Generalizing $f(n)=\int_0^\infty \frac{1}{e^{x^n}+1}=\left(1-2^{(n-1)/n}\right )\zeta(n^{-1})\Gamma(1+n^{-1})$

Question

I have come up with the following solution to this integral, but is just incomplete to my standards $$f(n)=\int_0^\infty \frac{1}{e^{x^n}+1}=\left(1-2^{(n-1)/n}\right )\zeta(n^{-1})\Gamma(1+n^{-1})$$

Seems to only work for $x\in\Bbb{N},x\gt 2$

This identity, therefore does not apply to $n=1$, and we all know that $f(1)=\ln 2$ because $\zeta(1)$ diverges.

So my question is this: How can you generalize the integral solution I gave to fit the case $n=1$?

Answer 1

In fact the formula does apply if you perform the limiting procedure correctly. Notice that

$$ 2^{(n-1)/n}-1\sim_{n\rightarrow 1}(n-1)\log(2)+\mathcal{O}(n-1)^2 $$

and

$$ -\zeta(n^{-1})\sim_{n\rightarrow 1} \frac{1}{n-1}+\mathcal{O}(1) $$

Therefore we obtain

$$ -\zeta(n^{-1})(2^{(n-1)/n}-1)\sim_{n\rightarrow 1}\log(2)+\mathcal{O}(n-1) $$

which yields together with $\Gamma(2)=1$ the desired result!

Answer 2

$$ \begin{align} \int_0^\infty\frac{\mathrm{d}x}{e^{x^n}+1} &=\frac1n\int_0^\infty\frac{x^{\frac1n-1}\,\mathrm{d}x}{e^x+1}\\ &=\frac1n\int_0^\infty x^{\frac1n-1}\sum_{k=1}^\infty(-1)^{k-1}e^{-kx}\,\mathrm{d}x\\ &=\frac1n\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^{1/n}}\int_0^\infty x^{\frac1n-1}e^{-x}\,\mathrm{d}x\\[3pt] &=\frac1n\eta\left(\frac1n\right)\Gamma\left(\frac1n\right)\tag{1} \end{align} $$ where $\eta(s)$ is the Dirichlet eta function: $$ \begin{align} \eta(s) &=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}\\[6pt] &=\left(1-2^{1-s}\right)\zeta(s)\tag{2} \end{align} $$ and as you say, $\eta(1)=\log(2)$.

See this recent answer.