# Cauchy's integral formula with special contour 4

suppose $$\gamma: [a,b] \rightarrow \mathbb{C}$$ is a path of integration with $$\gamma(a)=0, \gamma(b)=1 \ and \ \pm i \notin\gamma([a,b])$$ Show that, $$\int_{\gamma} \frac{1}{1+z^2} = \frac{\pi}{4} + k \pi$$

I would try to apply Cauchy's integral formula. Therefore i can split the integrand in partial fractions with singularities at $$\pm i$$ How do i have to choose my contour such that it fullfilles the conditions at the beginning.

• Maybe i have to construct a closed path of integration. out of $\gamma(a)=0, \gamma(b)=1$ ? – Steven33 Nov 24 '18 at 9:51

Let $$\Gamma$$ be another path: a straight line connecting from $$1$$ to $$0$$.

By residue theorem (let $$n$$ and $$m$$ be the winding numbers around $$z=+i$$ and $$z=-i$$ respectively),

\begin{align} \oint_{\gamma+\Gamma}\frac{dz}{1+z^2} &=2\pi i \left(n\operatorname*{Res}_{z=i}\frac1{1+z^2}+ m\operatorname*{Res}_{z=i}\frac1{1+z^2}\right) \\ &=\pi(-n+m) \\ &=\pi(m-n) \end{align}

Since $$\int_{\Gamma}\frac{dz}{1+z^2}=\int^0_1\frac{dt}{1+t^2}=\operatorname{arctan}t\bigg\vert^{t=0}_{t=1}=-\frac{\pi}4$$

Thus, $$\int_{\gamma}\frac{dz}{1+z^2}=\frac{\pi}4+(m-n)\pi$$

Doing without residue theorem:

Firstly, recognize that $$\int\frac{dz}{1+z^2}=\frac1{2i}\int\left(\frac1{z+i}-\frac1{z-i}\right)dz=\frac{\ln(z+i)-\ln(z-i)}{2i}$$

Thus, $$\int\frac{dz}{1+z^2}= \frac{\ln(z+i)-\ln(z-i)}{2i} \bigg\vert^{z=1}_{z=0}$$

Considering the multi-value-ness of $$\ln$$ you would obtain the desired result.

In essence, \begin{align} \frac{\ln(z+i)-\ln(z-i)}{2i}\bigg\vert^{z=1}_{z=0} &=\frac{\ln(1+i)-\ln(1-i)+\ln(i)-\ln(-i)}{2i}\\ &=\frac{\ln(\sqrt2e^{i(\pi/4+2m\pi)})-\ln(\sqrt2e^{i(-\pi/4+2n\pi)})}{2i}\\ &~~~~+\frac{\ln(e^{i(\pi/2+2p\pi)})-\ln(e^{i(-\pi/2+2q\pi)})}{2i}\\ &=i\frac{\pi/4+\pi/4+\pi/2+\pi/2+2N\pi}{2i}\\ &=\frac{\pi}4+N’\pi\\ \end{align} where $$N’$$ is an arbitrary integer.

• I am not allowed to uses the residue theorem. Therefore i wanted to split the integrand into partial fractions. How do u choose $\gamma_1, \gamma_2$? – Steven33 Nov 24 '18 at 10:26
• Dont i have to enclose my singularities with $\Gamma$ – Steven33 Nov 24 '18 at 10:39
• @Steven33 Please see my edited answer. – Szeto Nov 24 '18 at 12:03
• Thank you:) Do i have to parameterize the straight line connecting from 0 to 1 and put that into my definition of the contour integral, meaning: z=t $\forall t \in [0,1]$. Ok i see that makes not such a big difference.:) – Steven33 Nov 24 '18 at 12:42
• Thank you very much:) You made my day. Thank you:) I made a horrible mistake. – Steven33 Nov 24 '18 at 13:00