Dtjytyk

What loss function to use when labels are probabilities?

Direct Experience of Meditation

Stopping real property loss from eroding embankment

Why does this iterative way of solving of equation work?

Stop battery usage [Ubuntu 18]

Can a monk deflect thrown melee weapons?

When is phishing education going too far?

Why is "Captain Marvel" translated as male in Portugal?

How to say 'striped' in Latin

New Order #5: where Fibonacci and Beatty meet at Wythoff

Who can trigger ship-wide alerts in Star Trek?

Unexpected result with right shift after bitwise negation

Blender game recording at the wrong time

What LEGO pieces have "real-world" functionality?

What is the electric potential inside a point charge?

How to say that you spent the night with someone, you were only sleeping and nothing else?

Determine whether f is a function, an injection, a surjection

Statistical model of ligand substitution

How do I automatically answer y in bash script?

What was the last x86 CPU that did not have the x87 floating-point unit built in?

Simulating Exploding Dice

Writing Thesis: Copying from published papers

What do you call a plan that's an alternative plan in case your initial plan fails?

Is drag coefficient lowest at zero angle of attack?

Evaluate $int_{-infty}^{infty}frac{dx}{(1+x^2)^z}$ where $Re(z)>frac{1}{2}$

1

$begingroup$

I am dealing with the following complex integration:

$$int_{-infty}^{infty}frac{dx}{(1+x^2)^z} quadtext{ where }quadRe(z)>frac{1}{2}$$

But I do not know how exactly to deal with the $z$.

complex-analysis complex-integration

share|cite|improve this question

edited Mar 23 at 13:52

StubbornAtom

6,42731440

asked Mar 23 at 12:44

thdrthdr

62

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By exploiting symmetry and enforcing the substitution $x=tantheta$ you immediately get something manageable through Euler's Beta function.
  $endgroup$
  – Jack D'Aurizio
  Mar 23 at 13:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @thdr Please let me know how I can improve my answer. I really want to give you the best answer I can.
  $endgroup$
  – Mark Viola
  Mar 29 at 16:28
  

  
  
  
  

add a comment | 

1

$begingroup$

I am dealing with the following complex integration:

$$int_{-infty}^{infty}frac{dx}{(1+x^2)^z} quadtext{ where }quadRe(z)>frac{1}{2}$$

But I do not know how exactly to deal with the $z$.

complex-analysis complex-integration

share|cite|improve this question

edited Mar 23 at 13:52

StubbornAtom

6,42731440

asked Mar 23 at 12:44

thdrthdr

62

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By exploiting symmetry and enforcing the substitution $x=tantheta$ you immediately get something manageable through Euler's Beta function.
  $endgroup$
  – Jack D'Aurizio
  Mar 23 at 13:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @thdr Please let me know how I can improve my answer. I really want to give you the best answer I can.
  $endgroup$
  – Mark Viola
  Mar 29 at 16:28
  

  
  
  
  

add a comment | 

$begingroup$

I am dealing with the following complex integration:

$$int_{-infty}^{infty}frac{dx}{(1+x^2)^z} quadtext{ where }quadRe(z)>frac{1}{2}$$

But I do not know how exactly to deal with the $z$.

