How to use Mathematica to do a complex integrate with poles in real axis?
Clash Royale CLAN TAG#URR8PPP
$begingroup$
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 NIntegrate, then, Mathematica will give 0. + 3.14 I.
calculus-and-analysis complex
edited Mar 1 at 9:33
J. M. is slightly pensive♦
98.5k10308466
asked Feb 27 at 7:31
MPHYKEKMPHYKEK
864
$endgroup$
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 NIntegrate, then, Mathematica will give 0. + 3.14 I.
calculus-and-analysis complex
edited Mar 1 at 9:33
J. M. is slightly pensive♦
98.5k10308466
asked Feb 27 at 7:31
MPHYKEKMPHYKEK
864
$endgroup$
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 NIntegrate, then, Mathematica will give 0. + 3.14 I.
calculus-and-analysis complex
edited Mar 1 at 9:33
J. M. is slightly pensive♦
98.5k10308466
asked Feb 27 at 7:31
MPHYKEKMPHYKEK
864
$endgroup$
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 NIntegrate, then, Mathematica will give 0. + 3.14 I.
calculus-and-analysis complex
calculus-and-analysis complex
edited Mar 1 at 9:33
J. M. is slightly pensive♦
98.5k10308466
asked Feb 27 at 7:31
MPHYKEKMPHYKEK
864
edited Mar 1 at 9:33
J. M. is slightly pensive♦
98.5k10308466
asked Feb 27 at 7:31
MPHYKEKMPHYKEK
864
edited Mar 1 at 9:33
J. M. is slightly pensive♦
98.5k10308466
edited Mar 1 at 9:33
J. M. is slightly pensive♦
98.5k10308466
edited Mar 1 at 9:33
J. M. is slightly pensive♦
98.5k10308466
98.5k10308466
asked Feb 27 at 7:31
MPHYKEKMPHYKEK
864
asked Feb 27 at 7:31
MPHYKEKMPHYKEK
864
asked Feb 27 at 7:31
MPHYKEKMPHYKEK
864
864
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Try
Integrate[Exp[I z] 1/z, z, -Infinity, Infinity,PrincipalValue -> True]
(*I π*)
edited Feb 27 at 15:32
Dimitris
2,2511332
answered Feb 27 at 7:35
Ulrich NeumannUlrich Neumann
9,568617
$endgroup$
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 $.
answered Feb 27 at 8:13
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,44011029
$endgroup$
$begingroup$
This for me is the most satisfactory solution.
$endgroup$
– J. M. is slightly pensive♦
Mar 1 at 9:32
$begingroup$
@J.M.iscomputer-less Thank you for your comment.
$endgroup$
– Αλέξανδρος Ζεγγ
Mar 1 at 11:37
add a comment |
$begingroup$
If you are sure about your integral's behavior you can try
Integrate[Exp[I z] 1/z, z, -Infinity, Infinity,
GenerateConditions -> False]
(* I π *)
answered Feb 27 at 15:29
DimitrisDimitris
2,2511332
$endgroup$
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
);
);
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
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f192291%2fhow-to-use-mathematica-to-do-a-complex-integrate-with-poles-in-real-axis%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Try
Integrate[Exp[I z] 1/z, z, -Infinity, Infinity,PrincipalValue -> True]
(*I π*)
edited Feb 27 at 15:32
Dimitris
2,2511332
answered Feb 27 at 7:35
Ulrich NeumannUlrich Neumann
9,568617
$endgroup$
add a comment |
$begingroup$
Try
Integrate[Exp[I z] 1/z, z, -Infinity, Infinity,PrincipalValue -> True]
(*I π*)
edited Feb 27 at 15:32
Dimitris
2,2511332
answered Feb 27 at 7:35
Ulrich NeumannUlrich Neumann
9,568617
$endgroup$
add a comment |
$begingroup$
Try
Integrate[Exp[I z] 1/z, z, -Infinity, Infinity,PrincipalValue -> True]
(*I π*)
edited Feb 27 at 15:32
Dimitris
2,2511332
answered Feb 27 at 7:35
Ulrich NeumannUlrich Neumann
9,568617
$endgroup$
Try
Integrate[Exp[I z] 1/z, z, -Infinity, Infinity,PrincipalValue -> True]
(*I π*)
edited Feb 27 at 15:32
Dimitris
2,2511332
answered Feb 27 at 7:35
Ulrich NeumannUlrich Neumann
9,568617
edited Feb 27 at 15:32
Dimitris
2,2511332
edited Feb 27 at 15:32
Dimitris
2,2511332
edited Feb 27 at 15:32
Dimitris
2,2511332
2,2511332
answered Feb 27 at 7:35
Ulrich NeumannUlrich Neumann
9,568617
answered Feb 27 at 7:35
Ulrich NeumannUlrich Neumann
9,568617
answered Feb 27 at 7:35
Ulrich NeumannUlrich Neumann
9,568617
9,568617
add a comment |
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 $.
answered Feb 27 at 8:13
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,44011029
$endgroup$
$begingroup$
This for me is the most satisfactory solution.
