Prove : $\int_0^{\infty} \frac{x^6+1}{x^{12}+1}\mathrm dx=\frac{\pi}{\sqrt{6}}$

Question

I need to evaluate the following integral : $$I=\int_0^{\infty} \frac{x^6+1}{x^{12}+1}\mathrm dx$$

Simplifying denominator, we have :

$$I=\int_0^{\infty} \frac{x^6+1}{x^4+1}.\frac{1}{x^8-x^4+1}\mathrm dx$$

The first fraction is theoretically integrable by long division. But the second fraction does not seem promising : so integration by parts cannot be used. With such high exponents, I cannot come up with any algebraic or trigonometric substitutions.

How can I approach to solve this ? Is there any particular trick here which I'm missing ?

According to answer key, $$I=\frac{\pi}{√6}$$

Answer 1

Simplify the integral first and then partially fractionalize \begin{align} &\int_0^{\infty} \frac{x^6+1}{x^{12}+1}\overset{x\to 1/x}{dx}=\int_0^{\infty} \frac{x^{10}+x^4}{x^{12}+1}{dx}\\ =&\ \frac12 \int_0^{\infty} \frac{x^{10}+x^6+x^4+1}{x^{12}+1}{dx}=\frac12 \int_0^{\infty} \frac{x^6+1}{x^8-x^4 +1}dx \\ =&\ \frac{\sqrt3+1}{4\sqrt3} \int_0^{\infty} \frac{x^2+1}{x^4+\sqrt3x^2 +1}dx+ \frac{\sqrt3-1}{4\sqrt3} \int_0^{\infty} \frac{x^2+1}{x^4-\sqrt3x^2 +1}dx \\ =&\ \frac{\sqrt3+1}{4\sqrt3} \int_0^{\infty} \frac{d(x-\frac1x)}{(x-\frac1x)^2+2+\sqrt3} + \frac{\sqrt3-1}{4\sqrt3} \int_0^{\infty} \frac{d(x-\frac1x)}{(x-\frac1x)^2+2-\sqrt3}\\ =&\ \frac{\pi}{4\sqrt{3}}\left(\frac{\sqrt3+1}{\sqrt{2+\sqrt3}}+ \frac{\sqrt3-1}{\sqrt{2-\sqrt3}} \right)=\frac\pi{\sqrt6} \end{align}

Answer 2

According to comment of @KStargamer,

$$\int_0^\infty \frac{x^{s-1}}{x^{n}+1}dx=\frac{\pi}{n}\csc(\tfrac{s\pi}{n}).$$

So,

$${\int_0^{\infty} \frac{x^6}{x^{12}+1}\mathrm dx +\int_0^{\infty} \frac{1}{x^{12}+1}\mathrm dx\\ =\frac\pi{12}\csc(\tfrac{7\pi}{12})+\frac\pi{12}\csc(\tfrac{\pi}{12})\\ =\frac\pi{12}(\frac1{\cos15^o}+\frac1{\sin15^o})\\ =\frac\pi 3(\cos15^o+\sin15^o)\\ =\frac\pi 3\sqrt{1+\sin30^o}\\ =\frac{\pi}{\sqrt6}} $$

Answer 3

The integral can be solved by using countour integration: First notice that the function is even so you have $$ I=\frac{1}{2} \int_{-\infty}^\infty \frac{x^6+1}{x^{12}+1} $$ Now let's consider the interval $[-R,R]$ big enought and consider the domain $C$ formed by the interval and the semicircle $\gamma$ on the upper complex plane enclosing the interval $$ \int_{-R}^R \frac{x^6+1}{x^{12}+1}=\int_C \frac{x^6+1}{x^{12}+1}- \int_\gamma \frac{x^6+1}{x^{12}+1} $$ The integral over $C$ can be solved ny using the residue theorem. The roots of the denominator, i.e. the poles of the function are the twelth roots of $(-1)$ i.e.

$$ \exp\left(\pm{\frac{(2n+1)i\pi}{12}}\right) $$ for $n=0,1,2,3,4,5$. The roots on the upper complex plane are the one with positive $n$

