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

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7

2

$begingroup$

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 NIntegrate, then, Mathematica will give 0.+3.14 I.

calculus-and-analysis complex

share|improve this question

edited 3 hours ago

Glorfindel

2671311

asked 9 hours ago

MPHYKEKMPHYKEK

583

$endgroup$

add a comment | 

7

2

$begingroup$

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 NIntegrate, then, Mathematica will give 0.+3.14 I.

calculus-and-analysis complex

share|improve this question

edited 3 hours ago

Glorfindel

2671311

asked 9 hours ago

MPHYKEKMPHYKEK

583

$endgroup$

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 NIntegrate, then, Mathematica will give 0.+3.14 I.

calculus-and-analysis complex

share|improve this question

edited 3 hours ago

Glorfindel

2671311

asked 9 hours ago

MPHYKEKMPHYKEK

583

$endgroup$

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

share|improve this question

edited 3 hours ago

Glorfindel

2671311

asked 9 hours ago

MPHYKEKMPHYKEK

583

share|improve this question

edited 3 hours ago

Glorfindel

2671311

asked 9 hours ago

MPHYKEKMPHYKEK

583

edited 3 hours ago

Glorfindel

2671311

edited 3 hours ago

Glorfindel

2671311

asked 9 hours ago

MPHYKEKMPHYKEK

583

asked 9 hours ago

MPHYKEKMPHYKEK

583

3 Answers
3

active

oldest

votes

6

$begingroup$

Try

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

(*I π*)

share|improve this answer

edited 59 mins ago

dimitris

2,2041331

answered 8 hours ago

Ulrich NeumannUlrich Neumann

9,061516

$endgroup$

add a comment | 

5

$begingroup$

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 8 hours ago

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,3791929

$endgroup$

add a comment | 

2

$begingroup$

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 *)

share|improve this answer

answered 1 hour ago

dimitrisdimitris

2,2041331

$endgroup$

add a comment | 

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6

$begingroup$

Try

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

(*I π*)

share|improve this answer

edited 59 mins ago

dimitris

2,2041331

answered 8 hours ago

Ulrich NeumannUlrich Neumann

9,061516

$endgroup$

add a comment | 

6

$begingroup$

Try

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

(*I π*)

share|improve this answer

edited 59 mins ago

dimitris

2,2041331

answered 8 hours ago

Ulrich NeumannUlrich Neumann

9,061516

$endgroup$

add a comment | 

$begingroup$

Try

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

(*I π*)

share|improve this answer

edited 59 mins ago

dimitris

2,2041331

answered 8 hours ago

Ulrich NeumannUlrich Neumann

9,061516

$endgroup$

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

(*I π*)

share|improve this answer

edited 59 mins ago

dimitris

2,2041331

answered 8 hours ago

Ulrich NeumannUlrich Neumann

9,061516

edited 59 mins ago

dimitris

2,2041331

edited 59 mins ago

dimitris

2,2041331

answered 8 hours ago

Ulrich NeumannUlrich Neumann

9,061516

answered 8 hours ago

Ulrich NeumannUlrich Neumann

9,061516

5

$begingroup$

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 8 hours ago

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,3791929

$endgroup$

add a comment | 

5

$begingroup$

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 8 hours ago

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,3791929

$endgroup$

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 8 hours ago

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,3791929

$endgroup$

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

share|improve this answer

answered 8 hours ago

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,3791929

answered 8 hours ago

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,3791929

answered 8 hours ago

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,3791929

2

$begingroup$

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 *)

share|improve this answer

answered 1 hour ago

dimitrisdimitris

2,2041331

$endgroup$

add a comment | 

2

$begingroup$

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 *)

share|improve this answer

answered 1 hour ago

dimitrisdimitris

2,2041331

$endgroup$

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 *)

share|improve this answer

answered 1 hour ago

dimitrisdimitris

2,2041331

$endgroup$

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

 GenerateConditions -> False]

(* I π *)

share|improve this answer

answered 1 hour ago

dimitrisdimitris

2,2041331

answered 1 hour ago

dimitrisdimitris

2,2041331

answered 1 hour ago

dimitrisdimitris

2,2041331