- #1
mmzaj
- 107
- 0
I am looking for help with doing the following integral :
[tex]\frac{1}{2\pi i}\int_{1}^{\infty}\ln\left(\frac{1-e^{-2\pi i x}}{1-e^{2\pi i x}} \right )\frac{dx}{x\left(\ln x+z\right)}\;\;\;\;z\in \mathbb{C}[/tex]
i tried to transform it into a complex integral along a 'keyhole' contour, with a branch cut along the +ive real line [itex]\left[1,\infty\right)[/itex]. but then [itex]\;\ln x\;[/itex] would be transformed into [itex]\;\ln x+2\pi i\;[/itex] when doing the integral along [itex]\left(\infty,1\right]\;[/itex] which doesn't add up nicely to the portion along [itex]\left[1,\infty\right)[/itex]. any insights are appreciated.
[tex]\frac{1}{2\pi i}\int_{1}^{\infty}\ln\left(\frac{1-e^{-2\pi i x}}{1-e^{2\pi i x}} \right )\frac{dx}{x\left(\ln x+z\right)}\;\;\;\;z\in \mathbb{C}[/tex]
i tried to transform it into a complex integral along a 'keyhole' contour, with a branch cut along the +ive real line [itex]\left[1,\infty\right)[/itex]. but then [itex]\;\ln x\;[/itex] would be transformed into [itex]\;\ln x+2\pi i\;[/itex] when doing the integral along [itex]\left(\infty,1\right]\;[/itex] which doesn't add up nicely to the portion along [itex]\left[1,\infty\right)[/itex]. any insights are appreciated.