### Author Topic: FE-P6  (Read 1884 times)

#### Victor Ivrii

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##### FE-P6
« on: December 18, 2018, 06:22:02 AM »
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

• Jr. Member
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• Karma: 3
##### Re: FE-P6
« Reply #1 on: December 18, 2018, 12:16:13 PM »
$\int_{0}^{\infty} \frac{1}{1+z^n} dz= 2 \pi i \sum_{i=1}^{K} Res(f,z_i)$

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

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

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
=$e^{i(\frac{1}{n}+\frac{2k}{n})\pi}$
$I = \frac{e^{i(\frac{1}{n}+\frac{2k}{n})\pi}}{1-e^{i\alpha}}$
« Last Edit: December 18, 2018, 01:41:58 PM by Victor Ivrii »

#### Victor Ivrii

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##### Re: FE-P6
« Reply #2 on: December 18, 2018, 12:29:26 PM »
$k=?$

Need to simplify to a real number

#### Yifei Wang

• Jr. Member
• Posts: 10
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##### Re: FE-P6
« Reply #3 on: December 18, 2018, 01:21:42 PM »

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

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

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

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

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

« Last Edit: December 18, 2018, 01:44:15 PM by Victor Ivrii »

#### Victor Ivrii

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##### Re: FE-P6
« Reply #4 on: December 18, 2018, 01:43:13 PM »
Need to simplify to a real number

#### Victor Ivrii

• Elder Member
• Posts: 2563
• Karma: 0
##### FE-P6 Official
« Reply #5 on: December 20, 2018, 04:37:41 AM »
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*}