Fdtxjr

How does a predictive coding aid in lossless compression?

How could indestructible materials be used in power generation?

Short story with a alien planet, government officials must wear exploding medallions

Why didn't Miles's spider sense work before?

How dangerous is XSS?

How writing a dominant 7 sus4 chord in RNA ( Vsus7 chord in the 1st inversion)

Can a virus destroy the BIOS of a modern computer?

What type of content (depth/breadth) is expected for a short presentation for Asst Professor interview in the UK?

Method Does Not Exist error message

Avoiding the "not like other girls" trope?

Mathematica command that allows it to read my intentions

How do I handle a potential work/personal life conflict as the manager of one of my friends?

Why no variance term in Bayesian logistic regression?

How can saying a song's name be a copyright violation?

Why didn't Boeing produce its own regional jet?

What about the virus in 12 Monkeys?

Should I tell management that I intend to leave due to bad software development practices?

Could the museum Saturn V's be refitted for one more flight?

ssTTsSTtRrriinInnnnNNNIiinngg

Examples of smooth manifolds admitting inbetween one and a continuum of complex structures

CAST throwing error when run in stored procedure but not when run as raw query

Plagiarism or not?

Is there a hemisphere-neutral way of specifying a season?

Personal Teleportation: From Rags to Riches

Evaluate the integral: $int_0^inftyfractan^-1(tx)xleft(1+x^2right) mathrmdx$

Another polylog integralEvaluating $int_0^infty left[left(frac20152015+x+cdots +frac22+x+frac11+x-xright)^2016+1 right] ^-1mathrmdx$How do you evaluate $int_0^fracpi2 frac(sec x)^frac13(sec x)^frac13+(tan x)^frac13 , dx ?$What is $int_0^1 ln (1-x) ln left(ln left(frac1xright)right) , dx$?Integral $ int_0^infty ln xleft[ln left( fracx+12 right) - frac1x+1 - psi left( fracx+12 right) right] mathrmdx $Find: $int_0^pi x^2ln(sin(x))dx$A direct proof for $int_0^x frac- x ln(1-u^2)u sqrtx^2-u^2 , mathrmd u = arcsin^2(x)$A closed-form expression for $int_0^infty fracln (1+x^alpha) ln (1+x^-beta)x , mathrmd x$On the integrals $int_0^infty [1-x^n operatornamearccot^n(x)] , mathrmd x$Integral $int_0^infty fracx^2jmathrm dx(x^4+2ax^2+1)^n+1$

3

2

$begingroup$

I've been trying to evaluate the integral for a while now, and I've been unable to find it anywhere...
I tried substituting $tan^-1(tx)$ as $u$ but got nowhere... I have done dozens of other substitutions...
I've been told to use the properties of gaussian integrals and the gamma function, but can't seem to figure out a way to bring in $mathrme$...
A few hints that point in the right direction to think will really help!

It would be really really helpful if you could solve this using only the properties of gaussian and gamma functions!

definite-integrals

share|cite|improve this question

edited Mar 21 at 3:07

Mark Viola

134k1278176

asked Mar 20 at 22:03

Pratik ApshingePratik Apshinge

516

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Call this integral $f(t)$. Evaluate $f^prime (t)$ with partial fractions, then use $f(0)=0$.
  $endgroup$
  – J.G.
  Mar 20 at 22:06
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, I understand exactly what you are talking about...
  $endgroup$
  – Pratik Apshinge
  Mar 20 at 22:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How will the derivative of the function give me information about the function itself?
  $endgroup$
  – Pratik Apshinge
  Mar 20 at 22:17
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe you can integrate the derivative of the function.
  $endgroup$
  – Minus One-Twelfth
  Mar 20 at 22:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And, also, is there any possible way to do this with gaussian integrals or the gamma function?
  $endgroup$
  – Pratik Apshinge
  Mar 20 at 22:18
  

  
  
  
  

add a comment | 

3

2

$begingroup$

I've been trying to evaluate the integral for a while now, and I've been unable to find it anywhere...
I tried substituting $tan^-1(tx)$ as $u$ but got nowhere... I have done dozens of other substitutions...
I've been told to use the properties of gaussian integrals and the gamma function, but can't seem to figure out a way to bring in $mathrme$...
A few hints that point in the right direction to think will really help!

