Evaluate the Integral \(\int_{0}^{1} \frac{1}{1-\log_{2}x} dx\)

How to evaluate
\int\limits_{0}^{1} \frac{1}{1-\log_{2}x} dx
was a question posed at Mathematics Stack Exchange. Here is my solution.

Make successive substitutions \(z=1-\frac{\ln x}{\ln2}, \,y=z\ln2\)

\int\limits_{0}^{1} \frac{1}{1-\log_{2}x} dx &= 2\ln2 \int\limits_{1}^{\infty} \frac{1}{z} \mathrm{e}^{-z\ln2} dz \\
&= 2\ln2 \int\limits_{\ln2}^{\infty} \frac{\mathrm{e}^{-y}}{y} dy \\
&=-(2\ln2)\mathrm{Ei}(-\ln2) \approx 0.52495

\mathrm{Ei}(z) = -\int\limits_{-z}^{\infty} \frac{\mathrm{e}^{-t}}{t} dt
is the exponential integral function.

Leave a Reply

Your email address will not be published. Required fields are marked *