How Do You Integrate tan(x)ln(x) Over the Complex Plane?

[Frogeyedpeas]

This thing is killing me! Integration Help

So I was doing some research and came across the following indefinite integral:


∫tan(x)ln(x) dx

where the domain of x is the complex plane so this can be re-written as:

∫tan(z)ln(z) dz...

So I began solving the problem like so:

∫tan(x)ln(x) dx = -ln(cos(x))ln(x) + ∫ln(cos(x))/x dx (integration by parts)

∫tan(x)ln(x) dx = -ln(cos(x))ln(x) + ∫ln((e^ix + e^-ix)/2)/x dx (exponential definition of cos)

∫tan(x)ln(x) dx = -ln(cos(x))ln(x) + ∫ln((e^ix + e^-ix))/x dx - ln(2)ln(x) + C (properties of logarithms)

∫tan(x)ln(x) dx = -ln(cos(x))ln(x) + ∫ln((e^ix)^2 + 1)/x dx - ix - ln(2)ln(x) + C (property of logarithms along with combining terms in the initial fraction)

∫tan(x)ln(x) dx = -ln(cos(x))ln(x) + ∫ln((e^ix + i)(e^ix - i))/x dx - ix - ln(2)ln(x) + C (expanding the sum of squares)

Which leaves us with this as the remaining problem:

∫tan(x)ln(x) dx = -ln(cos(x))ln(x) + ∫ln((e^ix + i))/x dx + ∫ln((e^ix - i))/x dx - ix - ln(2)ln(x) + C

So how on Earth do you solve these two problems:

∫ln((e^ix + i)/x dx

∫ln(e^ix - i))/x dx

I tried using Wolfram Mathematica but it could not integrate this problem and I can't imagine how a discrete method would work on this.

 

[jackmell]

Maybe it doesn't have an elementary antiderivative. Can still do a nice job integrating it numerically as long as you go around the poles and branch point.

 

[Nebuchadnezza]

I believe strongly that this integral does not have a elementary antiderivative.

Although is it possible to find an exact value for the integral below? (no approximations)

I = \int_{0}^{1} \tan(x) \ln(x) \, dx