I get different answers when I try to evaluate the following integral when I use partial fraction decomposition and Cauchy's Integral Formula. The integral is, $$\color{blue}{\int _{-2}^{2}\frac{1}{z^4+1}dz}.$$
I would greatly appreciate any help finding where I go wrong, and what possible misunderstandings I have about either method. Thanks!
Here's my attempt of Cauchy's Integral Formula. First, I started by saying that, $$\color{blue}{\int _{-2}^{2}\frac{1}{z^4+1}dz} + \color{green}{\int _{C_R}\frac{1}{z^4+1}dz}=\color{red}{\int _{C}\frac{1}{z^4+1}dz},$$ where $C_R$ is the contour of the circle of radius 2 and center (0,0) from $z=2$ to $z=-2$ and $C$ is $C_R$ + the line segment from $z=-2$ to $z=2$. If we find the integrals on $C_R$ and $C$ then we can find the desired integral (in blue). First then, we can say that, $$\color{red}{\int _{C}\frac{1}{z^4+1}dz} = \int _{C_1}\frac{1}{z^4+1}dz + \int _{C_2}\frac{1}{z^4+1}dz,$$ where $C_1$ is goes from $z=2i$ to $z=0$ to $z=2$ and then along the circle of radius 2 back to $z=2i$, while $C_2$ is from $z=-2$ to $z=0$ to $z=2i$ and then along the circle back to $z=-2$.
First, the integral over $C_1$ can be calculated with Cauchy's Integral Formula like so: $$f_1(z_1) = \frac{1}{2\pi i}\int _{C_1} \frac{f_1(z)}{z-z_1}dz,$$ where $z_1=e^{i\pi/4}$ and $$f_1(z) = \frac{1}{(z-e^{3i\pi/4})(z-e^{4i\pi/4})(z-e^{7i\pi/4})}.$$ (We find $f_1(z)$ from that $f_1(z)/(z-z_1) = 1/(z^4+1)$.) Therefore, the value of the integral is, \begin{align*} 2\pi i f_1(z) &= 2\pi i \cdot \frac{1}{(\sqrt{2})(\sqrt{2}+\sqrt{2}i)(\sqrt{2}i)} \\ &= 2\pi i \cdot \frac{-i\sqrt{2}-\sqrt{2}}{(\sqrt{2})(4)(\sqrt{2})} \\ &= 2\pi i \cdot \frac{-1}{8}(1+i)\sqrt{2} \\ &= -\pi \frac{1}{4}(i-1)\sqrt{2}. \\ \end{align*} Similarly, we can calculate the integral around $C_2$ as, $$f_2(z_2) = \frac{1}{2\pi i}\int _{C_2} \frac{f_2(z)}{z-z_1}dz,$$ where $z_2=e^{3i\pi/4}$ and $$f_2(z) = \frac{1}{(z-e^{i\pi/4})(z-e^{4i\pi/4})(z-e^{7i\pi/4})}.$$ Thus the value of the integral over $C_2$ is, \begin{align*} 2\pi i f_2(z) &= 2\pi i \cdot \frac{1}{(-\sqrt{2})(\sqrt{2}i)(-\sqrt{2}+\sqrt{2}i)} \\ &= 2\pi i \cdot \frac{(-\sqrt{2}-\sqrt{2}i)(i)}{(-\sqrt{2})(-\sqrt{2})(4)} \\ &= 2\pi i \cdot -\frac{(\sqrt{2}+\sqrt{2}i)(i)}{8} \\ &= 2\pi i \cdot -\frac{\sqrt{2}i-\sqrt{2}}{8} \\ &= 2\pi i \cdot -\frac{1}{8}(i-1)\sqrt{2} \\ &= \pi \frac{1}{4}(1+i)\sqrt{2}. \\ \end{align*}
Then we calculate, $$\color{red}{\int _{C}\frac{1}{z^4+1}dz} = -\pi \frac{1}{4}(i-1)\sqrt{2} +\pi \frac{1}{4}(1+i)\sqrt{2},$$ $$\color{red}{\int _{C}\frac{1}{z^4+1}dz} = \frac{\pi}{\sqrt{2}}.$$
For the integral over $C_R$, we can perform the following steps, \begin{align*} \color{green}{\int _{C_R}\frac{1}{z^4+1}dz} &= \int _0^{\pi} \frac{16(4i)(e^{4i\theta})}{16e^{4i\theta}+1}d\theta \\ &= \ln (16e^{4i\theta}+1) \Big]_{0}^{\pi}.\\ &= 0. \end{align*}
Thus, $$\color{blue}{\int _{-2}^{2}\frac{1}{z^4+1}dz} + \color{green}{0}=\color{red}{\frac{\pi}{\sqrt{2}}},$$ $$\color{blue}{\int _{-2}^{2}\frac{1}{z^4+1}dz} =\color{red}{\frac{\pi}{\sqrt{2}}}.$$
