Integrating $\int_0^{\infty}\log_2(1+q)\exp(-\beta q) \,\mathrm{d}q$ I am trying to solve the following equation but not getting it correctly?
$$C = \int_0^{\infty}\log_2(1+q)\exp(-\beta q) \,\mathrm{d}q\tag{1}$$
where $\beta$ is constant and $q$ is integration variable.
Any help in this regard will be highly appreciated...
 A: Recall that $\log_2(x)= \displaystyle\frac{\ln(x)}{\ln(2)}$.
Integrating by parts:
$$\int\ln(1+q)e^{-\beta q} \,\mathrm{d}q = -\frac{1}{\beta}\ln(1+q)e^{-\beta q} + \int \frac{1}{\beta} \frac{e^{-\beta q}}{1+ q} dq = \\
= -\frac{1}{\beta}\ln(1+q)e^{-\beta q} + e^{\beta}\int  \frac{e^{-\beta (1+q)}}{\beta(1+ q)} dq.$$
Using the integration extrema:
$$\int_0^{+\infty}\ln(1+q)e^{-\beta q} \,\mathrm{d}q =
 \left[-\frac{1}{\beta}\ln(1+q)e^{-\beta q}\right]_{q=0}^{q=+\infty} + e^{\beta}\int_0^{+\infty}  \frac{e^{-\beta (1+q)}}{\beta(1+ q)} dq = \\
=
 [0-0] + e^{\beta}\int_0^{+\infty}  \frac{e^{-\beta (1+q)}}{\beta(1+ q)} dq.
$$
By substituting $s = \beta(1+q)$, we get:
$$\int_0^{+\infty}  \frac{e^{-\beta (1+q)}}{\beta(1+ q)} dq = \frac{1}{\beta}\int_{\beta}^{+\infty}\frac{e^{-s}}{s}ds = -\frac{1}{\beta}\mathrm{Ei}(-\beta),$$
where $\mathrm{Ei}(\cdot)$ is known as exponential integral.
Hence:
$$C =  -\displaystyle\frac{e^{\beta}}{\beta\ln(2)}\mathrm{Ei}(-\beta).$$
Unfortunately, since $\mathrm{Ei}$ cannot be explicitly computed (it is a special function). Then, numerical evaluation is required.
