Prove that $\int_0^\infty\frac{x^n}{1+e^{x-t}}\mathrm{d}x = \frac{t^{n+1}}{n+1} + o(t^n)$, when $t \to \infty,\,n\in\Bbb{R}^+$ I hae to prove that 
$$\int_0^\infty\frac{x^n}{1+e^{x-t}}\mathrm{d}x = \frac{t^{n+1}}{n+1} + o(t^n), \quad\text{ when } t \to \infty,\,n\in\Bbb{R}^+$$
where $o(\cdot)$ is the Little-o notation. 
What I have done so far:
Let $F_n(t) = \int_0^\infty\frac{x^n}{1+e^{x-t}}\mathrm{d}x$, then $F_0(t) =\int_0^\infty\frac{1}{1+e^{x-t}}\mathrm{d}x = \log(e^t+1)$. The derivative of $F_n(t)$ is:
\begin{align*}
F_n'(t) &= \int_0^\infty\frac{e^{x-t}x^n}{(1+e^{x-t})^2}\mathrm{d}x\\
&= -\int_0^\infty x^n\mathrm{d}\frac{1}{1+e^{x-t}}\\
&= \left.\frac{x^n}{1+e^{x-t}}\right|_0^\infty+\int_0^\infty \frac{1}{1+e^{x-t}}\mathrm{d}x^n\\
&= \int_0^\infty \frac{nx^{n-1}}{1+e^{x-t}}\mathrm{d}x\\
&= nF_{n-1}(t)
\end{align*}
 A: First, let's split up the integral:
$$
\begin{align}
\int_0^\infty\frac{x^n}{1+e^{x-t}}\mathrm{d}x
&=\int_{-t}^\infty\frac{(x+t)^n}{1+e^x}\mathrm{d}x\\
&=\color{#C00000}{\int_{-t}^0\frac{(x+t)^n}{1+e^x}\mathrm{d}x}+\color{#00A000}{\int_0^\infty\frac{(x+t)^n}{1+e^x}\mathrm{d}x}\tag{1}
\end{align}
$$
Note that the first integral on the right side of $(1)$ is
$$
\begin{align}
\color{#C00000}{\int_{-t}^0\frac{(x+t)^n}{1+e^x}\mathrm{d}x}
&=t^{n+1}\int_{-1}^0\frac{(1+x)^n}{1+e^{tx}}\mathrm{d}x\\
&=t^{n+1}\int_{-1}^0(1+x)^n\,\mathrm{d}x-\color{#0000FF}{t^{n+1}\int_{-1}^0\frac{(1+x)^n}{1+e^{tx}}e^{tx}\,\mathrm{d}x}\\
&=\frac{t^{n+1}}{n+1}+O\left(t^n\right)\tag{2}
\end{align}
$$
because
$$
\begin{align}
\color{#0000FF}{t^{n+1}\int_{-1}^0\frac{(1+x)^n}{1+e^{tx}}e^{tx}\,\mathrm{d}x}
&\le t^{n+1}\int_{-1}^0(1+x)^n\,e^{tx}\mathrm{d}x\\
&\le t^{n+1}\int_{-1}^0e^{nx}\,e^{tx}\mathrm{d}x\\
&\le\frac{t^{n+1}}{n+t}\\[6pt]
&=O\left(t^n\right)\tag{3}
\end{align}
$$
Furthermore, by dominated convergence
$$
\lim_{t\to\infty}\int_0^\infty\left(1+\frac xt\right)^n\,e^{-x}\,\mathrm{d}x
=1\tag{4}
$$
therefore, the second integral on the right side of $(1)$ is
$$
\begin{align}
\color{#00A000}{\int_0^\infty\frac{(x+t)^n}{1+e^x}\mathrm{d}x}
&\le t^n\int_0^\infty\left(1+\frac xt\right)^n\,e^{-x}\,\mathrm{d}x\\
&=O\left(t^n\right)\tag{5}
\end{align}
$$
Combining $(1)$, $(2)$, and $(5)$, we get
$$
\int_0^\infty\frac{x^n}{1+e^{x-t}}\mathrm{d}x
=\frac{t^{n+1}}{n+1}+O\left(t^n\right)\tag{6}
$$
A: We have:
$$ F_0(t) = \log(e^t+1) = t+\log(1+e^{-t}) = t+o(1) $$
and since your identity gives:
$$ F_n(t) = n\int_{0}^{t} F_{n-1}(u)\,du $$
we have:
$$ F_1(t) = \frac{t^2}{2}+o(t), $$
$$ F_2(t) = \frac{t^3}{3}+o(t^2), $$
and so on, so the claim holds by induction for any $n\in\mathbb{N}_{>0}$. Moreover, since:
$$ F_n(t) = \frac{t^{n+1}}{n+1}-\int_{0}^{t}\frac{(t-x)^n}{e^{x}+1}\,dx+\int_{0}^{+\infty}\frac{(t+x)^n}{e^x+1}\,dx\tag{1}$$
it is sufficient to estimate:
$$ \int_{0}^{+\infty}\frac{e^{nx/t}}{e^x+1}\,dx = \frac{t}{t-n}-\frac{t}{2t-n}+\frac{t}{3t-n}-\ldots=\log 2+O\left(\frac{n}{t}\right)$$
and prove that
$$ \lim_{t\to +\infty}\frac{1}{t^n}\int_{0}^{t}\frac{(t-x)^n}{e^x+1}\,dx=\lim_{t\to +\infty}\int_{0}^{t}\frac{(x/t)^n}{e^{t-x}+1}\,dx=\log 2,$$
that follows from the dominated convergence theorem, to have the claim for $n\in\mathbb{R}^+$, too.
