My First Problem

First of all, I just want to say that I love your problems, second I want to write a solution. Here we go.

By the definition of logarithms with base $b$ we know that $\log_b (a) = \frac{\ln a}{\ln b}$ . From this definition we can rewrite the integral as

$i \pi \int_{0}^{1} \frac{\ln x}{\ln i} \; dx \: \: \: \: \Rightarrow \: \: \: \: \frac{i \pi}{\ln i} \int_{0}^{1} \ln x \; dx$

From Complex Analysis we know that $\textrm{Log} (z) = \ln |z|+ i \textrm{Arg} (z), \; z \in \mathbb{C}$ . From this we can deduce that $\ln i = \frac{i \pi}{2}$ ; however, there are an infinite amount of arguments for this complex number, but I'll take the first principal solution ( $\ln i = \frac{i \pi}{2}, \: (n=1)$ ). Now the integral is

$2 \int_{0}^{1} \ln x \; dx$

By integration of parts we see that the anti-derivative of the natural logarithm is $x( \ln x -1 )$ . So the integral is now

$\int_{0}^{1} i \pi \log_i x \; dx \; = \; 2\left [ x( \ln x -1) \right ] |_{0}^{1}$

This creates problems when numerically evaluating this function. Here...

$2[1 \cdot ( \ln1-1 )- 0 \cdot ( \ln 0 -1)]$

I'd like to declare this as an undefined integral because of the $\ln 0 = \textrm{und. }$ but the assumption of the problem says that $0 \times \ln 0 =0$ so we then come to the conclusion from the above expression that the integral is

$\int_{0}^{1} i \pi \log_i x \; dx= \boxed{-2}$

$\int\limits_0^1{iπ log_i (x) dx}$

$= iπ \int\limits_0^1{log_i (x) dx}$

= $iπ[xlog_i (x)-\frac{x}{log (i)}]$

$i = {-1}^{\frac{1}{2}}$

$log (i) = \frac{1}{2} log (-1)$

From Euler's equation, $e^{iπ} + 1 = 0$ , $log (-1) = iπ$

$(iπ(1)log_i (1) - 2 \times (1)) - (iπ(0)log_i (0) - 2 \times (0))$

$= - 2$

Since areas are positive, $|-2| = 2$