$\int_0^\infty \frac{1}{1+x^ 9} \, dx$

$\int_0^\infty \frac{1}{1+x^9} \, dx$

I tried taking the integral of $\Gamma_R = [0,R] \cup \gamma_R \cup I_R$, where we see that \gamma_R is the circle parametrized by $z = Re^{it}$ with $t\in[0,\frac{\pi}{2}]$. And $I_R$ is the line form $iR$ to $0$

one can note the following three things:

1) $\int_0^\infty \frac{1}{1+x^ 9}\,dz$ = $\int_0^\infty \frac{1}{1+z^ 9}\,dz$. (Note that $z$ is on the $x$-axis)

2) $\int_{\gamma_R}\frac{1}{1+z^ 9} = 0$. (By the ML inequality)

the main problem is when i parametrize $I^{-}_{R}$ with $z(x) = xi$ and $x \in [0,R]$. The main problem is that i get that:

3) $\int_{I_R}\frac{1}{1+z^9} = -\int_{I_R^-}\frac{1}{1+z^9} = -\int_{0}^{R} \frac{i}{ix^9 + 1}$.

note that i want to work towards taking $R$ to infinity eventually and then equal the sum of the integrals to the sum of the residuals of $\Gamma_R$ times $2\pi i$. The main problem here is that my term in 3), does not look like the form $c \cdot \int_{0}^{\infty}\frac{1}{1+x^ 9}$ with c a constant value. If i can get this done i think i'm finished.

PS: The singular points i got where $e^{\frac{\pi i}{9}},e^{\frac{3 \pi i}{9}}$.

There is a simpler way, using Euler's $B$ function:

1) change the variable from $x$ to $y=x^9$. You'll get $\frac 1 9 \int \limits _0 ^\infty \frac {y^{- \frac 8 9}} {1+y} \mathbb{d}y$.

2) make a second change of variables: $t=\frac y {1+y}$; you'll get $\frac 1 9 \int \limits _0 ^1 t^{- \frac 8 9} (1-t)^ {- \frac 1 9} \mathbb{d}t$, which is $\frac 1 9 B(\frac 1 9, \frac 8 9)$.

3) finally, using that $B(x,y)=\frac {\Gamma(x) \Gamma(y)} {\Gamma(x+y)}$ and $\Gamma(x) \Gamma(1-x) = \frac \pi {\sin( \pi x)}$, you'll get $\frac \pi {9 \sin \frac \pi 9}$.

Let's solve a harder problem...$$\int_0^{\infty} {1 \over {1+x^p}} dx$$ Let $u=x^p$... $$\int_0^{\infty} {1 \over {1+x^p}} dx={1 \over p} \int_0^{\infty} {u^{{1 \over p}-1} \over {1+u}} du$$ The definition of the Beta function is... $$B(p,q)=\int_0^{\infty} {{t^{n-1} \over ({t+1})^{p+q}} dt}$$ Note the striking similarity and evaluate accordingly... $${1 \over p} \cdot B \left({1 \over p},1-{1 \over p} \right)$$ Use the Beta function's relation with the Gamma function to get... $${1 \over p} \cdot B \left({1 \over p},1-{1 \over p} \right)={1 \over p} \cdot {{\Gamma \left({1 \over p} \right) \cdot \Gamma \left(1-{1 \over p} \right)} \over {\Gamma (1)}}$$

Use Euler's reflection formula $\Gamma (x) \cdot \Gamma (1-x)= \pi \csc(\pi x)$. You should arrive at... $$\int_0^{\infty} {1 \over {1+x^p}} dx={\pi \over p} \cdot \csc \left( {\pi \over p }\right)$$ For $p=9$... $$\int_0^{\infty} {1 \over {1+x^9}} dx={\pi \over 9} \cdot \csc \left( {\pi \over 9} \right)=1.0206...$$

An easier way to evaluate the general case is to consider a sector surrounding one residue, rather than a whole circle because it's easier to compute the contour integral. To start, consider the function

$$f(z)=\frac 1{z^n+1}$$

Where $n$ is an integer. Now imagine a "pizza slice" contour of radius $R$ pictured below

As $R\to\infty$, the arc integral vanishes. Leaving us with

$$\oint\limits_{C}dz\, f(z)=\int\limits_0^{\infty}dx\, f(x)-e^{2\pi i/n}\int\limits_0^{\infty}dz\, f(z)$$

Here, observe that if you take out a factor of $e^{\pi i/n}$, you get$$e^{\pi i/n}(e^{-\pi i/n}-e^{\pi i/n})\int\limits_0^{\infty}dx\, f(x)=-2i e^{\pi i/n}\sin\left(\frac {\pi}n\right)\int\limits_0^{\infty}dx\, f(x)$$The contour integral as only one residue at $z=e^{\pi i/n}$. The residue is given by

$$\operatorname*{Res}_{z=e^{\pi i/n}}\frac 1{1+z^n}=\lim\limits_{z\to e^{\pi i/n}}\frac {z-e^{\pi i/n}}{z^n+1}=-\frac 1ne^{\pi i/n}$$Therefore$$-2ie^{\pi i/n}\sin\left(\frac {\pi}n\right)\int\limits_0^{\infty}dx\, f(x)=-\frac {2\pi i}ne^{\pi i/n}$$

Simplify both sides you see that

$$\int\limits_0^{\infty}\frac {dx}{1+x^n}=\frac {\pi}n\csc\left(\frac {\pi}n\right)$$

We can solve this using contour analysis by judiciously selecting our closed contour. We know that there are poles at $e^{i(2n-1)\pi/9}$ for $n=1,\dots,9$.

To that end, we let $C$ be the closed contour comprised of $(1)$ the straight line segment from $0$ to $R$, $(2)$ the circular arc from $R$ to $Re^{i2\pi/9}$, and $(3)$ the straight line segment from $Re^{i2\pi/9}$ to $0$.

Note that for $R>1$, $C$ encloses only the pole at $z=e^{i\pi/9}$ and that on the straight line segment from $Re^{i2\pi/9}$ to $0$, $z^9=(te^{i2\pi/9})^9=t^9$, for $t\in [0,R]$.

Applying the residue theorem for any $R>1$ reveals

$$\begin{align} \oint_C\frac{1}{1+z^9}\,dz &=\int_0^R \frac{1}{1+x^9}\,dx+\int_0^{2\pi/9}\frac{1}{1+R^9e^{i9\phi}}\,iRe^{i\phi}\,d\phi+\int_R^0 \frac{1}{1+(xe^{i2\pi/9})^9}\,e^{i2\pi/9}\,dx\\\\ &=(1-e^{i2\pi/9})\int_0^R \frac{1}{1+x^9}\,dx+\int_0^{2\pi/9}\frac{1}{1+R^9e^{i9\phi}}\,iRe^{i\phi}\,d\phi\\\\ &=2\pi i \text{Res}\left(\frac1{1+z^9},z=e^{i\pi/9}\right)\\\\ &=2\pi i \lim_{z\to e^{i\pi/9}}\left(\frac{z-e^{i\pi/9}}{1+z^9}\right)\\\\ &=2\pi i \left(\frac{1}{9e^{i8\pi/9}}\right) \end{align}$$

Letting $R\to \infty$, we find that

$$\int_0^\infty \frac{1}{1+x^9}\,dx=\frac{2\pi i }{9}\frac{1}{e^{i8\pi/9(1-e^{i2\pi/9})}}=\frac{\pi}{9\sin(\pi/9)}$$

as expected!