# Logarithmic Integral

1. Apr 11, 2010

### 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.

2. Apr 11, 2010

### phyzguy

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

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

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

3. Apr 11, 2010

### 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.