complex-analysis complex-integration

share|cite|improve this question

edited Mar 23 at 13:52

StubbornAtom

6,42731440

asked Mar 23 at 12:44

thdrthdr

62

$endgroup$

share|cite|improve this question

edited Mar 23 at 13:52

StubbornAtom

6,42731440

asked Mar 23 at 12:44

thdrthdr

62

share|cite|improve this question

edited Mar 23 at 13:52

StubbornAtom

6,42731440

asked Mar 23 at 12:44

thdrthdr

62

edited Mar 23 at 13:52

StubbornAtom

6,42731440

edited Mar 23 at 13:52

StubbornAtom

6,42731440

asked Mar 23 at 12:44

thdrthdr

62

asked Mar 23 at 12:44

thdrthdr

62

2 Answers
2

active

oldest

votes

1

$begingroup$

Using Schwinger parametrization one can write
$$int_{-infty}^{infty}frac{1}{(1+x^2)^z}dx=frac{1}{Gamma(z)}int_{-infty}^{infty}int_0^infty t^{z-1}e^{-t(1+x^2)}dtdx$$
Then switching the order of integration we have
$$frac{1}{Gamma(z)}int_0^{infty}t^{z-1}e^{-t}int_{-infty}^infty e^{-tx^2}dxdt$$
The second integral is a Gaussian integral and can be found by using the substitution $s=sqrt{t}xrightarrow ds=sqrt{t}dx$ which gives
$$int_{-infty}^infty e^{-tx^2}dx=frac1{sqrt{t}}int_{-infty}^infty e^{-s^2}ds=sqrt{frac{pi}t}$$
So the integral becomes
$$frac{1}{Gamma(z)}int_0^{infty}t^{z-1}e^{-t}sqrt{frac{pi}t}dt=frac{sqrt{pi}}{Gamma(z)}int_0^{infty}t^{z-1.5}e^{-t}dt$$
and then the integral left is the Gamma function - $Gamma(z-0.5)$ - because
$$int_0^infty t^{z-1}e^{-t}dt=Gamma(z)$$
Giving the final answer of
$$boxed{int_{-infty}^{infty}frac{1}{(1+x^2)^z}dx=sqrt{pi}frac{Gamma(z-0.5)}{Gamma(z)}}$$

share|cite|improve this answer

answered Mar 23 at 13:08

Peter ForemanPeter Foreman

7,2411319

$endgroup$

add a comment | 

0

$begingroup$

Let $f(z)$ be given by

$$f(z) =int_0^infty frac1{(1+x^2)^z},dx$$

Since the integrand is an even function of $x$ we have after enforcing the substitution $xmapsto sqrt x$

$$begin{align}
f(z)&=int_0^infty frac{x^{-1/2}}{(1+x)^z},dx\\
&=B(1/2,z-1/2)\\
&=sqrt pi ,frac{Gamma(z-1/2)}{Gamma(z)}
end{align}$$

share|cite|improve this answer

answered Mar 23 at 14:51

Mark ViolaMark Viola

134k1278177

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote and accept an answer as you see fit.
  $endgroup$
  – Mark Viola
  Mar 29 at 16:29
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3159275%2fevaluate-int-infty-infty-fracdx1x2z-where-rez-frac12%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

Using Schwinger parametrization one can write
$$int_{-infty}^{infty}frac{1}{(1+x^2)^z}dx=frac{1}{Gamma(z)}int_{-infty}^{infty}int_0^infty t^{z-1}e^{-t(1+x^2)}dtdx$$
Then switching the order of integration we have
$$frac{1}{Gamma(z)}int_0^{infty}t^{z-1}e^{-t}int_{-infty}^infty e^{-tx^2}dxdt$$
The second integral is a Gaussian integral and can be found by using the substitution $s=sqrt{t}xrightarrow ds=sqrt{t}dx$ which gives
$$int_{-infty}^infty e^{-tx^2}dx=frac1{sqrt{t}}int_{-infty}^infty e^{-s^2}ds=sqrt{frac{pi}t}$$
So the integral becomes
$$frac{1}{Gamma(z)}int_0^{infty}t^{z-1}e^{-t}sqrt{frac{pi}t}dt=frac{sqrt{pi}}{Gamma(z)}int_0^{infty}t^{z-1.5}e^{-t}dt$$
and then the integral left is the Gamma function - $Gamma(z-0.5)$ - because
$$int_0^infty t^{z-1}e^{-t}dt=Gamma(z)$$
Giving the final answer of
$$boxed{int_{-infty}^{infty}frac{1}{(1+x^2)^z}dx=sqrt{pi}frac{Gamma(z-0.5)}{Gamma(z)}}$$

share|cite|improve this answer

answered Mar 23 at 13:08

Peter ForemanPeter Foreman

7,2411319

$endgroup$

add a comment | 

1

$begingroup$

Using Schwinger parametrization one can write
$$int_{-infty}^{infty}frac{1}{(1+x^2)^z}dx=frac{1}{Gamma(z)}int_{-infty}^{infty}int_0^infty t^{z-1}e^{-t(1+x^2)}dtdx$$
Then switching the order of integration we have
$$frac{1}{Gamma(z)}int_0^{infty}t^{z-1}e^{-t}int_{-infty}^infty e^{-tx^2}dxdt$$
The second integral is a Gaussian integral and can be found by using the substitution $s=sqrt{t}xrightarrow ds=sqrt{t}dx$ which gives
$$int_{-infty}^infty e^{-tx^2}dx=frac1{sqrt{t}}int_{-infty}^infty e^{-s^2}ds=sqrt{frac{pi}t}$$
So the integral becomes
$$frac{1}{Gamma(z)}int_0^{infty}t^{z-1}e^{-t}sqrt{frac{pi}t}dt=frac{sqrt{pi}}{Gamma(z)}int_0^{infty}t^{z-1.5}e^{-t}dt$$
and then the integral left is the Gamma function - $Gamma(z-0.5)$ - because
$$int_0^infty t^{z-1}e^{-t}dt=Gamma(z)$$
Giving the final answer of
$$boxed{int_{-infty}^{infty}frac{1}{(1+x^2)^z}dx=sqrt{pi}frac{Gamma(z-0.5)}{Gamma(z)}}$$

