Integral of a singularity function

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I am reading the "Network Science" book (http://barabasi.com/networksciencebook/) and formula (6.8) is: $$p_k \sim \int\limits_0^1 {d\eta \frac{{C^* }}{\eta }\frac{1}{{k^{1 + C^* /\eta } }}} \sim \frac{{k^{ - (1 + C^* )} }}{{\ln k}} \hspace{20 mm} (6 . 8)$$

I spent quite some some but still cannot derive the result as inverse logarithmic correction of power law.

Could someone help to answer how the result is derived and what math theory is needed in order to solve such type of integral?

Many thanks! Jim

ps. As I am not a math guy, can we call the function inside the integral singular?

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Observe you have

\begin{align} \int^1_0 d\eta\ \frac{C}{\eta}k^{-C/\eta} = \frac{1}{\ln k}\int^1_0 \frac{C \ln k}{\eta}e^{-C/\eta \ln k}. \end{align} Set $u = (C \ln k)/\eta$, then it follows \begin{align} \int^1_0 \frac{C \ln k}{\eta}e^{-C/\eta \ln k}=C\ln k\int^\infty_{C\ln k} \frac{e^{-u}}{u}\ du \leq \int^\infty_{C\ln k}e^{- u}\ du = \frac{1}{k^C}. \end{align} Thus, we have \begin{align} \int^1_0 d\eta\ \frac{C}{\eta}\frac{1}{k^{1+C/\eta}} \leq \frac{1}{k^{1+C}\ln k} \end{align}