About an integral from MIT Integration Bee 2024

Question

Good evening,

I was interested in the third integral from the finals of the MIT Integration Bee 2024 :

$$I = \int_{-\infty}^{\infty} \frac{1}{x^4+x^3+x^2+x+1} \hspace{0.1cm} \mathrm{d}x$$

One way to solve this is to identify that the denominator is the fifth cyclotomic polynomial, so we can write :

$$I = \displaystyle\int_{-\infty}^{\infty} \frac{1-x}{(1-x)(x^4+x^3+x^2+x+1)} \hspace{0.1cm} \mathrm{d}x = \displaystyle\int_{-\infty}^{\infty} \frac{1-x}{1-x^5} \hspace{0.1cm} \mathrm{d}x$$

Then, by Chasles :

$$I = \displaystyle\int_{-\infty}^{0} \frac{1-x}{1-x^5} \hspace{0.1cm} \mathrm{d}x + \displaystyle\int_{0}^{\infty} \frac{1-x}{1-x^5} \hspace{0.1cm} \mathrm{d}x$$

Let $x\to -x$ in the first integral, we obtain :

$$I = \displaystyle\int_{0}^{\infty} \frac{1+x}{1+x^5} \hspace{0.1cm} \mathrm{d}x + \displaystyle\int_{0}^{\infty} \frac{1-x}{1-x^5} \hspace{0.1cm} \mathrm{d}x$$

And we use these two identities :

$$ \displaystyle\int_{0}^{\infty} \frac{x^{s-1}}{1+x^k} \hspace{0.1cm} \mathrm{d}x = \frac{\pi}{k} \csc\left( \frac{\pi s}{k} \right)$$

$$ \mathrm{PV} \displaystyle\int_{0}^{\infty} \frac{x^{s-1}}{1-x^k} \hspace{0.1cm} \mathrm{d}x = \frac{\pi}{k} \cot\left( \frac{\pi s}{k} \right)$$

As a consequence :

$$ I = \frac{\pi}{5} \left( \csc\left( \frac{\pi}{5} \right) + \csc\left( \frac{2\pi}{5} \right) + \cot\left( \frac{\pi}{5} \right) - \cot\left( \frac{2\pi}{5} \right)\right) = \frac{\pi}{5} \sqrt{10 + 2 \sqrt{5}}\approx 2.39026573179$$

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I have several questions :

1) I have never seen these two identities, apparently it comes from beta/gamma functions, does anyone have a reference with a proof ?

2) Is it possible to take another path ? With a contour integral for example ? I'm curious.

Answer 1

SHOWING THAT $\displaystyle \int_0^\infty \frac{x^{s-1}}{1+x^k}\,dx=\frac\pi k \csc(\pi s/k)$:

Note that we have for $0<s<k$

$$\begin{align} I(s,k)&=\int_0^\infty \frac{x^{s-1}}{1+x^k}\,dx\tag1\\\\ &=\frac1k\int_0^\infty \frac{x^{s/k-1}}{(1+x)}\,dx\tag2\\\\ &=B(s/k,1-s/k)\tag3\\\\ &=\Gamma(s/k)\Gamma(1-s/k)\tag4\\\\ &=\frac\pi k \csc(\pi s/k)\tag5 \end{align}$$

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NOTES:

In going from $(1)$ to $(2)$, we enforced the substitution $x\mapsto x^{1/k}$.

In going from $(2)$ to $(3)$, we used a well-known integral representation of the Beta function.

In going from $(3)$ to $(4)$, we exploited the relationship between the Beta function and the Gamma function.

And in arriving at $(5)$, we invoked Euler's reflection formula for the Gamma function.

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SHOWING THAT $\displaystyle \text{PV}\int_0^\infty \frac{x^{s-1}}{1-x^k}\,dx=\frac\pi k \cot(\pi s/k)$:

Let $I(s,k)$ denote the Principal Value integral

$$\begin{align} I(s,k)&=\text{PV}\int_0^\infty \frac{x^{s-1}}{1-x^k}\,dx\tag6 \end{align}$$

where $k\in \mathbb{N}_+$ and $1 \le s < k$. We will evaluate the integral in $(6)$ using contour integration.

