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.

  • $\begingroup$ Maybe i have to construct a closed path of integration. out of $ \gamma(a)=0, \gamma(b)=1$ ? $\endgroup$ – 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.

  • $\begingroup$ 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 $? $\endgroup$ – Steven33 Nov 24 '18 at 10:26
  • $\begingroup$ Dont i have to enclose my singularities with $ \Gamma$ $\endgroup$ – Steven33 Nov 24 '18 at 10:39
  • $\begingroup$ @Steven33 Please see my edited answer. $\endgroup$ – Szeto Nov 24 '18 at 12:03
  • $\begingroup$ 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.:) $\endgroup$ – Steven33 Nov 24 '18 at 12:42
  • $\begingroup$ Thank you very much:) You made my day. Thank you:) I made a horrible mistake. $\endgroup$ – Steven33 Nov 24 '18 at 13:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.