Is a closed form of this sum possible?

Question

In attempting to find a closed form of the integral $$I(s)=\int_{0}^1\ln(1+x^s)dx$$ I came across this sum $$G(s)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{sn+1}$$ and was wondering if a closed form expression is attainable.

Answer 1

It can be shown that, with use of the digamma function, $$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{s \, n + 1} = \frac{1}{s + 1} + \frac{1}{2 s} \, \left( \psi\left(1 + \frac{1}{2s}\right) - \psi\left( \frac{3}{2} + \frac{1}{2s}\right) \right).$$

Answer 2

$$\int\ln{(1+x^8)}dx = x\ln(x^8+1)-\sin{\frac\pi8}\ln(x^2-2x\sin{\frac\pi8}+1)+\sin{\frac\pi8}\ln(x^2+2x\sin{\frac\pi8}+1)\\+\cos{\frac\pi8}\ln(x^2-2x\cos{\frac\pi8}+1)+\cos{\frac\pi8}\ln(x^2+2x\cos{\frac\pi8}+1)-8x\\-2\sin{\frac\pi8}\tan^{-1}\bigg(\csc\frac\pi8\cdot(x+\cos\frac\pi8)\bigg)+2\cos{\frac\pi8}\tan^{-1}\bigg(\sec\frac\pi8\cdot(x-\sin\frac\pi8)\bigg)\\+2\cos{\frac\pi8}\tan^{-1}\bigg(\sec\frac\pi8\cdot(x+\sin\frac\pi8)\bigg)$$

The point is, it exists, but is essentially useless.

Answer 3

Unfortunately it is not elementary. The series fits the definition of another function:

$$ G=\frac{\Phi(-1,1,1+1/s)}{s} $$

Where $\Phi$ is the Lerch transcedent.