- #1
pseudogenius
- 7
- 0
I was trying to evaluate this integral,
[tex]\int\frac{dx}{\ln x}[/tex]
I substituted [tex]x=e^{i\theta}[/tex] and I get,
[tex]\int\frac{e^{i\theta}}{\theta}d\theta[/tex]
which is,
[tex]\int\frac{\cos \theta}{\theta}+i\frac{\sin \theta}{\theta} \ d\theta[/tex]
[tex]\int\frac{\cos \theta}{\theta} \ d\theta+i\int\frac{\sin \theta}{\theta} \ d\theta[/tex]
[tex]Ci(\theta)+i \ Si(\theta)[/tex]
[tex]Ci(\theta)[/tex] and [tex]Si(\theta)[/tex] are the cosine and sine integrals, respectively.
therefore,
[tex]\int\frac{dx}{\ln x}=Ci(-i\ln x)+i \ Si(-i\ln x)[/tex]
I was just asking if anybody has seen the logarithmic integral( [tex]li(x)[/tex] ) expressed this way.
[tex]\int\frac{dx}{\ln x}[/tex]
I substituted [tex]x=e^{i\theta}[/tex] and I get,
[tex]\int\frac{e^{i\theta}}{\theta}d\theta[/tex]
which is,
[tex]\int\frac{\cos \theta}{\theta}+i\frac{\sin \theta}{\theta} \ d\theta[/tex]
[tex]\int\frac{\cos \theta}{\theta} \ d\theta+i\int\frac{\sin \theta}{\theta} \ d\theta[/tex]
[tex]Ci(\theta)+i \ Si(\theta)[/tex]
[tex]Ci(\theta)[/tex] and [tex]Si(\theta)[/tex] are the cosine and sine integrals, respectively.
therefore,
[tex]\int\frac{dx}{\ln x}=Ci(-i\ln x)+i \ Si(-i\ln x)[/tex]
I was just asking if anybody has seen the logarithmic integral( [tex]li(x)[/tex] ) expressed this way.