It would be really really helpful if you could solve this using only the properties of gaussian and gamma functions!

definite-integrals

share|cite|improve this question

edited Mar 21 at 3:07

Mark Viola

134k1278176

asked Mar 20 at 22:03

Pratik ApshingePratik Apshinge

516

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Call this integral $f(t)$. Evaluate $f^prime (t)$ with partial fractions, then use $f(0)=0$.
  $endgroup$
  – J.G.
  Mar 20 at 22:06
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, I understand exactly what you are talking about...
  $endgroup$
  – Pratik Apshinge
  Mar 20 at 22:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How will the derivative of the function give me information about the function itself?
  $endgroup$
  – Pratik Apshinge
  Mar 20 at 22:17
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe you can integrate the derivative of the function.
  $endgroup$
  – Minus One-Twelfth
  Mar 20 at 22:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And, also, is there any possible way to do this with gaussian integrals or the gamma function?
  $endgroup$
  – Pratik Apshinge
  Mar 20 at 22:18
  

  
  
  
  

add a comment | 

$begingroup$

I've been trying to evaluate the integral for a while now, and I've been unable to find it anywhere...
I tried substituting $tan^-1(tx)$ as $u$ but got nowhere... I have done dozens of other substitutions...
I've been told to use the properties of gaussian integrals and the gamma function, but can't seem to figure out a way to bring in $mathrme$...
A few hints that point in the right direction to think will really help!

It would be really really helpful if you could solve this using only the properties of gaussian and gamma functions!

definite-integrals

share|cite|improve this question

edited Mar 21 at 3:07

Mark Viola

134k1278176

asked Mar 20 at 22:03

Pratik ApshingePratik Apshinge

516

$endgroup$

share|cite|improve this question

edited Mar 21 at 3:07

Mark Viola

134k1278176

asked Mar 20 at 22:03

Pratik ApshingePratik Apshinge

516

share|cite|improve this question

edited Mar 21 at 3:07

Mark Viola

134k1278176

asked Mar 20 at 22:03

Pratik ApshingePratik Apshinge

516

edited Mar 21 at 3:07

Mark Viola

134k1278176

edited Mar 21 at 3:07

Mark Viola

134k1278176

asked Mar 20 at 22:03

Pratik ApshingePratik Apshinge

516

asked Mar 20 at 22:03

Pratik ApshingePratik Apshinge

516

2 Answers
2

active

oldest

votes

4

$begingroup$

Let:
$$
I(t)=int_0^inftyfracarctan txx(1+x^2)dx.
$$
Then
$$
I'(t)=int_0^inftyfrac1(1+t^2x^2)(1+x^2)dx=
frac1t^2-1int_0^inftyleft[fract^21+t^2x^2-frac11+x^2right]dx\
=frac1t^2-1 left[tarctan(tx)-arctan xright]_0^infty=fracpi2fract^2-1=fracpi2frac1+1.
$$
Finally integrating the last expression over $t$ one obtains:
$$
I(t)=I(0)+fracpi2text sgn(t)log(|t|+1)=text sgn(t)fracpi2log(|t|+1).
$$

share|cite|improve this answer

edited Mar 21 at 7:05

answered Mar 20 at 22:45

useruser

6,19211031

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wouldn't $left[ttan^-1left(txright)-tan^-1xright]^ $ from 0 to infinity be 0?
  $endgroup$
  – Pratik Apshinge
  Mar 20 at 22:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @PratikApshinge No. It will be $(t-1)pi/2$.
  $endgroup$
  – user
  Mar 20 at 22:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm so sorry, made a calculation mistake!
  $endgroup$
  – Pratik Apshinge
  Mar 20 at 22:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No problem. You're welcome.
  $endgroup$
  – user
  Mar 20 at 22:56
  

  
  
