Estimate of complex integral Prove that
$$ 
\left|\int_c (2-\frac{e^z}{z-\log 2}) dz \right|  <\frac{2}{3}
$$
when C is the part of circle $\left| \frac{z}{\pi} -1 \right|^2 =2$ where $Re(z)\geq 0$.
($\log$ means natural logarithm)
This question was part of my exam couple of weeks ago.
My attempt was to go far left in complex plane and then estimate the integral using trianlg equality. However the singularity at $\log 2$ is a bit tricky for me because I've only done estimates like this with integrals that are analytic in the contour.
 A: I don't see a direct application of the standard estimates, so let's use a deviation. The circle $\lvert z-\pi\rvert = \pi\sqrt{2}$ intersects the imaginary axis in $\pm \pi i$. Let $\gamma$ be a (piecewise smooth) path in the left half-plane connecting $-\pi i$ and $\pi i$; we shall see later what kind of path might be convenient. Then $C-\gamma$ is a (piecewise smooth) closed path winding around $\log 2$ once, so
$$\int_{C-\gamma} 2 - \frac{e^z}{z-\log 2}\,dz = -2\pi i e^{\log 2} = -4\pi i.$$
Also, since $2$ has the primitive $2z$, we have
$$\int_\gamma 2\,dz = [2z]_{-\pi i}^{\pi i} = 4\pi i,$$
and therefore
$$\int_C 2 - \frac{e^z}{z-\log 2}\,dz = -4\pi i + \int_\gamma 2 - \frac{e^z}{z-\log 2}\,dz = -\int_\gamma \frac{e^z}{z-\log 2}\,dz.$$
It remains to choose $\gamma$ in such a way that
$$\left\lvert\int_\gamma \frac{e^z}{z-\log 2}\,dz\right\rvert < \frac{2}{3}$$
can be shown. For example, the polygonal path with vertices $-\pi i,\, -R-\pi i,\, -R +\pi i, \pi i$ for large enough $R$ does that.
