I am having a massive amount of trouble integrating this, I really have no clue how to get the answer in the book:
$$\int_{-\infty}^{\infty} \frac{x^4}{1+x^8}dx$$
I know I need to find the poles on this function, which is basically the value of $x^8 = -1$ or I could split it up like the following: $$x^8 + 1 = (x^4 + i)(x^4 - i)$$ With this I got the following poles (looking at $x^4 - i$), all have an order one: \begin{array} & x_1 = e^{\frac{i\pi}{8}} \\ x_2 = e^{\frac{3i\pi}{8}} \\ x_3 = e^{\frac{5i\pi}{8}} \\ x_4 = e^{\frac{7i\pi}{8}} \end{array} Now to calculate the residues I am trying to make use of the result that given a rational function $\frac{F}{G}$ such that both of them are analytic on a disk of radius $r$ with $G(z_0) = 0$ but $G'(z_0) \neq 0$ then we know that: $$Res(\frac{F}{G};z_0) = \frac{F(z_0)}{G'(z_0)}$$ With that I am having difficulty getting results that are useful, I maybe calculating wrong and I am not sure but I have been spending a long time on this one question. The solution in the book is $\frac{\pi}{4}[sin(\frac{3\pi}{8}]^{-1}$. I would like a step by step solution to this or at least some guidance because I really need to learn this and I am not sure how to do this properly. Thank you!
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{-\infty}^{\infty}{x^{4} \over 1 + x^{8}}\,\dd x:\ {\large ?}.\quad}$ Let's take a 'pizza slice' with angle $\ds{{2\pi \over 8} = {\pi \over 4}\ \mbox{and} \quad x_{1} = \expo{\ic\pi/8}}$
\begin{align} &\color{#00f}{\large\int_{-\infty}^{\infty}{x^{4} \over 1 + x^{8}}\,\dd x} = 2\bracks{{\pi\ic \over 4x_{1}^{3}}\, {\expo{-\ic\pi/8} \over \expo{-\ic\pi/8} + \expo{\ic\pi/8}}} = {\pi\ic \over 2x_{1}^{4}}\,{1 \over 2\cos\pars{\pi/8}} = {\pi \over 2}\,{1 \over 2\cos\pars{\pi/8}} \\[3mm]&={\pi \over 4}\,\sec\pars{\pi \over 8} =\color{#00f}{\large{1 \over 4}\root{4 - 2\root{2}}\,\pi} \end{align}