$endgroup$
– J. M. is slightly pensive♦
Mar 1 at 9:32
$begingroup$
@J.M.iscomputer-less Thank you for your comment.
$endgroup$
– Αλέξανδρος Ζεγγ
Mar 1 at 11:37
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 $.
answered Feb 27 at 8:13
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,44011029
$endgroup$
$begingroup$
This for me is the most satisfactory solution.
$endgroup$
– J. M. is slightly pensive♦
Mar 1 at 9:32
$begingroup$
@J.M.iscomputer-less Thank you for your comment.
$endgroup$
– Αλέξανδρος Ζεγγ
Mar 1 at 11:37
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 $.
answered Feb 27 at 8:13
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,44011029
$endgroup$
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 $.
answered Feb 27 at 8:13
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,44011029
answered Feb 27 at 8:13
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,44011029
answered Feb 27 at 8:13
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,44011029
answered Feb 27 at 8:13
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,44011029
4,44011029
$begingroup$
This for me is the most satisfactory solution.
$endgroup$
– J. M. is slightly pensive♦
Mar 1 at 9:32
$begingroup$
@J.M.iscomputer-less Thank you for your comment.
$endgroup$
– Αλέξανδρος Ζεγγ
Mar 1 at 11:37
add a comment |
$begingroup$
This for me is the most satisfactory solution.
$endgroup$
– J. M. is slightly pensive♦
Mar 1 at 9:32
$begingroup$
@J.M.iscomputer-less Thank you for your comment.
$endgroup$
– Αλέξανδρος Ζεγγ
Mar 1 at 11:37
$begingroup$
This for me is the most satisfactory solution.
$endgroup$
– J. M. is slightly pensive♦
Mar 1 at 9:32
$begingroup$
This for me is the most satisfactory solution.
$endgroup$
– J. M. is slightly pensive♦
Mar 1 at 9:32
$begingroup$
@J.M.iscomputer-less Thank you for your comment.
$endgroup$
– Αλέξανδρος Ζεγγ
Mar 1 at 11:37
$begingroup$
@J.M.iscomputer-less Thank you for your comment.
$endgroup$
– Αλέξανδρος Ζεγγ
Mar 1 at 11:37
add a comment |
$begingroup$
If you are sure about your integral's behavior you can try
Integrate[Exp[I z] 1/z, z, -Infinity, Infinity,
GenerateConditions -> False]
(* I π *)
answered Feb 27 at 15:29
DimitrisDimitris
2,2511332
$endgroup$
add a comment |
$begingroup$
If you are sure about your integral's behavior you can try
Integrate[Exp[I z] 1/z, z, -Infinity, Infinity,
GenerateConditions -> False]
(* I π *)
answered Feb 27 at 15:29
DimitrisDimitris
2,2511332
$endgroup$
add a comment |
$begingroup$
If you are sure about your integral's behavior you can try
Integrate[Exp[I z] 1/z, z, -Infinity, Infinity,
GenerateConditions -> False]
(* I π *)
answered Feb 27 at 15:29
DimitrisDimitris
2,2511332
$endgroup$
If you are sure about your integral's behavior you can try
Integrate[Exp[I z] 1/z, z, -Infinity, Infinity,
GenerateConditions -> False]
(* I π *)
answered Feb 27 at 15:29
DimitrisDimitris
2,2511332
answered Feb 27 at 15:29
DimitrisDimitris
2,2511332
answered Feb 27 at 15:29
DimitrisDimitris
2,2511332
answered Feb 27 at 15:29
DimitrisDimitris
2,2511332
2,2511332
add a comment |
add a comment |
Thanks for contributing an answer to Mathematica Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f192291%2fhow-to-use-mathematica-to-do-a-complex-integrate-with-poles-in-real-axis%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
