Mrthfx

What is the wife of a henpecked husband called?

Why do neural networks need so many training examples to perform?

Which communication protocol is used in AdLib sound card?

How should I handle players who ignore the session zero agreement?

How do I prevent a homebrew Grappling Hook feature from trivializing Tomb of Annihilation?

Crontab: Ubuntu running script (noob)

Existence of Riemann surface, holomorphic maps

Why is it that Bernie Sanders is always called a "socialist"?

What is a good reason for every spaceship to carry a weapon on board?

Why didn't Tom Riddle take the presence of Fawkes and the Sorting Hat as more of a threat?

TikZ graph edges not drawn nicely

Identify KNO3 and KH2PO4 at home

Should I reinstall Linux when changing the laptop's CPU?

How does one write from a minority culture? A question on cultural references

Why did Democrats in the Senate oppose the Born-Alive Abortion Survivors Protection Act (2019 S.130)?

I have trouble understanding this fallacy: "If A, then B. Therefore if not-B, then not-A."

any clues on how to solve these types of problems within 2-3 minutes for competitive exams

How much mayhem could I cause as a fish?

Why does magnet wire need to be insulated?

Why are the books in the Game of Thrones citadel library shelved spine inwards?

Separate environment for personal and development use under macOS

How do you voice extended chords?

A Missing Symbol for This Logo

Plausible reason for gold-digging ant

How to use Mathemaica to do a complex integrate with poles in real axis?

How to do this complex integration on the real line?Complex integrals and residuesIntegrate yields complex value, while after variable transformation the result is real. Bug?Complex numbers from two arrays with Real and Imaginary partshow to take complex conjugate of a real variableIntegral over real valued function becomes complexHow to compute my integralComputation of complex integral over circleReal Integrand, Complex Integral result?Using “if” to compare 2 expressions (Complex and real numbers)

4

$begingroup$

I want to use Mathematica to compute the following complex integral:

Integrate[Exp[I z ] 1/z, {z, -Infinity, Infinity}]

Mathematica reports that it does not converge on {-Infinity, Infinity}.

But, from the textbook, we know, the result is I Pi.

Of course, if I use the NItegrate, then, Mathematica will give 0.+3.14 I.

complex

share|improve this question

edited 1 hour ago

Αλέξανδρος Ζεγγ

4,3491929

asked 1 hour ago

MPHYKEKMPHYKEK

433

$endgroup$

add a comment | 

4

$begingroup$

I want to use Mathematica to compute the following complex integral:

Integrate[Exp[I z ] 1/z, {z, -Infinity, Infinity}]

Mathematica reports that it does not converge on {-Infinity, Infinity}.

But, from the textbook, we know, the result is I Pi.

Of course, if I use the NItegrate, then, Mathematica will give 0.+3.14 I.

complex

share|improve this question

edited 1 hour ago

Αλέξανδρος Ζεγγ

4,3491929

asked 1 hour ago

MPHYKEKMPHYKEK

433

$endgroup$

add a comment | 

$begingroup$

I want to use Mathematica to compute the following complex integral:

Integrate[Exp[I z ] 1/z, {z, -Infinity, Infinity}]

Mathematica reports that it does not converge on {-Infinity, Infinity}.

But, from the textbook, we know, the result is I Pi.

Of course, if I use the NItegrate, then, Mathematica will give 0.+3.14 I.

complex

share|improve this question

edited 1 hour ago

Αλέξανδρος Ζεγγ

4,3491929

asked 1 hour ago

MPHYKEKMPHYKEK

433

$endgroup$

Integrate[Exp[I z ] 1/z, {z, -Infinity, Infinity}]

share|improve this question

edited 1 hour ago

Αλέξανδρος Ζεγγ

4,3491929

asked 1 hour ago

MPHYKEKMPHYKEK

433

share|improve this question

edited 1 hour ago

Αλέξανδρος Ζεγγ

4,3491929

asked 1 hour ago

MPHYKEKMPHYKEK

433

edited 1 hour ago

Αλέξανδρος Ζεγγ

4,3491929

edited 1 hour ago

Αλέξανδρος Ζεγγ

4,3491929

asked 1 hour ago

MPHYKEKMPHYKEK

433

asked 1 hour ago

MPHYKEKMPHYKEK

433

2 Answers
2

active

oldest

votes

3

$begingroup$

Try

Integrate[Exp[I z] 1/z, {z, -Infinity, Infinity},PrincipalValue -> True]

(*I [Pi]*)

share|improve this answer

answered 1 hour ago

Ulrich NeumannUlrich Neumann

9,031516

$endgroup$

add a comment | 

2

$begingroup$

One can also consider using the residue theorem. The residue is readily obtained by

Residue[Exp[I z] 1/z, {z, 0}]

returning 1, which means that the integral is $ mathrm i pi $.

share|improve this answer

answered 1 hour ago

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,3491929

$endgroup$

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: "387"
};
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: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
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
},
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%2fmathematica.stackexchange.com%2fquestions%2f192291%2fhow-to-use-mathemaica-to-do-a-complex-integrate-with-poles-in-real-axis%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

3

$begingroup$

Try

Integrate[Exp[I z] 1/z, {z, -Infinity, Infinity},PrincipalValue -> True]

(*I [Pi]*)

share|improve this answer

answered 1 hour ago

Ulrich NeumannUlrich Neumann

9,031516

$endgroup$

add a comment | 

3

$begingroup$

Try

Integrate[Exp[I z] 1/z, {z, -Infinity, Infinity},PrincipalValue -> True]

(*I [Pi]*)

share|improve this answer

answered 1 hour ago

Ulrich NeumannUlrich Neumann

9,031516

$endgroup$

add a comment | 

$begingroup$

Try

Integrate[Exp[I z] 1/z, {z, -Infinity, Infinity},PrincipalValue -> True]

(*I [Pi]*)

share|improve this answer

answered 1 hour ago

Ulrich NeumannUlrich Neumann

9,031516

$endgroup$

Integrate[Exp[I z] 1/z, {z, -Infinity, Infinity},PrincipalValue -> True]

(*I [Pi]*)

share|improve this answer

answered 1 hour ago

Ulrich NeumannUlrich Neumann

9,031516

answered 1 hour ago

Ulrich NeumannUlrich Neumann

9,031516

answered 1 hour ago

Ulrich NeumannUlrich Neumann

9,031516

2

$begingroup$

One can also consider using the residue theorem. The residue is readily obtained by

Residue[Exp[I z] 1/z, {z, 0}]

returning 1, which means that the integral is $ mathrm i pi $.

share|improve this answer

answered 1 hour ago

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,3491929

$endgroup$

add a comment | 

2

$begingroup$

One can also consider using the residue theorem. The residue is readily obtained by

Residue[Exp[I z] 1/z, {z, 0}]

returning 1, which means that the integral is $ mathrm i pi $.

share|improve this answer

answered 1 hour ago

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,3491929

$endgroup$

add a comment | 

$begingroup$

One can also consider using the residue theorem. The residue is readily obtained by

Residue[Exp[I z] 1/z, {z, 0}]

returning 1, which means that the integral is $ mathrm i pi $.

share|improve this answer

answered 1 hour ago

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,3491929

$endgroup$

Residue[Exp[I z] 1/z, {z, 0}]

share|improve this answer

answered 1 hour ago

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,3491929

answered 1 hour ago

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,3491929

answered 1 hour ago

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,3491929