I was working on the following Integral:
$$I=\int_0^1\ln^2(t)\psi^8(-t)dt=\frac{7}{8}\zeta(3)$$
Where, $$\psi(q)=\sum_{n=-\infty}^{\infty}q^{n(2n+1)}$$ is a Ramanujan Theta Function.
Now it is well known that, $$q\psi^8(q)=\sum_{n=1}^{\infty}n^3\left[\frac{q^n}{1-q^{2n}}\right]$$
Using this we have,
$$I=-\sum_{n=1}^{\infty}(-1)^nn^3\int_0^1\ln^2(t)\frac{t^{n-1}}{1-t^{2n}}dt$$
And, $$\int_0^1\ln^2(t)\frac{t^{n-1}}{1-t^{2n}}dt=\frac{2}{n^3}\sum_{r=0}^{\infty}\frac{1}{(2r+1)^3}$$
Therefore, $$I=-2\left(\sum_{n=1}^{\infty}(-1)^{n}\right)\left(\sum_{r=0}^{\infty}\frac{1}{(1+2r)^{3}}\right)$$
Now I am not exactly sure what to do with the Summation on the Left, if we had to assign it a value then we could give using Geometric Series Formula, $$\sum_{n=1}^{\infty}(-1)^{n}=-\frac{1}{2}$$
Then it's proven that,
$$I=\frac{7}{8}\zeta(3)$$
I am not sure about that last step as the series switches between $-1$ and $0$.
After rechecking the steps I am confused how this situation arises.