Simplifying logarithmic integral $Li(x)$

Question

I was wondering if someone could suggest me a hint on how to obtain the following expression $$ Li(x) = \int_{0}^x \frac{1}{\log y} dy = \int_{1}^x (1 - \frac{1}{ y}) \frac{1}{ \log y} dy + \log \log x + \gamma, $$ where $\gamma$ is the Euler's constant. It is in the book I am reading without explanation, and I couldn't quite get it to work when I tried. Thank you very much!

Answer 1

Hint. You may just observe that $$ (\log | \log y|)'=\frac{ 1/y}{\log y}=\frac1{y\log y} $$ using $$ (\log |u|)' = \frac{u'}{u}. $$ Then, $$ \begin{align} Li(x) &= \int_{a_0}^x \frac{1}{\log y} dy \qquad \qquad (a_0>e)\\\\ &=\int_{a_0}^x (1 - \frac{1}{ y}) \frac{1}{ \log y} dy+\int_{a_0}^x \frac{1}{ y} \frac{1}{ \log y} dy\\\\ &=\int_{a_0}^x (1 - \frac{1}{ y}) \frac{1}{ \log y} dy + \log \log x + C \end{align} $$ and take $C=-\log \log a_0+\gamma$.

Answer 2

If the equality between integrals would be true, then by splitting your second integral, one would have $$ \int_0^x\frac{1}{y\log y}\,dy=\log\log x+\gamma. $$ However, the integral $$ \int_0^x\frac{1}{y\log y}\,dy $$ diverges (the singularity at zero is too strong).