Integral $\int_\tau^\infty e^{\frac{-g_m}{\bar\gamma_m}}\frac{dg_m}{1+Pg_m}$

Question

$$I=\int_\tau^\infty e^{\frac{-g_m}{\bar\gamma_m}}\frac{dg_m}{1+Pg_m}$$

As you know exponential integral define in [0 inf], but I want to calculate it in [thu inf]. I'm really appreciating everyone who can help me solve it. tnx

Answer 1

$$I=\int_\tau^\infty e^{-g/\gamma}\frac{dg}{1+Pg}$$ Let $g=\frac{x-1}{P}$ $$I=\frac{e^{1/\gamma P}}{P}\int_{1+\tau P}^\infty e^{-x/\gamma P}\frac{dx}{x}$$ Let $a=\frac{1}{\gamma P}$ supposing $\gamma>0$ and $P>0$ $$I=\frac{e^\alpha}{P}\int_{1+\tau P}^\infty e^{-a x}\frac{dx}{x}$$ $$I=\frac{e^\alpha}{P} \left[ \text{Ei}(-a x) \right]_{x=1+\tau P}^{x=\infty}$$ Ei is the special function named "Exponential Integral" : http://mathworld.wolfram.com/ExponentialIntegral.html

$\text{Ei}(-\infty)=0$ $$I=\frac{e^\alpha}{P} \left( 0-\text{Ei}\left(-a (1+\tau P)\right) \right)$$

$$\int_\tau^\infty e^{-g/\gamma}\frac{dg}{1+Pg} = - \frac{e^{1/(\gamma P)}}{P} \text{Ei}\left(-\frac{1+\tau P}{\gamma P}\right)$$