MHB Integral Challenge: Evaluating $\int_0^\infty \frac{x^2+2}{x^6+1} \, dx$

anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Evaluate $\displaystyle\int\limits_0^{\infty} \dfrac{x^2+2}{x^6+1} \, dx$.
 
Mathematics news on Phys.org
We have $$I= \int_0^\infty \frac{x^2+2}{x^6+1}\,\mathrm{dx}$$
Let $x \mapsto \frac{1}{x}$ then$$I= \int_0^\infty \frac{x^2+2x^4}{x^6+1}\,\mathrm{dx}$$
So that $$2I = 2\int_0^\infty \frac{1+x^2+x^4}{1+x^6}\,\mathrm{dx}$$

i.e. $$I = \int_0^\infty \frac{1+x^2+x^4}{1+x^6}\,\mathrm{dx}$$By partial fractions $$\begin{aligned} \frac{1+x^2+x^4}{1+x^6} &= \frac{1}{3}\cdot \frac{1}{1+x^2}+\frac{2}{3}\cdot \frac{x^2+1}{x^4-x^2+1} \\& = \frac{1}{3}\cdot \frac{1}{1+x^2} +\frac{2}{3}\cdot \frac{1+1/x^2}{x^2+\frac{1}{x^2}+1} \\& = \frac{1}{3}\cdot \frac{1}{1+x^2} +\frac{2}{3}\cdot \frac{1+1/x^2}{(x-1/x)^2+1}\end{aligned} $$
This along with the sub $u = x-\frac{1}{x}$ helps us finish off the integral,
$$\begin{aligned} I & = \frac{1}{3}\cdot \int_0^\infty \frac{1}{1+x^2}\, \mathrm{dx} +\frac{2}{3}\cdot \int_0^\infty\frac{1+1/x^2}{(x-1/x)^2+1}\,\mathrm{dx} \\& = \frac{1}{3}\cdot \frac{\pi}{2}+\frac{2}{3} \cdot \int_{-\infty}^{\infty} \frac{1}{u^2+1}\,\mathrm{du} \\& = \frac{\pi}{6}+\frac{2\pi}{3} \\& = \frac{5 \pi}{6}.\end{aligned} $$Therefore $$\int_0^\infty \frac{x^2+2}{x^6+1}\,\mathrm{dx}= \frac{5\pi}{6}.$$
 
Last edited:
Consider the function $\displaystyle f(z) = \frac{z^2+2}{z^6+1}$; $f$ has simple poles at $z = \pm i, \pm i^{\frac{1}{3}}, \pm i^{\frac{5}{3}}$, three of which lie in the upper half plane;
those are $z_1 = i, ~ z_2 = i^{\frac{1}{3}}, ~ z_3 = i^{\frac{5}{3}}$. The sum of the residues at these poles is given by

$\displaystyle \begin{aligned} \sum_{1 \le i \le 3} \text{res} f(z_i) & = \frac{1}{6} (-2 i - \sqrt{3})+\frac{1}{6}(-2 i +\sqrt{3})-\frac{i}{6} \\& = -\frac{5i}{6}\end{aligned} $​

Now, let $\Gamma$ be the upper half-plane semi-circle $|z| = a$ with radius $a$ in counterclockwise direction.

Since $z_1, z_2, z_3$ all lie in the upper half plane, by the residue theorem we have

$\displaystyle \begin{aligned} \oint_\Gamma f(z)\,dz & =2i \pi \sum_{1 \le i \le 3} \text{res} f(z_i) \\& = 2i \pi \cdot \frac{-5i}{6} \\& = \frac{5\pi}{3} \end{aligned}$​
But also
$\displaystyle \oint_\Gamma f(z)\,dz=\int_{-a}^{a} f(z) \,\mathrm{dz} +\int_{\text{Arc}} f(z)\,\mathrm{dz}$​

Therefore

$\displaystyle \int_{-a}^{a} f(z) \,\mathrm{dz} +\int_{\text{Arc}} f(z)\,\mathrm{dz} = \frac{5\pi}{3}$​

By taking $a \to \infty$, since

$ \displaystyle \bigg| \int_{\text{Arc}} f(z) \,\mathrm{dz} \bigg| \le \frac{a^2+2}{a^6-1}~ a \pi \longrightarrow 0$​
we have $\displaystyle \int_{\text{Arc}} f(z)\,\mathrm{dz} \to 0$ as $a \to \infty$, therefore

$\displaystyle \int_{-\infty}^{\infty}\frac{z^2+2}{z^6+1}\,\mathrm{dz} = \frac{5\pi}{3} $​

Therefore (since $f$ is an even function),

$\displaystyle \int_{0}^{\infty}\frac{z^2+2}{z^6+1}\,\mathrm{dz} = \frac{5\pi}{6}. $​
 
Last edited:
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top