Inspired in an integral representation for the Dirichlet Eta function (that is the alternating series of the Riemann Zeta function) I've calculated some integrals using Wolfram Alpha.
Example 1. It is the integral representation for the Dirichlet Eta function, $$\int_0^\infty\frac{x^{s-1}}{e^x+1}dx=2^{-s}(2^s-2)\zeta(s)\Gamma(s)$$ for $\Re s>0$ (you can see this integral, for example, in the corresponding Wikipedia's article for the Dirichlet Eta function), here $\Gamma(s)$ is the Gamma function. Precisely this was my departure point in my explorations of this exercise, when one types in Wolfram Alpha online calculator the code:
integrate x^(s-1)/(e^x+1)dx from x= 0 to x= infty
and you can do a comparison with the first formulas in cited Wikipedia article.
After I've decided do littles changes to get different results, I've calculated some cases with this online calculator of $$\int_0^\infty \frac{x^{\lambda(s-1)}}{e^x+1}dx$$ for positive integers $\lambda\geq 1$. I' ve observed (the same observation seem true for negative integers) that seem easy to state a closed form for all those cases.
Example 2. For example one has $$\int_0^\infty\frac{x^{3(s-1)}}{e^x+1}dx=8^{-s}(8^s-8)\zeta(3s-2)\Gamma(3s-2)$$ for $\Re s>\frac{2}{3}$.
In this exercise is not required the most rigurous justification, are only required the calculations and a brief justification with an explanation of where is defined the identity.
I've tried the change of variable $u=x^{\lambda}$ for the general case, and my deduction was that the integral is $\frac{1}{\lambda}\int_0^\infty\frac{u^{s-1}}{e^{u^{1/\lambda}+1}}u^{\frac{1}{\lambda}-1}du$; my intuition say to me that perhaps only with a change of variable I can not solve the problem, I believe that I need different observations. I know that when one obtain the closed form, then one is able to discuss the convergence of the identity, that is where is defined, from the convergence of those factors in the identity.
Question . Can you state for example the case $\lambda=2$? Thus I am asking: Let $s=\sigma+it$ the complex variable, provide us the calculations to get $$\int_0^\infty\frac{x^{2s-2}}{e^x+1}dx=4^{-s}(4^s-4)\zeta(2s-1)\Gamma(2s-1),$$ for $\Re s>\frac{1}{2}$. Thanks in advance.
This is the code integrate x^(2s-2)/(e^x+1)dx from x= 0 to x= infty.
Also I don't know if it is easy to relate some case with $\lambda$ a positive integer with the identity for $\mu$ a negative integer, I say if using perhaps functionals equations, one can relate integrals corrosponding to positive $\lambda$'s with other integrals corresponding with negative integer (integer or rational?, I don't know) $\tilde{\lambda}$'s. If you want, you can add a comment about this last paragrpah. Thanks.
As soon as $\text{Re}(\alpha) > -1$ we have $$I(\alpha)=\int_{0}^{+\infty}\frac{x^\alpha}{e^x+1}\,dx = \sum_{n\geq 1}(-1)^{n+1}\int_{0}^{+\infty}x^\alpha e^{- n x}\,dx=\Gamma(\alpha+1)\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^{\alpha+1}} $$ hence $$ I(\alpha) = \Gamma(\alpha+1)\cdot\eta(\alpha+1)=\color{red}{\Gamma(\alpha+1)\cdot (1-2^{-\alpha})\cdot\zeta(\alpha+1)} $$ and your question is just related with the case $\alpha=\lambda(s-1)$. It is interesting to notice that by computing $I(0)=\log 2$ we have that $s=1$ is a simple pole of $\zeta(s)$ with residue $1$.