Proceeding, let $\varepsilon>0$ and $R>1+\varepsilon$. Let $J(s,k)$ denote the contour integral

$$\begin{align} J(s,k)&=\oint_C \frac{z^{s-1}}{1-z^k}\,dz\\\\ &=\int_0^{1-\varepsilon}\frac{x^{s-1}}{1-x^k}\,dx+\int_\pi^0 \frac{(1+\varepsilon e^{i\phi})^{s-1}}{1-(1+\varepsilon e^{i\phi})^k}\,i\varepsilon e^{i\phi}\,d\phi+\int_{1+\varepsilon}^R \frac{x^{s-1}}{1-x^k}\,dx\\\\ &+\int_0^{3\pi/k} \frac{(Re^{i\phi})^{s-1}}{1-(Re^{i\phi}))^k}\,iRe^{i\phi}\,d\phi+\int_R^0 \frac{(te^{i3\pi/k})^{s-1}}{1-(te^{i3\pi/k})^k}\,e^{i3\pi/k}\,dt\tag7 \end{align}$$

where $C$ is the classical wedge contour, modified by a semi-circular deformation around $z=1$. In $(7)$, we have choosen to cut the plane along the negative real axis and are on the principal branch of the complex logarithm.

Note that the only singularity enclosed by $C$ is the simple pole at $z=e^{i2\pi/k}$. Hence, the Residue Theorem guarantees that

$$\begin{align} J(s,k)&=2\pi i \text{Res}\left(\frac{z^{s-1}}{1-z^k}, z=e^{i2\pi/k}\right)\\\\ &=-2\pi i \frac{e^{i2\pi s/k}}{k}\tag8 \end{align}$$

In addition, we let $\varepsilon\to0$ and $R\to \infty$. Then, we see that

$$\begin{align} \lim_{\varepsilon\to0\\R\to\infty}J(s,k)&=I(s,k)+i\pi/k-e^{i3\pi s/k}\underbrace{\int_0^\infty \frac{t^{s-1}}{1+t^k}\,dt}_{=\frac\pi k\csc(\pi s/k)\,\,\text{using}\,\,(5)}\\\\ &=I(s,k)+i\pi/k-e^{i3\pi s/k} \left(\frac\pi k \csc(\pi s/k)\right)\tag9 \end{align}$$

Putting $(8)$ and $(9)$ together and exploiting trigonometric identities of $\sin(3x)$, $\cos(3x)$, $\sin(2x)$, and $\cos(2x)$, we find that

$$I(s,k)=\frac\pi k \cot(\pi s/k)$$

as was to be shown!

Answer 2

For the question (2), we can use power series after splitting the interval into two. $$ \begin{aligned} (PV)\int_0^{\infty} \frac{x^{s-1}}{1-x^k} d x&=\int_0^1 \frac{x^{s-1}}{1-x^k} d x+\int_1^{\infty} \frac{x^{s-1}}{1-x^k} d x \\ = & \int_0^1 \frac{x^{s-1}}{1-x^k} d x+\int_0^1 \frac{\frac{1}{x^{s-1}}}{1-\frac{1}{x^k}} \frac{d x}{x^2} \quad \textrm{ via } x\mapsto \frac{1}{x} \\ = & \int_0^1 \frac{x^{s-1}-x^{k-s-1}}{1-x^k} d x \\ = & \sum_{n=0}^{\infty} \int_0^1\left(x^{s-1}-x^{k-s-1}\right) x^{n k} d x \\ = & \sum_{n=0}^{\infty}\left(\frac{1}{n k+s}-\frac{1}{(n+1) k-s}\right) \\ = & \frac{1}{k} \sum_{n=0}^{\infty}\left(\frac{1}{n+\frac{s}{k}}+\frac{1}{-n-1+\frac{s}{k}}\right)\\=& \frac{1}{k} \sum_{n \in \mathbb{Z}} \frac{1}{n+\frac{s}{k}}\\=& \frac{\pi}{k} \cot \left(\frac{\pi s}{k}\right) \end{aligned} $$ where the last answer uses the identity: $\pi \cot (\pi z)=\sum_{n\in \mathbb{Z}} \frac{1}{z+n}$, where $n \in \mathbb{Z}.$

Answer 3

Another possible path for the integral: substitute $x=\dfrac{1-y}{1+y}$, then decompose into partial fractions.

