Dtjytyk

How big is a MODIS 250m pixel in reality?

How to create the Curved texte?

In a future war, an old lady is trying to raise a boy but one of the weapons has made everyone deaf

Existence of subset with given Hausdorff dimension

Sailing the cryptic seas

What did Alexander Pope mean by "Expletives their feeble Aid do join"?

How to simplify this time periods definition interface?

How to read the value of this capacitor?

How to explain that I do not want to visit a country due to personal safety concern?

Did Ender ever learn that he killed Stilson and/or Bonzo?

Does someone need to be connected to my network to sniff HTTP requests?

Life insurance that covers only simultaneous/dual deaths

How do I hide Chekhov's Gun?

Do the common programs (for example: "ls", "cat") in Linux and BSD come from the same source code?

Recruiter wants very extensive technical details about all of my previous work

Interplanetary conflict, some disease destroys the ability to understand or appreciate music

Is it possible to upcast ritual spells?

If I can solve Sudoku can I solve Travelling Salesman Problem(TSP)? If yes, how?

A link redirect to http instead of https: how critical is it?

Gravity magic - How does it work?

How can I track script which gives me "command not found" right after the login?

What are the naunces between the use of 訊く instead of 聞く in the following sentence?

Why doesn't the EU now just force the UK to choose between referendum and no-deal?

What options are left, if Britain cannot decide?

How to show that $int_{C_R} frac{e^{ialpha z }}{z^2+1}dz to 0$ as $Rto infty$?

Calculate $ int_{mathbb{R}} frac{dx}{x^4+1}$ using the residues theorem.Show that $intlimits_{-infty}^infty f(t)dt=0$ where $fin H^infty(mathbb{H})$Prove that $ frac{1}{2pi i}int_{C_r}frac{e^{lambda t}}{lambda^{k+1}}dlambda =frac{t^k}{k!}$Using Residues to calculate $int_{-infty}^{infty} frac{dx}{(1+x^2)^{n+1}}$How to compute $int_0^{infty} frac{sqrt{x}}{x^2-1}mathrm dx$Evaluate the complex integral $int_{C_R}frac{z^3}{(z-1)(z-4)^2}$show that $int_{C_R} frac{z e^{iz}}{1+z^2}dz$ tends to zeroIntegral of $int_{-infty}^{infty} left(frac{1}{alpha + ix} + frac{1}{alpha - ix}right)^2 , dx$How to obtain the following estimates of $int_{1}^{t} frac{ds}{(t-s)^{1/2}s^{alpha}}$Calculate $int_{-infty}^{infty} frac{sin(z)}{z}dz$

2

1

$begingroup$

Suppose $C_R$ is a semicircle centered at $0$ with radius $R>0$. How can we show that the integral $int_{C_R} frac{e^{ialpha z }}{z^2+1}dz$ tends to zero as $Rto infty$?

My idea is to estimate the bound of the integrand, but it seems that it only works when $alpha ge 0$.

It is clear that $left| frac{{1}}{z^2+1} right|le frac{{1}}{R^2-1}$. Then we want to show that $|e^{ialpha z}| = |e^{-alpha y}|le 1$ for $y>0$.

What is the difference between the two parametric representations, $Re^{it}$ and $Re^{-it}$ ($0le tle pi$)? Note that $alpha$ can be either positive or negative.

