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

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

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5

1

$begingroup$

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 22 mins ago

Glorfindel

2671311

asked 6 hours ago

MPHYKEKMPHYKEK

483

$endgroup$

add a comment | 

5

1

$begingroup$

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 22 mins ago

Glorfindel

2671311

asked 6 hours ago

MPHYKEKMPHYKEK

483

$endgroup$

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 22 mins ago

Glorfindel

2671311

asked 6 hours ago

MPHYKEKMPHYKEK

483

$endgroup$

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

share|improve this question

edited 22 mins ago

Glorfindel

2671311

asked 6 hours ago

MPHYKEKMPHYKEK

483

share|improve this question

edited 22 mins ago

Glorfindel

2671311

asked 6 hours ago

MPHYKEKMPHYKEK

483

edited 22 mins ago

Glorfindel

2671311

edited 22 mins ago

Glorfindel

2671311

asked 6 hours ago

MPHYKEKMPHYKEK

483

asked 6 hours ago

MPHYKEKMPHYKEK

483

2 Answers
2

active

oldest

votes

4

$begingroup$

Try

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

(*I [Pi]*)

share|improve this answer

answered 5 hours ago

Ulrich NeumannUlrich Neumann

9,041516

$endgroup$

add a comment | 

3

$begingroup$

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 5 hours ago

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,3591929

$endgroup$

add a comment | 

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4

$begingroup$

Try

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

(*I [Pi]*)

share|improve this answer

answered 5 hours ago

Ulrich NeumannUlrich Neumann

9,041516

$endgroup$

add a comment | 

4

$begingroup$

Try

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

(*I [Pi]*)

share|improve this answer

answered 5 hours ago

Ulrich NeumannUlrich Neumann

9,041516

$endgroup$

add a comment | 

$begingroup$

Try

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

(*I [Pi]*)

share|improve this answer

answered 5 hours ago

Ulrich NeumannUlrich Neumann

9,041516

$endgroup$

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

(*I [Pi]*)

share|improve this answer

answered 5 hours ago

Ulrich NeumannUlrich Neumann

9,041516

answered 5 hours ago

Ulrich NeumannUlrich Neumann

9,041516

answered 5 hours ago

Ulrich NeumannUlrich Neumann

9,041516

3

$begingroup$

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 5 hours ago

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,3591929

$endgroup$

add a comment | 

3

$begingroup$

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 5 hours ago

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,3591929

$endgroup$

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 5 hours ago

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,3591929

$endgroup$

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

share|improve this answer

answered 5 hours ago

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,3591929

answered 5 hours ago

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,3591929

answered 5 hours ago

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,3591929