  
  

add a comment | 

2

$begingroup$

We can write $arctan(tx)=int_0^t fracx1+y^2x^2,dy$. Proceeding, we find that for $t>0$

$$beginalign
int_0^infty fracarctan(tx)x(1+x^2),dx&=int_0^infty frac11+x^2int_0^t frac11+y^2x^2,dy,dx\\
&overbrace=^textFubiniint_0^t int_0^infty frac1(1+x^2)(1+y^2x^2),dx,dy\\
&=int_0^t fracpi/2y+1,dy\\
&=fracpi2 log(1+t)
endalign$$

Inasmuch as the integral of interest is an odd function of $t$, we have

$$int_0^infty fracarctan(tx)x(1+x^2),dx=begincasesfracpi 2log(1+t)&, t>0\\0&,t=0\\-fracpi2 log(1-t)&,t<0endcases$$

share|cite|improve this answer

edited Mar 21 at 0:48

answered Mar 20 at 22:23

Mark ViolaMark Viola

134k1278176

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

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%2f3156061%2fevaluate-the-integral-int-0-infty-frac-tan-1txx-left1x2-right%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

4

$begingroup$

Let:
$$
I(t)=int_0^inftyfracarctan txx(1+x^2)dx.
$$
Then
$$
I'(t)=int_0^inftyfrac1(1+t^2x^2)(1+x^2)dx=
frac1t^2-1int_0^inftyleft[fract^21+t^2x^2-frac11+x^2right]dx\
=frac1t^2-1 left[tarctan(tx)-arctan xright]_0^infty=fracpi2fract^2-1=fracpi2frac1+1.
$$
Finally integrating the last expression over $t$ one obtains:
$$
I(t)=I(0)+fracpi2text sgn(t)log(|t|+1)=text sgn(t)fracpi2log(|t|+1).
$$

share|cite|improve this answer

edited Mar 21 at 7:05

answered Mar 20 at 22:45

useruser

6,19211031

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wouldn't $left[ttan^-1left(txright)-tan^-1xright]^ $ from 0 to infinity be 0?
  $endgroup$
  – Pratik Apshinge
  Mar 20 at 22:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @PratikApshinge No. It will be $(t-1)pi/2$.
  $endgroup$
  – user
  Mar 20 at 22:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm so sorry, made a calculation mistake!
  $endgroup$
  – Pratik Apshinge
  Mar 20 at 22:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No problem. You're welcome.
  $endgroup$
  – user
  Mar 20 at 22:56
  

  
  
  
  

add a comment | 

4

$begingroup$

Let:
$$
I(t)=int_0^inftyfracarctan txx(1+x^2)dx.
$$
Then
$$
I'(t)=int_0^inftyfrac1(1+t^2x^2)(1+x^2)dx=
frac1t^2-1int_0^inftyleft[fract^21+t^2x^2-frac11+x^2right]dx\
=frac1t^2-1 left[tarctan(tx)-arctan xright]_0^infty=fracpi2fract^2-1=fracpi2frac1+1.
$$
Finally integrating the last expression over $t$ one obtains:
$$
I(t)=I(0)+fracpi2text sgn(t)log(|t|+1)=text sgn(t)fracpi2log(|t|+1).
$$

share|cite|improve this answer

edited Mar 21 at 7:05

answered Mar 20 at 22:45

useruser

6,19211031

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wouldn't $left[ttan^-1left(txright)-tan^-1xright]^ $ from 0 to infinity be 0?
  $endgroup$
  – Pratik Apshinge
  Mar 20 at 22:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @PratikApshinge No. It will be $(t-1)pi/2$.
  $endgroup$
  – user
  Mar 20 at 22:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm so sorry, made a calculation mistake!
  $endgroup$
  – Pratik Apshinge
  Mar 20 at 22:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No problem. You're welcome.
  $endgroup$
  – user
  Mar 20 at 22:56
  

  
  