calculus complex-analysis multivariable-calculus

share|cite|improve this question

edited Mar 10 at 21:35

user398843

asked Mar 10 at 20:38

user398843user398843

689216

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  if $z=x+iy$ then $|e^{ialpha z}|=|e^{ialpha x}cdot e^{-alpha y}|=frac{1}{e^{alpha y}}$
  $endgroup$
  – rtybase
  Mar 10 at 20:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @rtybase I saw that, but the issue is that $alpha$ can be negative and $e^{-alpha y}>1$ for $y>0$ in this case.
  $endgroup$
  – user398843
  Mar 10 at 20:56
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed it can, so you can't rely on $|e^{ialpha z}| leq 1$.
  $endgroup$
  – rtybase
  Mar 10 at 21:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  just divide the problem into two cases according to the sign of $alpha$ and use the corresponding, appropriate half circles. Then, you should be able to get a single answer by using $|alpha|.$In fact, it should be $dfrac{pi}{e^{|alpha|}}$
  $endgroup$
  – dezdichado
  Mar 10 at 21:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user398843 like, if $alpha>0$, then use the parametrization: $Re^{it}$, so as to have: $e^{ialpha z} = e^{ialpha Re^{it}} = e^{ialpha Rcos t}e^{-Rsin t}.$
  $endgroup$
  – dezdichado
  Mar 10 at 22:06
  

  
  
  
  

 | 
show 2 more comments

2

1

$begingroup$

Suppose $C_R$ is a semicircle centered at $0$ with radius $R>0$. How can we show that the integral $int_{C_R} frac{e^{ialpha z }}{z^2+1}dz$ tends to zero as $Rto infty$?

My idea is to estimate the bound of the integrand, but it seems that it only works when $alpha ge 0$.

It is clear that $left| frac{{1}}{z^2+1} right|le frac{{1}}{R^2-1}$. Then we want to show that $|e^{ialpha z}| = |e^{-alpha y}|le 1$ for $y>0$.

What is the difference between the two parametric representations, $Re^{it}$ and $Re^{-it}$ ($0le tle pi$)? Note that $alpha$ can be either positive or negative.

calculus complex-analysis multivariable-calculus

share|cite|improve this question

edited Mar 10 at 21:35

user398843

asked Mar 10 at 20:38

user398843user398843

689216

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  if $z=x+iy$ then $|e^{ialpha z}|=|e^{ialpha x}cdot e^{-alpha y}|=frac{1}{e^{alpha y}}$
  $endgroup$
  – rtybase
  Mar 10 at 20:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @rtybase I saw that, but the issue is that $alpha$ can be negative and $e^{-alpha y}>1$ for $y>0$ in this case.
  $endgroup$
  – user398843
  Mar 10 at 20:56
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed it can, so you can't rely on $|e^{ialpha z}| leq 1$.
  $endgroup$
  – rtybase
  Mar 10 at 21:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  just divide the problem into two cases according to the sign of $alpha$ and use the corresponding, appropriate half circles. Then, you should be able to get a single answer by using $|alpha|.$In fact, it should be $dfrac{pi}{e^{|alpha|}}$
  $endgroup$
  – dezdichado
  Mar 10 at 21:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user398843 like, if $alpha>0$, then use the parametrization: $Re^{it}$, so as to have: $e^{ialpha z} = e^{ialpha Re^{it}} = e^{ialpha Rcos t}e^{-Rsin t}.$
  $endgroup$
  – dezdichado
  Mar 10 at 22:06
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

Suppose $C_R$ is a semicircle centered at $0$ with radius $R>0$. How can we show that the integral $int_{C_R} frac{e^{ialpha z }}{z^2+1}dz$ tends to zero as $Rto infty$?

My idea is to estimate the bound of the integrand, but it seems that it only works when $alpha ge 0$.

It is clear that $left| frac{{1}}{z^2+1} right|le frac{{1}}{R^2-1}$. Then we want to show that $|e^{ialpha z}| = |e^{-alpha y}|le 1$ for $y>0$.

What is the difference between the two parametric representations, $Re^{it}$ and $Re^{-it}$ ($0le tle pi$)? Note that $alpha$ can be either positive or negative.

calculus complex-analysis multivariable-calculus

share|cite|improve this question

edited Mar 10 at 21:35

user398843

asked Mar 10 at 20:38

user398843user398843

689216

$endgroup$

share|cite|improve this question

edited Mar 10 at 21:35

user398843

asked Mar 10 at 20:38

user398843user398843

689216

share|cite|improve this question

edited Mar 10 at 21:35

user398843

asked Mar 10 at 20:38

user398843user398843

689216

asked Mar 10 at 20:38

user398843user398843

689216

asked Mar 10 at 20:38

user398843user398843

689216