Table of Integrals, Series, and Products, 8th Edition

$\begin{array}{ll}{\displaystyle {\int }_0^{\infty }\frac{x^{v-1}{e}^{-\mu x}}{1-e^{-bx}}\text{d}x=\frac{1}{b^v}\Gamma \left(v\right)\zeta \left(v,\frac{\mu }{b}\right)}\hfill & \left[\mathrm{Re}b>0,\mathrm{Re}\mu >0,\mathrm{Re}v>1\right]\hfill \end{array}$

ET I 144(10)

8. 

$\begin{array}{ll}{\displaystyle {\int }_0^{\infty }\frac{x^{n-1}{e}^{-px}}{1+e^x}\text{d}x=\left(n-1\right)!{\displaystyle \sum _{k=1}^{\infty }\frac{{\left(-1\right)}^{k-1}}{{\left(p+k\right)}^n}}}\hfill & \left[P>-1,n=1,2,\mathrm{\dots }\right]\hfill \end{array}$

ET I 144(10)BI (83)(9)

9. 

${\displaystyle {\int }_0^{\infty }\frac{x{e}^{-x}\text{d}x}{e^x-1}\text{d}x=\frac{{\pi }^2}{6}-1}$

(cf. 4.231 3)BI (82)(1)

10. 

$\begin{array}{ll}{\displaystyle {\int }_0^{\infty }\frac{x{e}^{-2x}\text{d}x}{e^{-x}+1}\text{d}x}\hfill & =1-\frac{{\pi }^2}{12}\hfill \end{array}$

(cf.4.251 6BI (82)(2)

11. 

$\begin{array}{ll}{\displaystyle {\int }_0^{\infty }\frac{x{e}^{-3x}}{e^{-x}+1}\text{d}x}\hfill & =\frac{{\pi }^2}{12}-\frac{3}{4}\hfill \end{array}$

(cf.4.251 5)BI (82)(3)

12.¹¹ 

${\displaystyle {\int }_0^{\infty }\frac{x{e}^{-\left(2n-1\right)x}}{1+e^{}}}$