$\int_0^\infty \frac{1}{1+x^4}dx$ using the Residue Theorem

complex-analysiscontour-integrationresidue-calculussingularity

I'm trying to evaluate the integral

$$\int_0^\infty \frac{1}{1+x^4}dx $$

using the Residue Theorem.

My approach:

Let's consider

$$\oint_\Gamma f$$

with $f(z)=\frac{1}{1+z^4}$ and $\Gamma = \Gamma_1 + \Gamma_2$, where:

  • $\Gamma_1:[-R,R]\rightarrow \mathbb{C}$, with $\Gamma_1(t)=t$
  • $\Gamma_2:[0, \pi] \rightarrow \mathbb{C}$, with $\Gamma_2(t)=Re^{it}$

So basically $\Gamma$ is the semicircle centers in the origin with imaginary part greater or equal to zero.

First we need to find the isolated singularities $\alpha_i$ of the function $f$. This singularities are the solution of the equation $1 + z^4 = 0$:

enter image description here

Let's call:

  • $\alpha_1 = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}$
  • $\alpha_2 = -\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}$
  • $\alpha_3 = -\frac{\sqrt{2}}{2} – i \frac{\sqrt{2}}{2}$
  • $\alpha_4 = \frac{\sqrt{2}}{2} – i \frac{\sqrt{2}}{2}$

So now we have:

$$\oint_\Gamma f = 2 \pi i \sum_i \text{Res}(f,\alpha_i) \text{Ind}_\Gamma(\alpha_i)$$

All singularities are poles of order one, with:

$$\text{Res}(f,\alpha_i)=\frac{1}{1+4\alpha_i^3}$$

So we end up with:

$$\oint_\Gamma f = 2 \pi i \sum_i \frac{\text{Ind}_\Gamma(\alpha_i)}{1+4\alpha_i^3} $$

Because of the shape of our curve $\Gamma$ we have that $\text{Ind}_\Gamma(\alpha_3)=\text{Ind}_\Gamma(\alpha_4)=0$

So we end up with:

$$\oint_\Gamma f = 2 \pi i \underbrace{\left( \frac{1}{1+4\alpha_1^3} + \frac{1}{1+4\alpha_2^3} \right)}_{:=\xi} $$

Now we can work on the left side of this expression:

$$\int_{\Gamma_1} f + \int_{\Gamma_2} f = 2 \pi i \xi$$

We have that:

$$\int_{\Gamma_1}f = \int_{-R}^R \frac{1}{1 + t^4} dt$$

And we also know that
$$\begin{align}
\int_{\Gamma_2}f &\leq \int_0^\pi \left|\frac{Rie^{it}}{1 + R^4e^{4it}} \right| dt
\\
\\ &= \int_0^\pi \frac{R}{\left|1 + R^4e^{4it}\right|} dt
\\
\\ &= \int_0^\pi \frac{1}{\left| \frac{1}{R} + R^3e^{4it}\right|} dt
\end{align}$$

If we let $R \to \infty$ we have that $\int_{\Gamma_1} f = \int_{-\infty}^\infty \frac{dt}{1 + t^4}$ and $\int_{\Gamma_2} f = 0$ and because $\int_{-\infty}^\infty \frac{dt}{1 + t^4} = 2 \int_{0}^\infty \frac{dt}{1 + t^4}$, we end up with:

$$\int_{0}^\infty \frac{dt}{1 + t^4} = \pi i \xi$$

The thing is that $\pi i \xi$ is a complex number, so what did I do wrong?

Best Answer

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[15px,#ffd]{\int_{0}^{\infty}{\dd x \over 1 + x^{4}}}}$

I'll consider $\ds{\int_{\mathcal{P}}{\dd z \over 1 + z^{4}}}$ where $\ds{\mathcal{P}}$ is a contour in the first quadrant. Namely, $$ \mathcal{P} \equiv \pars{0,R}\cup R\expo{\ic\pars{0,\pi/2}}\cup\pars{R\expo{\ic\pi/2},0} \,,\qquad R > 1 $$ There is just one single pole inside $\ds{\mathcal{P}}$: $\ds{\expo{\ic\pi/4}}$.
Then, \begin{align} \int_{\mathcal{P}}{\dd z \over 1 + z^{4}} & = \left. 2\pi\ic\,{1 \over 4z^{3}}\right\vert_{\ z\ =\ \exp\pars{\ic\pi/4}} = \left. {1 \over 2}\,\pi\ic\,{z \over z^{4}} \right\vert_{\ z\ =\ \exp\pars{\ic\pi/4}} = -\,{1 \over 2}\,\pi\ic\expo{\ic\pi/4} \\[5mm] & = \int_{0}^{R}{\dd x \over 1 + x^{4}}\ +\ \overbrace{\int_{0}^{\pi/2}{R\expo{\ic\theta}\ic\,\dd\theta \over 1 + R^{4}\expo{4\ic\theta}}} ^{\ds{\stackrel{\mrm{as}\ R\ \to\ \infty}{\LARGE \to}\ \color{red}{\large 0}}}\ +\ \int_{R}^{0}{\expo{\ic\pi/2}\dd r \over 1 + \pars{r\expo{\ic\pi/2}}^{4}} \\[5mm] & \stackrel{R\ \to\ \infty}{\to}\,\,\, \pars{1 - \ic}\int_{0}^{\infty}{\dd x \over 1 + x^{4}} \\[5mm] \implies & \int_{0}^{\infty}{\dd x \over 1 + x^{4}} = -\,{1 \over 2}\,{\pi\ic\expo{\ic\pi/4} \over 1 - \ic} = \bbox[15px,#ffd,border:1px solid navy]{{\root{2} \over 4}\,\pi}\ \approx\ 1.1107 \end{align}
Note that $\ds{0 < \verts{\int_{0}^{\pi/2}{R\expo{\ic\theta}\ic\,\dd\theta \over 1 + R^{4}\expo{4\ic\theta}}}_{\ R\ > 1} < {1 \over 2}\,\pi\,{R \over R^{4} - 1}}$
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