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