  
  

add a comment | 

$begingroup$

Let:
$$
I(t)=int_0^inftyfracarctan txx(1+x^2)dx.
$$
Then
$$
I'(t)=int_0^inftyfrac1(1+t^2x^2)(1+x^2)dx=
frac1t^2-1int_0^inftyleft[fract^21+t^2x^2-frac11+x^2right]dx\
=frac1t^2-1 left[tarctan(tx)-arctan xright]_0^infty=fracpi2fract^2-1=fracpi2frac1+1.
$$
Finally integrating the last expression over $t$ one obtains:
$$
I(t)=I(0)+fracpi2text sgn(t)log(|t|+1)=text sgn(t)fracpi2log(|t|+1).
$$

share|cite|improve this answer

edited Mar 21 at 7:05

answered Mar 20 at 22:45

useruser

6,19211031

$endgroup$

share|cite|improve this answer

edited Mar 21 at 7:05

answered Mar 20 at 22:45

useruser

6,19211031

answered Mar 20 at 22:45

useruser

6,19211031

answered Mar 20 at 22:45

useruser

6,19211031

2

$begingroup$

We can write $arctan(tx)=int_0^t fracx1+y^2x^2,dy$. Proceeding, we find that for $t>0$

$$beginalign
int_0^infty fracarctan(tx)x(1+x^2),dx&=int_0^infty frac11+x^2int_0^t frac11+y^2x^2,dy,dx\\
&overbrace=^textFubiniint_0^t int_0^infty frac1(1+x^2)(1+y^2x^2),dx,dy\\
&=int_0^t fracpi/2y+1,dy\\
&=fracpi2 log(1+t)
endalign$$

Inasmuch as the integral of interest is an odd function of $t$, we have

$$int_0^infty fracarctan(tx)x(1+x^2),dx=begincasesfracpi 2log(1+t)&, t>0\\0&,t=0\\-fracpi2 log(1-t)&,t<0endcases$$

share|cite|improve this answer

edited Mar 21 at 0:48

answered Mar 20 at 22:23

Mark ViolaMark Viola

134k1278176

$endgroup$

add a comment | 

2

$begingroup$

We can write $arctan(tx)=int_0^t fracx1+y^2x^2,dy$. Proceeding, we find that for $t>0$

$$beginalign
int_0^infty fracarctan(tx)x(1+x^2),dx&=int_0^infty frac11+x^2int_0^t frac11+y^2x^2,dy,dx\\
&overbrace=^textFubiniint_0^t int_0^infty frac1(1+x^2)(1+y^2x^2),dx,dy\\
&=int_0^t fracpi/2y+1,dy\\
&=fracpi2 log(1+t)
endalign$$

Inasmuch as the integral of interest is an odd function of $t$, we have

$$int_0^infty fracarctan(tx)x(1+x^2),dx=begincasesfracpi 2log(1+t)&, t>0\\0&,t=0\\-fracpi2 log(1-t)&,t<0endcases$$

share|cite|improve this answer

edited Mar 21 at 0:48

answered Mar 20 at 22:23

Mark ViolaMark Viola

134k1278176

$endgroup$

add a comment | 

$begingroup$

We can write $arctan(tx)=int_0^t fracx1+y^2x^2,dy$. Proceeding, we find that for $t>0$

$$beginalign
int_0^infty fracarctan(tx)x(1+x^2),dx&=int_0^infty frac11+x^2int_0^t frac11+y^2x^2,dy,dx\\
&overbrace=^textFubiniint_0^t int_0^infty frac1(1+x^2)(1+y^2x^2),dx,dy\\
&=int_0^t fracpi/2y+1,dy\\
&=fracpi2 log(1+t)
endalign$$

Inasmuch as the integral of interest is an odd function of $t$, we have

$$int_0^infty fracarctan(tx)x(1+x^2),dx=begincasesfracpi 2log(1+t)&, t>0\\0&,t=0\\-fracpi2 log(1-t)&,t<0endcases$$

share|cite|improve this answer

edited Mar 21 at 0:48

answered Mar 20 at 22:23

Mark ViolaMark Viola

134k1278176

$endgroup$

share|cite|improve this answer

edited Mar 21 at 0:48

answered Mar 20 at 22:23

Mark ViolaMark Viola

134k1278176

answered Mar 20 at 22:23

Mark ViolaMark Viola

134k1278176

answered Mar 20 at 22:23

Mark ViolaMark Viola

134k1278176