Compute the following integral $$\int_{-1}^{0} \! \frac{i}{1+it} \, \mathrm{d}t$$ using a determination of the complex logarithm.
Rewriting I get $$\int_{-1}^{0} \! \frac{i}{1+it} \, \mathrm{d}t = i\int_{-1}^{0} \! \frac{1}{1+it} \, \mathrm{d}t$$ So I need a function $F$ such that $$F' = \frac{1}{1+it}$$ Let $l_{\pi}$ be a such a determination of the logarithm in $\mathbb{C}/\mathbb{R}^-$. We have $l_{\pi}' = \frac{1}{z}$. Define $F(z) = l_{\pi}(1+z)$ and hence $F'(z) = \frac{1}{1+z}$.
Taking the integral $$i\int_{-i}^{0} \! \frac{1}{1+z} \, \mathrm{d}t = i(F(0) – F(-i)) = \\ = i(l_{\pi}(1) – l_{\pi}(1-i)) = -(Arg(1) – Arg(1-i)) = -\left(0-\left(-\frac{1}{4}\pi\right)\right) = -\frac{1}{4} \pi i$$
But according to online computation tools this solution is not correct. I'm not quite sure if I got the determination correct.
Tanks for any hints.
Best Answer
The denominator of the integrand varies form $1-i$ to $1$ meaning that we can write the $\text{Ln}$ function as following $$\ln z=\ln r+i \theta\quad,\quad \theta \in (-\pi,\pi]$$ therefore $$\int_{-1}^{0} \! \frac{i}{1+it} \, \mathrm{d}t=\ln 1-\ln (1-i)=-\ln \sqrt 2e^{-\frac{\pi}{4}}=-\dfrac{\ln 2}{2}+i\dfrac{\pi}{4}$$