Solving $\int_{\partial \mathcal{D}(0,\frac{1}{2})} \dfrac{\log(1+z)}{(4z-1)^2}dz$ in the context of complex analysis

Question

We are given the integral \begin{align} \int_{\partial \mathcal{D}(0,\frac{1}{2})} \dfrac{\log(1+z)}{(4z-1)^2}dz \end{align} where $\partial D(0,1/2)$ is the normal counterclockwise circle center $0$ and radius $1/2$.

How can we compute it using Complex Analysis techniques?

Using the obvious parameterization of a circle in complex analysis, where we set \begin{align} \gamma(t) = a + re^{i \phi} \overbrace{\mapsto}^{a=0, \ r=1/2} \dfrac{e^{i \phi}}{2}, \ \phi \in \left[ 0,2\pi \right]. \end{align}

Putting it though on the original and after some computations yield \begin{align} \int_{0}^{2\pi} \dfrac{\log \left( \dfrac{2+e^{i\phi}}{2} \right)}{(2e^{i\phi}-1)^2}d\phi \end{align} which obviously does not solve in closed form.

Any ideas?

Final answer: I took the inspiration from @Bongo's post and also this post.

I am trying to bring the original integral to the form \begin{align} \dfrac{\Phi(z)}{(z-z_0)^m} \end{align} where $\Phi(z) = \dfrac{\log (1+z)}{16}$ and the only pole at $z_0 = 1/4$ order $m=2$.

We already know that the residue at given point is \begin{align} \mathrm{Res}_{z=z_0}(f) = \dfrac{\Phi^{(m-1)}(z_0)}{(m-1)!}. \end{align}

For $m=2$ and $z_0 = 1/4$ we can have \begin{align} \Phi'(z) = \dfrac{1}{16} \left( \dfrac{1}{1+z} \right) \overbrace{\mapsto}^{z_0=1/4} \dfrac{1}{20}. \end{align}

So by the residue theorem \begin{align} \int_{\partial \mathcal{D}(0,\frac{1}{2})} \dfrac{\log(1+z)}{(4z-1)^2}dz &= 2\pi i \sum \left( \text{residues} \right)\\ &= 2\pi i \dfrac{1}{20}\\ &= \dfrac{\pi i}{10}. \end{align}

Answer 1

With the Cauchy's differentiation formula the integral it's solved immediately: $$f'(a)=\frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{(z-a)^2}\text dz.$$ log($1+z$) is holomorphic in $\mathcal D(0,\frac{1}{2}+\varepsilon)$. $$\oint _{\partial \mathcal D(0,\frac{1}{2})}\frac{\log(1+z)}{(4z-1)^2}=\frac{2\pi i}{2\pi i}\oint _{\partial \mathcal D(0,\frac{1}{2})}\frac{\log(1+z)}{16(z-\frac{1}{4})^2}=$$ $$=\frac{2\pi i}{16}\cdot \frac{1}{2\pi i}\oint _{\partial \mathcal D(0,\frac{1}{2})}\frac{\log(1+z)}{(z-\frac{1}{4})^2}=\frac{2\pi i}{16}\cdot \left( log(1+a) \right)_{a=1/4}'=$$ $$=\frac{2\pi i}{16}\cdot\left( \frac{1}{1+a} \right)_{a=1/4}= \frac{2\pi i}{16}\cdot\frac{4}{5}=\frac{\pi i}{10}.$$

Answer 2

You just write the function as $\dfrac{log(1+z)}{16(z-\dfrac{1}{4})^{2}}$.

We set $z-\dfrac{1}{4}=w$ and we expand the

$log(1+z)=z-\dfrac{z^{2}}{2}+\dfrac{z^{3}}{3}-..... $and setting

$z=w+\dfrac{1}{4}$ we get the coefficient of $\dfrac{1}{w}$ (the Residue) as

$\dfrac{1}{16}(1-\dfrac{1}{4}+\dfrac{1}{4^{2}}-...)=\dfrac{1}{16}.\dfrac{4}{5}=\dfrac{1}{20}.$

By the residue theorem, since the pole is in $(0,1/2)$ we get

$I=2\pi\,i\,\dfrac{1}{20}=\dfrac{\pi\,i}{10}$