How to integrate this "$\int_0^\infty \frac{\frac{5}{4}x^3-\frac{93}{32}x^2+\frac{95}{24}x}{1+e^{\beta (x-x_0)}}$" I do some project about  harmonic oscillator potential in QM and need to find this term, but I don't know how to integrate this one. When, $\beta >0$
 A: This is not an easy problem which requires special function since, for $\beta >0$ and $n \geq0$,
$$\int_0^\infty \frac{x^n}{1+e^{\beta  (x-a )}}\,dx=- \frac{\Gamma (n+1) }{ \beta ^{n+1}}\,\,\text{Li}_{n+1}\left(-e^{ \beta a}\right)$$
If $a=0$ this would give for the integral
$$\frac{190 \pi ^2 \beta ^2-2511 \beta  \zeta (3)+42 \pi ^4}{576 \beta ^4}$$
If the integral was for $a$ to $\infty$, letting $x=y+a$ would lead to integrals
$$\int_0^\infty \frac{y^n}{1+e^{\beta  y}}\,dy=\frac{ \left(2^n-1\right)  \zeta (n+1) \Gamma (n+1)}{2^n\,\beta^{n+1} }$$ and your numerator would be
$$\frac{1}{96} \left(120 a^3-279 a^2+380 a\right)+\frac{1}{48} \left(180 a^2-279
   a+190\right) y+\left(\frac{15 a}{4}-\frac{93}{32}\right) y^2+\frac{5
   }{4}y^3$$ and the result of the integral would be
$$\frac{a \left(120 a^2-279 a+380\right) \log (2)}{96 \beta }+\frac{\pi ^2 \left(180
   a^2-279 a+190\right)}{576 \beta ^2}+\frac{9 (40 a-31) \zeta (3)}{64 \beta
   ^3}+\frac{7 \pi ^4}{96 \beta ^4}$$
Edit
Concerning the term
$$\text{Li}_{n+1}\left(-e^{ \beta a}\right)$$ if $a$ is small, we could expand it as
$$-\left(1-2^{-n}\right) \zeta (n+1)-a \beta  2^{-n} \left(2^n-2\right) \zeta (n)-a^2
   \beta ^2 2^{-n-1} \left(2^n-4\right) \zeta (n-1)-\frac{1}{3} a^3 \left(\beta ^3
   2^{-n-1} \left(2^n-8\right) \zeta (n-2)\right)+O\left(a^4\right)$$
