More generalized on the previous question: A improper integral with Glaisher-Kinkelin constant Show that : $\displaystyle\int_0^\infty\frac{\text{e}^{-ax}}{x^2}\left(\frac{1}{1-\text{e}^{-x}}-\frac{1}{x}-\frac12\right)^2\text{d}x=\frac{a}{36}\left(3+a^2+3\left(3-6a+2a^2\right)\ln a\right)+\zeta '\left(-1,a\right)-2\zeta '\left(-2,a\right) \\$ $\displaystyle\int_0^\infty\text{e}^{-ax}\left(\frac{1}{1-\text{e}^{-x}}-\frac{1}{x}\right)^3\text{d}x=\frac{23}{24}-\frac{13a}{4}+\frac{3a^2}{2}-\frac32(a-1)\ln 2+\frac{a^2}{2}\ln a+\frac32(1-a)\ln \pi+3(a-1)\ln \Gamma(a)-\frac12(2-3a+a^2)\psi(a)-6\zeta'(-1,a)$ Assuming $\displaystyle a>0$.
Asked
Active
Viewed 130 times
3
-
1Woosh that's ugly. – L. F. Feb 18 '13 at 02:15