share|cite|improve this answer

answered Mar 23 at 13:08

Peter ForemanPeter Foreman

7,2411319

$endgroup$

add a comment | 

$begingroup$

Using Schwinger parametrization one can write
$$int_{-infty}^{infty}frac{1}{(1+x^2)^z}dx=frac{1}{Gamma(z)}int_{-infty}^{infty}int_0^infty t^{z-1}e^{-t(1+x^2)}dtdx$$
Then switching the order of integration we have
$$frac{1}{Gamma(z)}int_0^{infty}t^{z-1}e^{-t}int_{-infty}^infty e^{-tx^2}dxdt$$
The second integral is a Gaussian integral and can be found by using the substitution $s=sqrt{t}xrightarrow ds=sqrt{t}dx$ which gives
$$int_{-infty}^infty e^{-tx^2}dx=frac1{sqrt{t}}int_{-infty}^infty e^{-s^2}ds=sqrt{frac{pi}t}$$
So the integral becomes
$$frac{1}{Gamma(z)}int_0^{infty}t^{z-1}e^{-t}sqrt{frac{pi}t}dt=frac{sqrt{pi}}{Gamma(z)}int_0^{infty}t^{z-1.5}e^{-t}dt$$
and then the integral left is the Gamma function - $Gamma(z-0.5)$ - because
$$int_0^infty t^{z-1}e^{-t}dt=Gamma(z)$$
Giving the final answer of
$$boxed{int_{-infty}^{infty}frac{1}{(1+x^2)^z}dx=sqrt{pi}frac{Gamma(z-0.5)}{Gamma(z)}}$$

share|cite|improve this answer

answered Mar 23 at 13:08

Peter ForemanPeter Foreman

7,2411319

$endgroup$

share|cite|improve this answer

answered Mar 23 at 13:08

Peter ForemanPeter Foreman

7,2411319

answered Mar 23 at 13:08

Peter ForemanPeter Foreman

7,2411319

answered Mar 23 at 13:08

Peter ForemanPeter Foreman

7,2411319

0

$begingroup$

Let $f(z)$ be given by

$$f(z) =int_0^infty frac1{(1+x^2)^z},dx$$

Since the integrand is an even function of $x$ we have after enforcing the substitution $xmapsto sqrt x$

$$begin{align}
f(z)&=int_0^infty frac{x^{-1/2}}{(1+x)^z},dx\\
&=B(1/2,z-1/2)\\
&=sqrt pi ,frac{Gamma(z-1/2)}{Gamma(z)}
end{align}$$

share|cite|improve this answer

answered Mar 23 at 14:51

Mark ViolaMark Viola

134k1278177

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote and accept an answer as you see fit.
  $endgroup$
  – Mark Viola
  Mar 29 at 16:29
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $f(z)$ be given by

$$f(z) =int_0^infty frac1{(1+x^2)^z},dx$$

Since the integrand is an even function of $x$ we have after enforcing the substitution $xmapsto sqrt x$

$$begin{align}
f(z)&=int_0^infty frac{x^{-1/2}}{(1+x)^z},dx\\
&=B(1/2,z-1/2)\\
&=sqrt pi ,frac{Gamma(z-1/2)}{Gamma(z)}
end{align}$$

share|cite|improve this answer

answered Mar 23 at 14:51

Mark ViolaMark Viola

134k1278177

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote and accept an answer as you see fit.
  $endgroup$
  – Mark Viola
  Mar 29 at 16:29
  

  
  
  
  

add a comment | 

$begingroup$

Let $f(z)$ be given by

$$f(z) =int_0^infty frac1{(1+x^2)^z},dx$$

Since the integrand is an even function of $x$ we have after enforcing the substitution $xmapsto sqrt x$

$$begin{align}
f(z)&=int_0^infty frac{x^{-1/2}}{(1+x)^z},dx\\
&=B(1/2,z-1/2)\\
&=sqrt pi ,frac{Gamma(z-1/2)}{Gamma(z)}
end{align}$$

share|cite|improve this answer

answered Mar 23 at 14:51

Mark ViolaMark Viola

134k1278177

$endgroup$

share|cite|improve this answer

answered Mar 23 at 14:51

Mark ViolaMark Viola

134k1278177

answered Mar 23 at 14:51

Mark ViolaMark Viola

134k1278177

answered Mar 23 at 14:51

Mark ViolaMark Viola

134k1278177