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

$\begin{array}{ll}{\displaystyle {\int }_0^{\infty }\left(1-\frac{x+2}{2x}\left(1-e^{-x}\right)\right)e^{-qx}\frac{\text{d}x}{x}=-1+\left(q+\frac{1}{2}\right)\text{ln}}\frac{q+1}{q}\hfill & \left[q>0\right]\hfill \end{array}$

BI (89)(23)

2. 

${\displaystyle {\int }_0^{\infty }\left(\frac{e^{-x}-1}{x}+\frac{1}{1+x}\right)}\frac{\text{d}x}{x}=C-1$

BI (92)(16)

3. 

$\begin{array}{ll}\left(e^{-px}-\frac{1}{1+a^2{x}^2}\right)\frac{\text{d}x}{x}=-C+\text{ln}\frac{a}{p}\hfill & \left[p>0\right]\hfill \end{array}$

BI (92)(11)

3.443

1. 

$\begin{array}{ll}{\displaystyle {\int }_0^{\infty }\left(\frac{e^{-x}{p}^2}{2}-\frac{p}{x}+\frac{1-e^{-px}}{x^2}\right)\frac{\text{d}x}{x}}=\frac{p^2}{2}\text{ln}p-\frac{3}{4}{p}^2\hfill & [p>0]\hfill \end{array}$

BI (89)(32)

2. 

$\begin{array}{ll}{\displaystyle {\int }_0^{\infty }\frac{{(1-e^{-px})}^n{e}^{-qx}}{x^3}\text{d}x=\frac{1}{2}{\displaystyle \sum _{k=2}^n{(-1)}^{k-1}\left(\begin{array}{l}n\hfill \\ k\hfill \end{array}\right){(q+kp)}^2\text{ln}(q+kp)}}\hfill & [n>2,q>0,pn+q>0]\hfill \end{array}$

(cf.4.268 4) BI (89)(30)

3. 

$\begin{array}{l}{\displaystyle {\int }_0^{\infty }{(1-e^{-p}}^{}}\hfill \end{array}$