Is there a closed form for the integral $\int_{0}^{\infty} \frac{x^{n}}{e^{x}+1} d x$ where $n\in R^+.$

Question

When I first met the integral $\int_{0}^{\infty} \frac{x}{e^{x}+1} d x$, I started to investigate it in terms of power series. After evaluating the integral using integration by parts, I write a post in Quora

$$ \int_{0}^{\infty} \frac{x}{e^{x}+1} d x =\frac{1}{2}\zeta(2)= \frac{\pi^{2}}{12}. \tag{1} $$

After a while, I encountered another post which use one more partner integral $$\int_{0}^{\infty}\frac{x^{2}}{e^{x}-1}dx $$and found that\begin{equation} \int_{0}^{\infty}\frac{x^{2}}{e^{x}+1}dx=\frac{3}{2}\zeta\left( 3\right) \tag{2} \end{equation}

These two results urges me to generalize it as

$$\int_{0}^{\infty}\frac{x^{n}}{e^{x}+1}dx=K\zeta\left( n+1\right) \tag{3} $$

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Proof: \begin{aligned} I_{n} &=\int_{0}^{\infty} \frac{x^{n} e^{-x}}{1+e^{x}} d x \\ &=\int_{0}^{\infty} e^{-x} x^{n} \sum_{k=0}^{\infty}(-1)^{k} e^{-k x} d x\\&= =\sum_{k=0}^{\infty}(-1)^{k} \underbrace{\int_{0}^{\infty} x^{n} e^{-(k+1) x} d x}_{J_{k}} \end{aligned}

Applying integration by parts on $J_n$ yields a reduction formula

$$ \begin{aligned} J_{k} &=-\frac{1}{k+1} \int_{0}^{\infty} x^{n} d\left(e^{-(k+1) x}\right) \\ &=-\frac{1}{k+1}\left[\frac{x^{n}}{e^{(k+1) x}}\right]_{0}^{\infty}+\frac{n}{k+1} \int_{0}^{\infty} x^{n-1} e^{-(k+1) x} d x \\ &=\frac{n}{k+1} J_{k-1} \end{aligned} $$

Using the formula $n$ times gives $$ J_{k}=\frac{n}{k+1} \cdot \frac{n-1}{k+1} \cdots \frac{1}{k+1} J_{0}=\frac{n !}{(k+1)^{n+1}} $$ Then we can generalize successfully that $$ \begin{aligned} I_{n} &=\sum_{k=0}^{\infty} \frac{(-1)^{k} n !}{(k+1)^{n+1}} \\ &=n ! \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{n+1}} \\&=\boxed{\left(1-\frac{1}{2^{n}}\right) n !\zeta(n+1)} \end{aligned} $$

My Question

How about when $n$ is not a natural number? Your opinions and alternate solutions are warmly welcome.

Answer 1

After submitting the question, I feel that I can use the Gamma function instead of power series and get the answer faster.

Converting the integral $$J_{k} =\int_{0}^{\infty} x^{n} e^{-(k+1) x} d x$$

into a Gamma Function by letting $t=(k+1) x$, then

\begin{aligned}J_{k} &=\int_{0}^{\infty}\left(\frac{t}{k+1}\right)^{n} e^{-t} d t \\ &=\frac{1}{(k+1)^{n}} \int_{0}^{\infty} t^{n} e^{-t} d t \\ &=\frac{1}{(k+1)^{n}} \Gamma(n+1) \end{aligned} Hence $$ \boxed{I_n=\left(1-\frac{1}{2^{n}}\right) \Gamma(n+1) \zeta(n+1),} $$ where $n$ is not necessarily a natural number but $\Re(n)>0$.