Finer asymptotic estimate of an integral I'm studying the asymptotic behaviour for large $n\in \mathbb N$ of $\displaystyle \int_1^\infty \frac{1}{1+t^{n+1}}$
Using the substitution $u=(n+1)\ln(t)$, $$\displaystyle \int_1^\infty \frac{1}{1+t^{n+1}}=\frac{1}{n+1}\int_0^\infty\exp(\frac{u}{n+1})\frac{1}{1+\exp(u)}du$$
And, using dominated convergence, one has $\displaystyle \int_0^\infty\exp(\frac{u}{n+1})\frac{1}{1+\exp(u)}du=\ln(2)+o(1)$, so that $$\displaystyle \int_1^\infty \frac{1}{1+t^{n+1}}=\frac{\ln(2)}{n}+o(\frac{1}{n})$$
I'm looking to refine that estimate, that is to say get a sharper estimate of $\displaystyle \int_1^\infty \frac{1}{1+t^{n+1}}-\frac{\ln(2)}{n}$.
It amounts to finding an estimate of $$\int_0^\infty \frac{1}{1+e^u}(\exp(\frac{u}{n+1})-1)du$$ but that is harder.
 A: As Lucian remarked,
$$ \int_0^\infty \dfrac{dt}{1+t^{n+1}} = \dfrac{\pi}{n+1} \csc\left(\dfrac{\pi}{n+1}\right) $$
That is asymptotically 
$$ 1+{\frac {{\pi }^{2}}{6{n}^{2}}}-{\frac {{\pi }^{2}}{3{n}^{3}}
}+{\frac {{\pi }^{2} \left( 7\,{\pi }^{2}+180 \right) }{360\,{n}^{4}}}
 + \ldots$$
Now we have to subtract the integral from $0$ to $1$.  For $0 < t < 1$, 
$$ \eqalign{\int_0^1 \dfrac{dt}{1+t^{n+1}} &= \int_0^1 \sum_{k=0}^\infty (-1)^k t^{k(n+1)} \; dt\cr
&= \sum_{k=0}^\infty \dfrac{(-1)^k}{1+k(n+1)}\cr
&= \dfrac{\Psi\left(\dfrac{n+2}{2(n+1)}\right) - \Psi\left(\dfrac{1}{2(n+1)}\right)}{2(n+1)}\cr
&= 1-{\frac {\ln  \left( 2 \right) }{n}}+{\frac {{\pi }^{2}+12\ln 
 \left( 2 \right) }{12{n}^{2}}}+{\frac {-2\,{\pi }^{2}-9\,\zeta 
 \left( 3 \right) -12\;\ln  \left( 2 \right) }{12{n}^{3}}}+\dfrac{7\,{\pi }^{4}+180\,{\pi }^{2}+720\,\ln  \left( 2 \right) +1620\,\zeta 
 \left( 3 \right)
}{720\;{n}^{4}}+\ldots
}$$
So the final result is
$$ {\frac {\ln  \left( 2 \right) }{n}}+{\frac {{\pi }^{2}-12\,\ln 
 \left( 2 \right) }{12{n}^{2}}}+{\frac {-2\,{\pi }^{2}+12\,\ln 
 \left( 2 \right) +9\,\zeta  \left( 3 \right) }{12{n}^{3}}}+{\frac {7\,
{\pi }^{4}+180\,{\pi }^{2}-720\,\ln  \left( 2 \right) -1620\,\zeta 
 \left( 3 \right) }{720\,{n}^{4}}}+\ldots
$$
