FE-P6

[Victor Ivrii]

Calculate for real $n>1$
$$
I:= \int_0^\infty\frac{dx}{1+x^n}.
$$

Hint:  Consider
$$
\int_\gamma \frac{dz}{1+z^n}
$$
with with an arc of radius $R\to \infty$ and an angle $\alpha=\frac{2\pi}{n}$. Express the integral over the second straight segment through integral over the first one.

[Yifei Wang]

$\int_{0}^{\infty} \frac{1}{1+z^n} dz= 2 \pi i \sum_{i=1}^{K} Res(f,z_i)$

$z^n = -1 = e^{i\pi}$

$z = e^{i(\frac{1}{n}+\frac{2k}{n})\pi}$

we can decompose the integral into three parts, where $C_1$ is the straight line on x-axis, $C_R$ is the curve with radius R and $C_2$ being the line on the second quadrant

    $\int_{C}$  = $\int_{C_1}$  + $\int_{C_R}$  + $\int_{C_2}$

for $C_R$:
  let x = z $dx = dz$
  $\int_{C_R}$ = $\int_{0}^{R} \frac{1}{1+z^n} dz$
  as $\lim_{R\to\infty} $ = $\int_{0}^{\infty} \frac{1}{1+z^n} dz$
  = $2 \pi R *\frac{\alpha}{2 \pi}* \mid \frac{1}{1+z^n} \mid_{Max}$
  = $\alpha R \mid \frac{1}{1+z^n} \mid_{Max}$
  $\leq \alpha R \mid \frac{1}{1+R^n} \mid$
  $\leq \alpha R \mid \frac{1}{R^n} \mid$
  as $\lim_{R\to\infty} $
  $\leq \alpha R \mid \frac{1}{R^n} \mid$ = 0
  $\int_{0}^{R} \frac{1}{1+z^n} dz = 0$
 
for $C_1$:
  let $x = t e^{i\theta}, dx = e^{i\theta}dt $ $ t \in [0,R] $ and $\theta$ be the angle between the line and x-axis
  $\int_{C_1}$ = $\int_{0}^{R} \frac{1}{1+x^n} dz$
  as $\lim_{R\to\infty} $ = $\int_{0}^{\infty} \frac{1}{1+x^n} dx$
  = $\int_{0}^{\infty} \frac{e^{i\theta}}{1+{t e^{i\theta}}^n} dt$
  = $\int_{0}^{\infty} \frac{e^{i\theta}}{1+t^n e^{ni\theta}}dt$
  Since $\theta$ = 0
  = $\int_{0}^{\infty} \frac{1dt}{1+t^n}$ = $I$
  = $I$
 
    for $C_2$:
  let $x = t e^{i\alpha}, dx = e^{i\alpha}dt $ $ t \in [R,0] $
  $\int_{C_1}$ = $\int_{R}^{0} \frac{1}{1+x^n} dz$
  as $\lim_{R\to\infty} $ = $\int_{\infty}^{0} \frac{1}{1+x^n} dz$
  =$\int_{\infty}^{0} \frac{e^{i\alpha}}{1+{te^{i\alpha}}^n} dt$
  =$\int_{\infty}^{0} \frac{e^{i\alpha}}{1+{t^ne^{i\alpha n}}} dt$
  = -$e^{i\alpha}\int_{0}^{\infty} \frac{1}{1+{t^ne^{i\alpha n}}} dt$
  = -$e^{i\alpha}I$
 
 
  $\int_{C}$ = $I$ -$e^{i\alpha}I$
  = $1-e^{i\alpha}$I
  =$e^{i(\frac{1}{n}+\frac{2k}{n})\pi}$
  $I = \frac{e^{i(\frac{1}{n}+\frac{2k}{n})\pi}}{1-e^{i\alpha}}$

[Victor Ivrii]

$k=?$

Need to simplify to a real number

[Yifei Wang]

sorry for reposting, the loading for edit seems like taking forever.

$\int_{0}^{\infty} \frac{1}{1+z^n} dz$ = $2 \pi i$ $\sum_{i=1}^{n} Res(f,z_i)$

$z^n = -1 = e^i\pi$

$z = e^{i(\frac{1}{n}+\frac{2k}{n})\pi}$

we can decompose the integral into three parts, where $C_1$ is the straight line on the x-axis, $C_R$ is the curve with radius R and $C_2$ being the line on the second quadrant


