:

We know that for $|t|\leq\pi$, $$t^2=\frac{\pi^2}3+4\sum_{n\geq1}\frac{(-1)^n}{n^2}\cos nt$$ Solving for the sum then integrating both sides from $0$ to $x$, $$\sum_{n\geq1}\frac{(-1)^n}{n^3}\sin(nx)=\frac{x^3}{12}-\frac{\pi^2x}{12}$$ Then plugging in $x=\frac1{m^2}$ for integer $m\geq1$, $$\sum_{n\geq1}\frac{(-1)^n}{n^3}\sin\bigg(\frac{n}{m^2}\bigg)=\frac{1}{12m^6}-\frac{\pi^2}{12m^2}$$ then applying $\sum_{m\geq1}$ on both sides, $$\sum_{m\geq1}\sum_{n\geq1}\frac{(-1)^n}{n^3}\sin\bigg(\frac{n}{m^2}\bigg)=\frac1{12}\zeta(6)-\frac{\pi^2}{12}\zeta(2)$$ We simplify to reach our conclusion $$\sum_{m\geq1}\sum_{n\geq1}\frac{(-1)^n}{n^3}\sin\bigg(\frac{n}{m^2}\bigg)=\frac{\pi^6}{11340}-\frac{\pi^4}{72}$$

Answer 1

After you so elegant solution, I am quite ashamed to have done it the following way.

$$\sum_{n=1}^\infty\frac{(-1)^n}{n^3}\sin(nx)=\frac{i}{2} \left(\text{Li}_3\left(-e^{-i x}\right)-\text{Li}_3\left(-e^{i x}\right)\right)$$

Replace $x=\frac 1 {m^2}$ and numerically compute $$\frac{i}{2}\sum_{m=1}^\infty \left(\text{Li}_3\left(-e^{-\frac i {m^2} }\right)-\text{Li}_3\left(-e^{\frac i {m^2}}\right)\right)$$

It takes a long time to arrive to $-1.268125453640218644419$.

Update

I watched the news, I read the newspaper, I had a heavy breakfast and came back to my computer $$-1.26812545364021864441879478985797582963162172461540$$

Edit

I tried to understand why this computation was taking so much computing time.

Let us consider $$a_m=\frac{i}{2} \left(\text{Li}_3\left(-e^{-\frac i {m^2} }\right)-\text{Li}_3\left(-e^{\frac i {m^2}}\right)\right)$$ For large values of $m$, we have $$a_m=-\frac{\pi ^2}{12 m^2}+\frac{1}{12 m^6}+O\left(\frac{1}{m^{14}}\right)$$ So $$\sum_{m=1}^\infty a_m\simeq\sum_{m=1}^p a_m+\sum_{m=p+1}^\infty\left(-\frac{\pi ^2}{12 m^2}+\frac{1}{12 m^6} \right)=\sum_{m=1}^p a_m+\frac{\psi ^{(5)}(p+1)-120 \pi ^2 \psi ^{(1)}(p+1)}{1440}$$ and the last term is almost $$\frac{\psi ^{(5)}(p+1)-120 \pi ^2 \psi ^{(1)}(p+1)}{1440}\simeq -\frac{\pi^2}{12p}\left(1-\frac{1}{2 p}+\frac{1}{6 p^2}+O\left(\frac{1}{p^4}\right) \right)$$ In other words, for $k$ significant figures, we need to add a lot terms.

This explains that.