Contour Integral help with residue theorem

Question

$$ K = \int_{0}^{\infty}\frac{1}{x^{4}+x^{2}+1}dx $$

I am supposed to use contour integration to solve this, but I can't even determine the singularities. The denominator doesn't have any that I can see. From what I understand about contour integration, isn't the solution $ 2\pi i\sum residues $ ? Perhaps I am missing something?

Answer 1

You can also deal with this integral without using residue theory.

Let

\begin{align*} I &=\int \frac{1}{1+x^{2}+x^{4}} \ dx \\ &= \frac{1}{2} \cdot \int \frac{2}{1+x^{2}+x^{4}} \ dx \\ &= \frac{1}{2} \cdot \int \frac{(1-x^{2}) + (1+x^{2})}{1+x^{2}+x^{4}} \ dx \\ &=\frac{1}{2} \cdot \int\frac{1-x^{2}}{1+x^{2}+x^{4}} \ dx+ \frac{1}{2} \cdot \int\frac{(1+x^{2})}{1+x^{2}+x^{4}} \ dx \\ &= I_{1} + I_{2} \end{align*}

Now to evaluate $I_{1}$ do the following

\begin{align*} I_{1} &=-\frac{1}{2}\cdot\int \frac{x^{2}-1}{x^{4}+x^{2}+1} \ dx \\ &= -\frac{1}{2}\cdot \int\frac{1-\frac{1}{x^2}}{x^{2}+1+\frac{1}{x^2}} \ dx \\ &=-\frac{1}{2} \cdot \int \frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^{2}-1} = -\frac{1}{2} \cdot \int \frac{dt}{t^{2}-1} \end{align*}

Answer 2

Observe that: $\dfrac{1}{x^4+x^2+1}=\dfrac{x^2-1}{x^6-1}$, now you can use substitution and then integration by parts.

Answer 3

Define

$$f(z)=\frac1{z^4+z^2+1}$$

and we have

$$z^4+z^2+1=0\implies z^2_{1,2}=\frac{-1\pm3 i}{2}=e^{\pm\left(\frac{2\pi i}3+2k\pi i\right)}\implies z_k=\pm e^{\pm\frac{\pi i}3\left(1+3k\right)}\;,\;\;k=0,1$$

Thus, our function has two simple poles on the upper half hemisphere (this is all we shall need), with residues

$$I=\text{Res}_{z=e^{\pi i/3}}(z-e^{\pi i/3})f(z)\stackrel{\text{l'H}}=\lim_{z\to e^{\pi i/3}}\frac1{4z^3+2z}=\frac1{4e^{\pi i}+2e^{\pi i/3}}=\frac1{-3+\sqrt3\,i}=-\frac14-\frac{\sqrt3}{12}i$$

$$II=\text{Res}_{z=e^{2\pi i/3}}(z-e^{2\pi i/3})f(z)\stackrel{\text{l'H}}=\lim_{z\to e^{2\pi i/3}}\frac1{4z^3+2z}=\frac1{4e^{2\pi i}+2e^{2\pi i/3}}=\frac1{3+\sqrt3\,i}=\frac14-\frac{\sqrt3}{12}i$$

Thus, taking the contour:

$$C_R:=[-R,R]\cup\gamma_R:=\{\,z\in\Bbb C\;;\;|z|=R\;,\;\text{Im}(z)\ge 0\,\}\;,\;\;\Bbb R\ni R>>0$$

We get by Cauchy's Integral Theorem

$$\frac\pi{\sqrt3}=2\pi i(-\frac{\sqrt3}6i)=2\pi i(I+II)=\oint\limits_{C_R}f(z)\,dz=\int\limits_{-R}^R\frac{dx}{x^4+x^2+1}+\int\limits_{\gamma_R}f(z)\,dz$$

But by the evaluation theorem we have that

$$\left|\;\int\limits_{\gamma_R}f(z)\,dz\;\right|\le\frac1{R^4-R^2-1}\cdot\pi R\xrightarrow[R\to\infty]{}0$$

So we're left with

$$\int\limits_{-\infty}^\infty\dfrac{dx}{x^4+x^2+1}=\lim_{R\to\infty}\int\limits_{-R}^R\frac{dx}{x^4+x^2+1}=\frac\pi{\sqrt3}$$

Now use our function is even and divide through by $\,2\,$ to get the wanted integral ...

Answer 4

The singularities of the denominator are in the complex plane:

$$z^4+z^2+1 = 0 \implies z^2 = \frac{-1 \pm i \sqrt{3}}{2}$$

In polar form, $z^2 = e^{i 2 \pi/3}$ or $e^{i 4 \pi/3}$. Taking square roots, the poles are at $z_1=e^{i \pi/3}$, $z_2=e^{i 2 \pi/3}$, $z_3 = -e^{i \pi/3}$, $z_4=-e^{i 2 \pi/3}$. Note the particular branch of the square root chosen here.

In this case, the integral stated is equal to $i 2 \pi$ times the sum of the residues inside the integration contour. What contour? Well, we had actually considered

$$\oint_C \frac{dz}{z^4+z^2+1}$$

where $C$ is a semicircular contour in the upper-half plane of radius $R$. In the limit as $R \to \infty$, the integral over the circular arc portion goes to zero. To see this, parametrize this portion of the contour by $z=R e^{i \phi}$ and get

$$i R \int_0^{\pi} d\phi \, e^{i \phi} \frac{1}{R^4 e^{i 4 \phi}+R^2 e^{i 2 \phi} + 1} \sim \frac{1}{R^3} \quad (R \to \infty)$$

To compute the integral using residues, we must identify which residues are inside the integration contour. In this case, only $z_1$ and $z_2$ are inside, so we need only consider those.

Residue calculation is simplified by the following fact: when $f(z) = p(z)/q(z)$, the residue of a simple pole of $f$ at $z=z_0$ is $f(z_0)/(q'(z_0)$. Using this fact, we have the sum of the residues being

$$\frac{1}{4 (e^{i \pi/3})^3 + 2 e^{i \pi/3}} + \frac{1}{4 (e^{i 2 \pi/3})^3 + 2 e^{i 2 \pi/3}}$$

I leave the algebra to the reader; the result I get is

$$K = \frac{\pi}{2 \sqrt{3}}$$