$$\begin{align*} \int_{-\infty}^\infty \frac{1-x}{1-x^5} \, dx &= 2 \int_{-1}^1 \frac{y^2+2y+1}{y^4+10y^2+5} \\ &= 4 \int_0^1 \frac{y^2+1}{y^4+10y^2+5} \, dy \\ &= \frac2{\sqrt5} \int_0^1 \left(\frac{2+\sqrt5}{5+2\sqrt5+y^2} - \frac{2-\sqrt5}{5-2\sqrt5+y^2}\right) \, dy \end{align*}$$

Answer 4

Since it is a fast exam question, there is an easy way to calculate residues:

$f(z)=\frac{z-1}{z^5-1}$. Let $w=e^{\tfrac{2\pi i}5}$. Then by L'Hospital rule $$r_1=Res_{z=w}f(z)=\frac{z-1}{5z^4}\vert_{z=w}=\frac{w^2-w}5$$ $$r_2=Res_{z=w^2}f(z)=\frac{z-1}{5z^4}\vert_{z=w^2}=\frac{w^4-w^2}5 $$ Hence, $$I=2\pi i(r_1+r_2)=\frac{4\pi}5\sin72^\circ=\frac{\pi\sqrt2\sqrt{5+\sqrt5}}{5}$$

Answer 5

The prosaic way to proceed is also possible, we just factor the denominator: $$ f(x)= x^4 + x^3 + x^2 + x + 1 =\left(x^2 +\frac 12 x+1\right)^2 -\frac 54x^2 =\prod\left(x^2 +\frac {1\pm \sqrt 5}2 x+1\right) \ . $$ It is invariated by the Galois substitution $\sqrt 5\to-\sqrt 5$. The partial fraction decomposition is, also taking care of Galois invariance $$ \begin{aligned} \frac 1{f(x)} &= \frac 1{\sqrt 5} \left( \frac{x+\frac 12(1+\sqrt 5)}{x^2+\frac 12(1+\sqrt 5)+1} - \frac{x+\frac 12(1-\sqrt 5)}{x^2+\frac 12(1-\sqrt 5)+1} \right) \\ &= \frac1{\sqrt 5}\sum_{a=(1\pm\sqrt5)/2} \pm\frac{x+a}{x^2 +ax+1} \ . \end{aligned} $$ We have for a parameter $a$ the value $\int_{-M}^M\frac{x+a}{x^2 +ax+1}\; dx = \int_{-M}^M\frac{x+\frac 12a}{x^2 +ax+1}\; dx +\frac a2\int_{-M}^M\frac1{x^2 +ax+1}\; dx $.

• The first piece leads to $\frac 12\ln(x^2+ax+1)$ taken from $-M$ to $M$. In our case, we make a difference for the values $a=\frac 12(1\pm\sqrt 5)$. We obtain a difference of logarithms, thus a logarithm of the fraction $\frac{x^2+(1+\sqrt 5)x/2+1}{x^2+(1-\sqrt 5)x/2+1}$, and now the limit for $\pm M\to\infty$ exists and is zero.

• The second piece remains, and it is $\displaystyle\frac a2\int_{\Bbb R}\frac 1{x^2 +ax+1}\; dx=\frac {\pi a}{\sqrt {4-a^2}}$. For $a=\frac 12(1\pm\sqrt 5)$ we than compute:

  $a^2/(4-a^2)=\frac 15(5\pm 2\sqrt 5)$,

  then taking the square root, and taking care of the sign of $a$,

  $a/\sqrt{4-a^2}=\color{blue}{\pm} \frac 1{\sqrt5}\cdot\sqrt{5\pm 2\sqrt 5}$, so putting all together, the result is $$ \frac \pi{\sqrt 5} \cdot\frac 1{\sqrt 5} \left( \sqrt {5+ 2\sqrt 5} \color{blue}{+} \sqrt{5- 2\sqrt 5} \right) =\bbox[yellow]{\ \frac\pi 5\sqrt{10+2\sqrt 5}\ } \ . $$ Note: To see the last equality, we compute $\left( \sqrt {5+ 2\sqrt 5} \color{blue}{+} \sqrt{5- 2\sqrt 5} \right)^2=5+5\color{blue}{+} 2\sqrt{25-20}=10\color{blue}{+} 2\sqrt 5$.