Here's my attempt at partial fraction decomposition. \begin{align} \color{blue}{\int_{-2}^{2} \frac{1}{z^4+1}dz} &= \int_{-2}^{2} \frac{\sqrt{2}z-2}{(4)(-z^2+\sqrt{2}z-1)}dz + \int_{-2}^{2} \frac{\sqrt{2}z+2}{(4)(z^2+\sqrt{2}z+1)}dz \\ &= \int_{-2}^{2} \frac{\sqrt{2}z-2}{(4)(-z^2+\sqrt{2}z-1)}dz + \int_{-2}^{2} \frac{\sqrt{2}z+2}{(4)(z^2+\sqrt{2}z+1)}dz. \\ \end{align} For the integral on the left: \begin{align} \frac{1}{4}\int_{-2}^{2} \frac{\sqrt{2}z-2}{(-z^2+\sqrt{2}z-1)}dz &= \frac{1}{4} \int_{-2}^{2} \frac{\sqrt{2}-2z}{\sqrt{2}(-z^2+\sqrt{2}z-1)} dz \\ &- \frac{1}{4}\int_{-2}^{2}\frac{1}{-z^2+\sqrt{2}z-1}dz. \\ \end{align} For the integral in the first term, we can use substitution: \begin{align} \frac{1}{4} \int_{-2}^{2} \frac{\sqrt{2}-2z}{\sqrt{2}(-z^2+\sqrt{2}z-1)} dz &= \frac{1}{4\sqrt{2}} \int_{-5-2\sqrt{2}}^{-5+2\sqrt{2}} \frac{1}{u}du \\ &= -\frac{1}{4\sqrt{2}} \ln(u)\Big]_{-5-2\sqrt{2}}^{-5+2\sqrt{2}} \\ &= -\frac{1}{4\sqrt{2}} (\ln(-5+2\sqrt{2}) - \ln(-5-2\sqrt{2})). \end{align} For the integral in the second term: \begin{align} \frac{1}{4}\int_{-2}^{2}\frac{1}{-z^2+\sqrt{2}z-1}dz &= \frac{1}{4}\int_{-2}^{2}\frac{1}{-\left(z-\frac{1}{\sqrt{2}}\right)^2-\frac{1}{2}}dz\\ &= \frac{1}{4}\int_{-2-\frac{1}{\sqrt{2}}}^{2-\frac{1}{\sqrt{2}}}\frac{1}{-u^2-\frac{1}{2}}du\\ &= -\frac{1}{2}\int_{-2-\frac{1}{\sqrt{2}}}^{2-\frac{1}{\sqrt{2}}}\frac{1}{2u^2+1}du\\ &= -\frac{1}{2} \cdot \frac{1}{\sqrt{2}}\tan^{-1}(\sqrt{2}u)\Big]_{-2-\frac{1}{\sqrt{2}}}^{2-\frac{1}{\sqrt{2}}}\\ &= -\frac{1}{2\sqrt{2}}(\tan^{-1}(2\sqrt{2}-1)-\tan^{-1}(-2\sqrt{2}-1))\\ \end{align} Together that is: $$-\frac{1}{4\sqrt{2}} (\ln(-5+2\sqrt{2}) - \ln(-5-2\sqrt{2}))+\frac{1}{2\sqrt{2}}(\tan^{-1}(2\sqrt{2}-1)-\tan^{-1}(-2\sqrt{2}-1)).$$
For the integral on the right, I get the following: \begin{align} \frac{1}{4}\int_{-2}^{2} \frac{\sqrt{2}z+2}{(z^2+\sqrt{2}z+1)}dz &= \frac{1}{4\sqrt{2}} (\ln(5+2\sqrt{2}) - \ln(5-2\sqrt{2}))+\frac{1}{2\sqrt{2}}(\tan^{-1}(2\sqrt{2}-1)-\tan^{-1}(-2\sqrt{2}-1)). \end{align}
Ok, having hashed that all out, my question is: what was done correctly? Checking with Wolfram Alpha, I tentatively believe that (barring typos) the partial fraction decomposition is correct. So where did I go wrong in the first method? Or is the second method wrong somewhere?
Thank you in advance!
Your computation of$$\int_{C_R}\frac1{z^4+1}\,\mathrm dz\label{a}\tag1$$is wrong. It's like computing $\displaystyle\int_\gamma\frac1z\,\mathrm dz$, where $\gamma(\theta)=e^{i\theta}$ ($\theta\in[0,2\pi]$), by doing\begin{align}\int_\gamma\frac1z\,\mathrm dz&=\int_0^{2\pi}\frac{ie^{i\theta}}{e^{i\theta}}\,\mathrm d\theta\\&=\left[\log(e^{i\theta})\right]_{\theta=0}^{\theta=2\pi}\\&=0.\end{align}This is wrong, of course; actually\begin{align}\int_\gamma\frac1z\,\mathrm dz&=\int_0^{2\pi}\frac{ie^{i\theta}}{e^{i\theta}}\,\mathrm d\theta\\&=\int_0^{2\pi}i\,\mathrm d\theta\\&=2\pi i.\end{align}And the error is the same that you made, which was to assume that there is a global differentiable logarithm function. There is no such function.
In fact, \eqref{a} is equal to$$\frac{\log\left(\frac1{17}\left(33-20 \sqrt{2}\right)\right)+2\arctan\left(\frac{2\sqrt{2}}3\right)}{2\sqrt{2}}.$$