Not Catalan's Constant

$\begin{aligned} \sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^3} &=& 1 - \dfrac1{3^3} + \dfrac1{5^3} - \dfrac1{7^3} + \cdots \\ &=& \left( 1 + \dfrac1{5^3} + \dfrac1{9^3} + \dfrac1{13^3} + \cdots \right) - \left( \dfrac1{3^3} + \dfrac1{7^3} + \dfrac1{11^3} + \cdots \right) \\ &=& \sum_{n=0}^\infty \dfrac1{(4n+1)^3} - \sum_{n=1}^\infty \dfrac1{(4n-1)^3} \\ &=& \sum_{n=0}^\infty \dfrac1{(4n+1)^3} - \sum_{n=0}^\infty \dfrac1{(4n-1)^3} - 1 \\ &=& \dfrac 1{4^3} \left [ \sum_{n=0}^\infty \dfrac1{(n+1/4)^3} - \sum_{n=0}^\infty \dfrac1{(n-1/4)^3} \right ] - 1 \\ &=& \dfrac1{4^3} \left [-\dfrac{\psi_2(1/4)}{2} + \dfrac{\psi_2(-1/4)}{2} \right ] - 1\\ &=&\dfrac{\pi^3}{32} \end{aligned}$

For the last step reasoning is: use recursive formula for polygamma function and then use reflection formula for polygamma function and simplify.

Let $D_N$ be the square contour with vertices at $\epsilon N + i\eta N$ for $\epsilon,\eta = -1,1$ . The function $f(z) = \pi \sec \pi z$ is meromorphic, even and antiperiodic with (anti)period $1$ . Moreover $|f(z)| \le \pi$ for all $z \in D_N$ for all $N \in \mathbb{N}$ . Thus $\frac{1}{2\pi i}\int_{D_N} \frac{f(z)}{z^3}\,dz \; = \; \mathrm{Res}_{z=0} \frac{f(z)}{z^3} + \sum_{n=0}^{N-1}\left(\mathrm{Res}_{z=n+\frac12} + \mathrm{Res}_{z = -n-\tfrac12}\right) \frac{f(z)}{z^3}$ for any $N \in \mathbb{N}$ . Letting $N \to \infty$ , and using the evenness of $f(z)$ , we deduce that $0 \; = \; \mathrm{Res}_{z=0} \frac{f(z)}{z^3} + 2\sum_{n=0}^\infty\mathrm{Res}_{z=n+\frac12}\frac{f(z)}{z^3}$ But this tells us that $0 \; = \; \tfrac12\pi^3 - 2\sum_{n=0}^\infty \frac{(-1)^n}{(n+\frac12)^3}$ and hence that $\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3} \; = \; \tfrac{1}{32}\pi^3$ making the answer $3 +32 = \boxed{35}$ .