So If you call $p_n(x)$ the polynomial resulting by the polynomial division $$ p_n(x)= \frac{x^{12}+1}{x-\exp\left({\frac{(2n+1)i\pi}{12}}\right)} $$ You have $$ \int_C \frac{x^6+1}{x^{12}+1}=2\pi i \sum_{n=0}^5 \frac{\exp\left({\frac{(2n+1)i\pi}{2}}\right)+1}{p_n\left(\exp\left({\frac{(2n+1)i\pi}{12}}\right)\right)} $$ Now it is only a matter of doing the polynomial divisions and evaluating them. As in the wikipedia link you can see that the integral over the semicircle $\gamma$ goes to zero when $R \to \infty$ So $$ I=\pi i \sum_{n=0}^5 \frac{\exp\left({\frac{(2n+1)i\pi}{2}}\right)+1}{p_n\left(\exp\left({\frac{(2n+1)i\pi}{12}}\right)\right)} $$ EDIT

As said in the comment below evaluating the residue is simple if we note that, said $c_n=\exp\left({\frac{(2n+1)i\pi}{12}}\right)$ $$ \lim_{x \to c_n}\frac{x^{12}+1}{x-c_n}=\lim_{x \to c_n} 12x^{11} $$ So $$p_n\left(\exp\left({\frac{(2n+1)i\pi}{12}}\right)\right)=12 \exp\left({\frac{11(2n+1)i\pi}{12}}\right)=\frac{-12}{\exp\left({\frac{(2n+1)i\pi}{12}}\right)} $$ And the residue becomes $$ I=-\frac{\pi i}{12} \sum_{i=0}^5 \left( \exp\left({\frac{7(2n+1)i\pi}{12}}\right)+\exp\left({\frac{(2n+1)i\pi}{12}}\right) \right) $$

Answer 4

$\newcommand{\d}{\,\mathrm{d}}\newcommand{\res}{\operatorname{Res}}$A different method via contour integration which avoids potentially painful polynomial division. Take $x=t^{1/6}$; consider: $$I=\frac{1}{6}\int_0^\infty\frac{t+1}{t^2+1}\cdot t^{-5/6}\d t$$The integrand is, when taking the logarithm given by $0\le\arg<2\pi$, extensible to a meromorphic function on $\Bbb C\setminus[0,\infty)$ which satisfies $zf(z)\to0$ uniformly as $z\to\infty$. Thus by taking a Hankel contour we find (branch-jump in the logarithm as it crosses the positive real axis): $$\left(1-\exp\left(2\pi i\cdot\frac{-5}{6}\right)\right)I=2\pi i(\res(f;i)+\res(f;-i))$$Since the poles at $\pm i$ are simple it's easy to get the residues and see: $$\begin{align}I&=\frac{\pi\cdot\exp(5\pi i/6)}{6\cdot\sin(5\pi/6)}\left(\frac{i+1}{2i}\exp\left(-\frac{5\pi i}{12}\right)-\frac{1-i}{2i}\exp\left(-\frac{15\pi i}{12}\right)\right)\\&=\frac{\pi}{6\sin(5\pi/6)}(\cos(5\pi/12)+\sin(5\pi/12))\\&=\frac{\pi\sqrt{2}}{3}\cdot\sin(2\pi/3)\\&=\frac{\pi}{\sqrt{6}}\end{align}$$Using standard trigonometric identities and special values to clean up the last few expressions. A very key point is that $\arg(-i)$ is not $-\pi/2$ here; by my definition of the logarithm used to calculate $z^{-5/6}$, the argument is $3\pi/2$. This is necessary so that the branch cut lies on the domain of integration, $[0,\infty)$.

Answer 5

\begin{align}J&=\int_0^\infty \frac{1+x^6}{1+x^{12}}dx\\ &\overset{u=x^{12}}=\frac{1}{12}\int_0^\infty \frac{u^{-\frac{11}{12}}}{1+u}du+\frac{1}{12}\int_0^\infty \frac{u^{-\frac{5}{12}}}{1+u}du\\ &=\frac{1}{12}B\left(\frac{1}{12},1-\frac{1}{12}\right)+\frac{1}{12}B\left(\frac{7}{12},1-\frac{7}{12}\right)\\ &=\frac{1}{12}\frac{\Gamma\left(\frac{1}{12}\right)\Gamma\left(1-\frac{1}{12}\right)}{\Gamma(1)}+\frac{1}{12}\frac{\Gamma\left(\frac{7}{12}\right)\Gamma\left(1-\frac{7}{12}\right)}{\Gamma(1)}\\ &=\frac{\pi}{12\sin\left(\frac{\pi}{12}\right)}+\frac{\pi}{12\sin\left(\frac{7\pi}{12}\right)}\\ &=\frac{\pi}{12\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)}+\frac{\pi}{12\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)}\\ &=\frac{\pi\sqrt{2}}{6(\sqrt{3}-1)}+\frac{\pi\sqrt{2}}{6(\sqrt{3}+1)}\\ &=\frac{(\sqrt{3}+1)\pi\sqrt{2}}{12}+\frac{(\sqrt{3}-1)\pi\sqrt{2}}{12}\\ &=\frac{2\pi\sqrt{6}}{12}=\boxed{\frac{\pi}{\sqrt{6}}} \end{align}

NB: $B$ is the Beta Euler function.

