Gsyfuy

A curious equality of integrals involving the prime counting function?

How to deal with possible delayed baggage?

Non-Cancer terminal illness that can affect young (age 10-13) girls?

How to politely refuse in-office gym instructor for steroids and protein

False written accusations not made public - is there law to cover this?

TikZ graph edges not drawn nicely

Do "fields" always combine by addition?

How do I append a character to the end of every line in an excel cell?

In Linux what happens if 1000 files in a directory are moved to another location while another 300 files were added to the source directory?

Square Root Distance from Integers

Why don't key signatures indicate the tonic?

Why is Agricola named as such?

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

How can I play a serial killer in a party of good PCs?

How do you funnel food off a cutting board?

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

Is "the fire consumed everything on its way" correct?

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

Why did Luke use his left hand to shoot?

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

What makes papers publishable in top-tier journals?

What is the difference between "...", '...', $'...', and $"..." quotes?

Am I a Rude Number?

How to not let the Identify spell spoil everything?

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 2 hours 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 2 hours 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 2 hours 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 2 hours ago

MPHYKEKMPHYKEK

433

share|improve this question

edited 1 hour ago

Αλέξανδρος Ζεγγ

4,3491929

asked 2 hours ago

MPHYKEKMPHYKEK

433

edited 1 hour ago

Αλέξανδρος Ζεγγ

4,3491929

edited 1 hour ago

Αλέξανδρος Ζεγγ

4,3491929

asked 2 hours ago

MPHYKEKMPHYKEK

433

asked 2 hours 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 2 hours 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 2 hours 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 2 hours 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 2 hours 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 2 hours ago

Ulrich NeumannUlrich Neumann

9,031516

answered 2 hours ago

Ulrich NeumannUlrich Neumann

9,031516

answered 2 hours 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