    $\int_{C}$  = $\int_{C_1}$  + $\int_{C_R}$  + $\int_{C_2}$

for $C_R$:
  let x = z $dx = dz$
  $\int_{C_R}$ = $\int_{0}^{R} \frac{1}{1+z^n} dz$
  as $\lim_{R\to\infty} $ = $\int_{0}^{\infty} \frac{1}{1+z^n} dz$
  = $2 \pi R *\frac{\alpha}{2 \pi}* \mid \frac{1}{1+z^n} \mid_{Max}$
  = $\alpha R \mid \frac{1}{1+z^n} \mid_{Max}$
  $\leq \alpha R \mid \frac{1}{1+R^n} \mid$
  $\leq \alpha R \mid \frac{1}{R^n} \mid$
  as $\lim_{R\to\infty} $
  $\leq \alpha R \mid \frac{1}{R^n} \mid$ = 0
  $\int_{0}^{R} \frac{1}{1+z^n} dz = 0$
 
 $----------------------------------------------------$
for $C_1$:
  let $x = t e^{i\theta}, dx = e^{i\theta}dt $ $ t \in [0,R] $ and $\theta$ be the angle between the line and x-axis
  $\int_{C_1}$ = $\int_{0}^{R} \frac{1}{1+x^n} dz$
  as $\lim_{R\to\infty} $ = $\int_{0}^{\infty} \frac{1}{1+x^n} dx$
  = $\int_{0}^{\infty} \frac{e^{i\theta}}{1+{t e^{i\theta}}^n} dt$
  = $\int_{0}^{\infty} \frac{e^{i\theta}}{1+t^n e^{ni\theta}}dt$
  Since $\theta$ = 0
  = $\int_{0}^{\infty} \frac{1dt}{1+t^n}$ = $I$
  = $I$
 
 
  $-------------------------------------------------------$
  for $C_2$:
  let $x = t e^{i\alpha}, dx = e^{i\alpha}dt $ $ t \in [R,0] $
  $\int_{C_1}$ = $\int_{R}^{0} \frac{1}{1+x^n} dz$
  as $\lim_{R\to\infty} $ = $\int_{\infty}^{0} \frac{1}{1+x^n} dz$
  =$\int_{\infty}^{0} \frac{e^{i\alpha}}{1+{te^{i\alpha}}^n} dt$
  =$\int_{\infty}^{0} \frac{e^{i\alpha}}{1+{t^ne^{i\alpha n}}} dt$
  = -$e^{i\alpha}\int_{0}^{\infty} \frac{1}{1+{t^ne^{i\alpha n}}} dt$
  = -$e^{i\alpha}I$
 
 
  $\int_{C}$ = $I$ -$e^{i\alpha}I$
  = $1-e^{i\alpha}$I
  =$e^{i(\frac{1}{n}+\frac{2k}{n})\pi}$
  $I = \frac{e^{i(\frac{1}{n}+\frac{2k}{n})\pi}}{1-e^{i\alpha}}$


since we are in the principal branch, we let K = 0 Then you do not need $k$ at all anywhere

$I = \frac{e^{i(\frac{1}{n})\pi}}{1-e^{i\alpha}}$

[Victor Ivrii]

Need to simplify to a real number

[Victor Ivrii]

There is a single pole of $f(z)=\frac{1}{1+z^n}$ inside $\Gamma$, namely $z=e^{i\pi/n}$, which is a simple pole and the residue is $\frac{1}{(1+z^n)'}\bigr|_{z=e^{i\pi/n}}= \frac{1}{nz^{n-1}}\bigr|_{z=e^{i\pi/n}}=-\frac{1}{n}e^{i\pi/n}$.

Therefore due to the residue theorem $I_R+J_R+K_R= -\frac{2}{n}\pi i \times  e^{i\alpha/2}$, where $K_R$ is an integral over an arc and $J_R$ is an integral over the second straight segment.

Then
$$
J_R=\int _{R}^0 \frac{e^{i\alpha}\,dt }{1+e^{in\alpha}t^n}=-e^{i\alpha}I_R
$$
with $I_R=\int_0^R\frac{dx}{1+x^n}$.

On the other hand,
$$
|K_R|=|\int_0^\alpha \frac{iRe^{it}\,dt}{1+e^{itn}R^n }|\le \frac{R}{R^n-1}\int_0^\alpha\, dt=\frac{\alpha R}{R^n-1}\to 0
\qquad\text{as }\ \ R\to \infty.$$


Then $(1-e^{i\alpha}) I =-\frac{2}{n}\pi i \times  e^{i\alpha/2}$ and
\begin{align*}
I =& -\frac{2}{n}\pi i \times  \frac{e^{i\alpha/2}}{1-e^{i\pi \alpha}}=
&&-\frac{2}{n}\pi i \times  \frac{1}{e^{-i\alpha/2}-e^{i\pi \alpha/2}}=\\
&-\frac{2}{n}\pi i \times  \frac{1}{-2i\sin(\alpha/2)}=
&&\frac{\pi}{n\sin(\pi /n)}.
\end{align*}