Answer 6

Let us define $$ t = \frac{1}{1+x^{12}}$$

Then

$$ x = (\frac{1-t}{t})^{\frac{1}{12}}$$

i.e.

$$ x^6 = (\frac{1-t}{t})^{\frac{1}{2}}$$

This implies that

$$ dx = \frac{-1}{12} \cdot (1-t)^{\frac{-11}{12}} \cdot t^{\frac{-13}{12}}dt$$

After changing the bounds of integration and swapping the upper and lower limits for a negative sign, we get:

$$\ I = \int_0^1 t \cdot ((\frac{1-t}{t})^{\frac{1}{2}} +1) \cdot \frac{1}{12}\cdot (1-t)^{\frac{-11}{12}} \cdot t^{\frac{-13}{12}}dt$$

Splitting the integral into two, we get:

$$\ I = \int_0^1 t \cdot ((\frac{1-t}{t})^{\frac{1}{2}}) \cdot \frac{1}{12}\cdot (1-t)^{\frac{-11}{12}} \cdot t^{\frac{-13}{12}}dt + \int_0^1 t \cdot (1) \cdot \frac{1}{12}\cdot (1-t)^{\frac{-11}{12}} \cdot t^{\frac{-13}{12}}dt$$

Let

$$\ I_1 = \int_0^1 t \cdot ((\frac{1-t}{t})^{\frac{1}{2}}) \cdot \frac{1}{12}\cdot (1-t)^{\frac{-11}{12}} \cdot t^{\frac{-13}{12}}dt $$

and

$$\ I_2 = \int_0^1 t \cdot (1) \cdot \frac{1}{12}\cdot (1-t)^{\frac{-11}{12}} \cdot t^{\frac{-13}{12}}dt$$

For $\ I_1$, we get:

$$\ I_1 = \int_0^1 t \cdot ((\frac{1-t}{t})^{\frac{1}{2}}) \cdot \frac{1}{12}\cdot (1-t)^{\frac{-11}{12}} \cdot t^{\frac{-13}{12}}dt $$

$$ = \frac{1}{12} \cdot \int_0^1 t^{\frac{-7}{12}} \cdot (1-t)^{\frac{-5}{12}}dt $$

$$\ I_1 = \frac{1}{12} \cdot \beta(\frac{5}{12}, \frac{7}{12}) = \frac{1}{12} \cdot \Gamma(\frac{5}{12})\cdot \Gamma(\frac{7}{12}) = \frac{1}{12} \cdot \frac{\pi}{\sin(\frac{5\pi}{12})}$$

Where $\beta(x,y)$ is the (Eulerian) Beta function, which is defined as:

$$\beta(m,n) = \int_0^1 t^{m-1} \cdot (1-t)^{n-1} dt$$

We used some special properties of this function, which are:

$$\beta(x,y) = \frac{\Gamma(x)\cdot \Gamma(y)}{\Gamma(x+y)}$$

And Euler's Reflection formula, which states that

$$\Gamma(z)\cdot \Gamma(1-z) = \frac{\pi}{sin(\pi z)}$$

Similarly, $I_2$ evaluates to:

$$\ I_2 = \frac{1}{12} \cdot \beta(\frac{1}{12}, \frac{11}{12}) = \frac{1}{12} \cdot \frac{\pi}{\sin(\frac{\pi}{12})}$$

Adding $I_1$ and $I_2$, we have:

$$\ I = I_1 + I_2 = \frac{\pi}{12} \cdot (\csc(\frac{5\pi}{12}) + \csc(\frac{\pi}{12}))$$

$$= \frac{\pi}{12} \cdot(2\sqrt2) \cdot (\frac{1}{\sqrt3-1}+\frac{1}{\sqrt3+1})$$

And we finally get our result:

$$\ I = \frac{\pi}{\sqrt6}$$

In fact, you can use this technique to prove that for $m > n+1$ and $n >0$, we have:

$$\ I(m,n) = \int_0^{\infty} \frac{x^n}{1+x^m} dx = \frac{\pi}{m} \cdot \csc(\frac{\pi(n+1)}{m})$$