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How to show that $int_C_R frace^ialpha z z^2+1dz to 0$ as $Rto infty$?

Calculate $ int_mathbbR fracdxx^4+1$ using the residues theorem.Show that $intlimits_-infty^infty f(t)dt=0$ where $fin H^infty(mathbbH)$Prove that $ frac12pi iint_C_rfrace^lambda tlambda^k+1dlambda =fract^kk!$Using Residues to calculate $int_-infty^infty fracdx(1+x^2)^n+1$How to compute $int_0^infty fracsqrtxx^2-1mathrm dx$Evaluate the complex integral $int_C_Rfracz^3(z-1)(z-4)^2$show that $int_C_R fracz e^iz1+z^2dz$ tends to zeroIntegral of $int_-infty^infty left(frac1alpha + ix + frac1alpha - ixright)^2 , dx$How to obtain the following estimates of $int_1^t fracds(t-s)^1/2s^alpha$Calculate $int_-infty^infty fracsin(z)zdz$

2

1

$begingroup$

Suppose $C_R$ is a semicircle centered at $0$ with radius $R>0$. How can we show that the integral $int_C_R frace^ialpha z z^2+1dz$ 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 $left| frac1z^2+1 right|le frac1R^2-1$. 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 $|e^ialpha z|=|e^ialpha xcdot e^-alpha y|=frac1e^alpha y$
  $endgroup$
  – 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 $dfracpie^$
  $endgroup$
  – dezdichado
  Mar 10 at 21:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user398843 like, if $alpha>0$, then use the parametrization: $Re^it$, so as to have: $e^ialpha z = e^ialpha Re^it = e^ialpha Rcos te^-Rsin t.$
  $endgroup$
  – 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 $int_C_R frace^ialpha z z^2+1dz$ 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 $left| frac1z^2+1 right|le frac1R^2-1$. 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 $|e^ialpha z|=|e^ialpha xcdot e^-alpha y|=frac1e^alpha y$
  $endgroup$
  – 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 $dfracpie^$
  $endgroup$
  – dezdichado
  Mar 10 at 21:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user398843 like, if $alpha>0$, then use the parametrization: $Re^it$, so as to have: $e^ialpha z = e^ialpha Re^it = e^ialpha Rcos te^-Rsin t.$
  $endgroup$
  – 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 $int_C_R frace^ialpha z z^2+1dz$ 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 $left| frac1z^2+1 right|le frac1R^2-1$. 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