A special integral

We have -

$I = \int_{2006}^{2020} \frac {\ln^{i}(\psi^{i})}{\psi} d\psi = \int_{2006}^{2020} \psi^{-1} e^{\ln(\ln^{i}(\psi^{i}))} d\psi = \int _{2006}^{2020} \psi^{-1} e^{i \cdot \ln(i \cdot \ln\psi)} d\psi= \int_{2006}^{2020} \psi^{-1} e^{i \cdot (\ln(\ln\psi) + \frac {i\pi}{2})} d\psi$

$= e^{\frac {-\pi}{2}}\left(\int_{2006}^{2020} \psi^{-1} \cos(\ln(\ln\psi)) d\psi + i\int_{2006}^{2020} \psi^{-1} \sin(\ln(\ln\psi)) d\psi\right)$

By Euler's formula. Note that we can write $\ln(i\cdot \ln(\psi))$ in its polar form to take the natural log . Substituting $\beta = \ln(\psi)$ gives us $\psi d\beta = d\psi$ . So, our original integral after integrating by parts twice (on both integrals) simplifies to -

$= e^{\frac {-\pi}{2}} \left( \int_{2006}^{2020} \cos(\ln \beta) d\beta+ i \int_{2006}^{2020} \sin(\ln\beta) d\beta \right) = e^{\frac {-\pi}{2}} \left( \beta\left(\frac {\cos(\ln \beta) + \sin(\ln \beta)}{2}\right) \Bigg |_{\ln(2006)}^{\ln(2020)} +i\beta\left(\frac {\sin(\ln \beta) - \cos(\ln \beta)}{2}\right)\Bigg|_{\ln(2006)}^{\ln(2020)}\right)$ Evaluating both the functions at the given bounds and multiplying by $e^{\frac {-\pi}{2}}$ gives us -

$\boxed {I \approx 0.00064 + 0.0013i}$

Note -

Let $r$ be the length of the complex vector drawn from the origin to the complex number and $\theta$ be the angle it makes with the real axis. Then, $\ln(a + bi) = \ln(r\cos \theta + ir\sin \theta) = \ln(r(\cos \theta + i\sin \theta)) = \ln(r \cdot e^{i\theta}) = \ln(r) + i\theta$ . In our case, $r = \ln(\psi)$ and since real part is equal to zero, $\theta = \frac {\pi}{2}$ .