Evaluating Logarithmic Integral with Cosine and Sine Integrals

[pseudogenius]

I was trying to evaluate this integral,


\int\frac{dx}{\ln x}


I substituted x=e^{i\theta} and I get,

\int\frac{e^{i\theta}}{\theta}d\theta

which is,

\int\frac{\cos \theta}{\theta}+i\frac{\sin \theta}{\theta} \ d\theta

\int\frac{\cos \theta}{\theta} \ d\theta+i\int\frac{\sin \theta}{\theta} \ d\theta

Ci(\theta)+i \ Si(\theta)

Ci(\theta) and Si(\theta) are the cosine and sine integrals, respectively.

therefore,

\int\frac{dx}{\ln x}=Ci(-i\ln x)+i \ Si(-i\ln x)


I was just asking if anybody has seen the logarithmic integral( li(x) ) expressed this way.

 

[phyzguy]

Interesting. The Mathematica functions site (http://functions.wolfram.com/GammaBetaErf/LogIntegral/27/01/0003/) lists the following identity:

\text{li}(z)=\text{Ci}(i \log (z))-i\text{ Si}(i \log (z))-\log (i<br /> \log (z))+\frac{1}{2} \left(\log (\log (z))-\log<br /> \left(\frac{1}{\log (z)}\right)\right)

This contains your two terms but has some added terms. I'm not sure why.

 

[hamster143]

Pay attention to limits of integration.

Your first integral is from 0 to x, but your substitution means that your second integral is now a contour integral over from -i \ln{x} to i\infty. So, you can't use cosine and sine integrals already. An added difficulty is that 1/ln(x) has a singularity at x=1, and, consequently, e^{i\theta}/{\theta} has a singularity at 0. You need to think how to do the contour integral so that the result matches the Cauchy principal value of your original integral.