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Variable dimensional integrals

About high dimensional integralsAbout calculating IntegralsMultiple integrals where the number of integrals is aribtraryDirac delta in integralsComputing integrals in MathematicaHow does one evaluate this variable number of multiple integrals?Evaluating integrals of derivativesDefining derivatives of integralssymbolic Integration of a long function while retaining constantsEvaluating multiple integrals

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}

4

$begingroup$

I need to Integrate a function, but let the number of dimensions vary.

Integrate[Product[Exp[-( t[[i]] )^2], {i,1,4}], options]

with

t = {u,v,x,y}

and

options = {#,-Infinity,Infinity}&/@{u,v,x,y}

but the double braces on options doesn't allow this to work. I get this for options:

{{u, -Infinity, Infinity}, {v, -Infinity, Infinity}, {x,
-Infinity, Infinity}, {y, -Infinity, Infinity}}

If I want a general list of variables, can these be added as variables into integrate, with the corresponding limits? I need a variable dimension, simply taken from the length of a list.

calculus-and-analysis syntax

share|improve this question

asked 8 hours ago

Alexander Kartun-GilesAlexander Kartun-Giles

48722 silver badges1111 bronze badges

$endgroup$

add a comment | 

4

$begingroup$

I need to Integrate a function, but let the number of dimensions vary.

Integrate[Product[Exp[-( t[[i]] )^2], {i,1,4}], options]

with

t = {u,v,x,y}

and

options = {#,-Infinity,Infinity}&/@{u,v,x,y}

but the double braces on options doesn't allow this to work. I get this for options:

{{u, -Infinity, Infinity}, {v, -Infinity, Infinity}, {x,
-Infinity, Infinity}, {y, -Infinity, Infinity}}

If I want a general list of variables, can these be added as variables into integrate, with the corresponding limits? I need a variable dimension, simply taken from the length of a list.

calculus-and-analysis syntax

share|improve this question

asked 8 hours ago

Alexander Kartun-GilesAlexander Kartun-Giles

48722 silver badges1111 bronze badges

$endgroup$

add a comment | 

$begingroup$

I need to Integrate a function, but let the number of dimensions vary.

Integrate[Product[Exp[-( t[[i]] )^2], {i,1,4}], options]

with

t = {u,v,x,y}

and

options = {#,-Infinity,Infinity}&/@{u,v,x,y}

but the double braces on options doesn't allow this to work. I get this for options:

{{u, -Infinity, Infinity}, {v, -Infinity, Infinity}, {x,
-Infinity, Infinity}, {y, -Infinity, Infinity}}

If I want a general list of variables, can these be added as variables into integrate, with the corresponding limits? I need a variable dimension, simply taken from the length of a list.

calculus-and-analysis syntax

share|improve this question

asked 8 hours ago

Alexander Kartun-GilesAlexander Kartun-Giles

48722 silver badges1111 bronze badges

$endgroup$

Integrate[Product[Exp[-( t[[i]] )^2], {i,1,4}], options]

t = {u,v,x,y}

options = {#,-Infinity,Infinity}&/@{u,v,x,y}

share|improve this question

asked 8 hours ago

Alexander Kartun-GilesAlexander Kartun-Giles

48722 silver badges1111 bronze badges

share|improve this question

asked 8 hours ago

Alexander Kartun-GilesAlexander Kartun-Giles

48722 silver badges1111 bronze badges

asked 8 hours ago

Alexander Kartun-GilesAlexander Kartun-Giles

48722 silver badges1111 bronze badges

asked 8 hours ago

Alexander Kartun-GilesAlexander Kartun-Giles

48722 silver badges1111 bronze badges

1 Answer
1

active

oldest

votes

6

$begingroup$

You can use Sequence to eliminate a set of brackets:

t = {u, v, x, y};

Integrate[

    Product[

        Exp[-( t[[i]] )^2],

        {i,1,4}

    ],

    Sequence @@ ({#,-Infinity,Infinity}&/@t)

]

π^2

However, it is simpler to use a Region specification instead:

Integrate[Exp[-p.p], p ∈ FullRegion[4]]

π^2

share|improve this answer

answered 8 hours ago

Carl WollCarl Woll

85.9k33 gold badges110110 silver badges220220 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much. Is p a vector, hence FullRegion[4]?
  $endgroup$
  – Alexander Kartun-Giles
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @AlexanderKartun-Giles Yes, p is a symbolic vector.
  $endgroup$
  – Carl Woll
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok I will look that up. Thank you.
  $endgroup$
  – Alexander Kartun-Giles
  8 hours ago
  

  
  
  
  

add a comment | 

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6

$begingroup$

You can use Sequence to eliminate a set of brackets:

t = {u, v, x, y};

Integrate[

    Product[

        Exp[-( t[[i]] )^2],

        {i,1,4}

    ],

    Sequence @@ ({#,-Infinity,Infinity}&/@t)

]

π^2

However, it is simpler to use a Region specification instead:

Integrate[Exp[-p.p], p ∈ FullRegion[4]]

π^2

share|improve this answer

answered 8 hours ago

Carl WollCarl Woll

85.9k33 gold badges110110 silver badges220220 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much. Is p a vector, hence FullRegion[4]?
  $endgroup$
  – Alexander Kartun-Giles
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @AlexanderKartun-Giles Yes, p is a symbolic vector.
  $endgroup$
  – Carl Woll
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok I will look that up. Thank you.
  $endgroup$
  – Alexander Kartun-Giles
  8 hours ago
  

  
  
  
  

add a comment | 

6

$begingroup$

You can use Sequence to eliminate a set of brackets:

t = {u, v, x, y};

Integrate[

    Product[

        Exp[-( t[[i]] )^2],

        {i,1,4}

    ],

    Sequence @@ ({#,-Infinity,Infinity}&/@t)

]

π^2

However, it is simpler to use a Region specification instead:

Integrate[Exp[-p.p], p ∈ FullRegion[4]]

π^2

share|improve this answer

answered 8 hours ago

Carl WollCarl Woll

85.9k33 gold badges110110 silver badges220220 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much. Is p a vector, hence FullRegion[4]?
  $endgroup$
  – Alexander Kartun-Giles
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @AlexanderKartun-Giles Yes, p is a symbolic vector.
  $endgroup$
  – Carl Woll
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok I will look that up. Thank you.
  $endgroup$
  – Alexander Kartun-Giles
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

You can use Sequence to eliminate a set of brackets:

t = {u, v, x, y};

Integrate[

    Product[

        Exp[-( t[[i]] )^2],

        {i,1,4}

    ],

    Sequence @@ ({#,-Infinity,Infinity}&/@t)

]

π^2

However, it is simpler to use a Region specification instead:

Integrate[Exp[-p.p], p ∈ FullRegion[4]]

π^2

share|improve this answer

answered 8 hours ago

Carl WollCarl Woll

85.9k33 gold badges110110 silver badges220220 bronze badges

$endgroup$

t = {u, v, x, y};

Integrate[

    Product[

        Exp[-( t[[i]] )^2],

        {i,1,4}

    ],

    Sequence @@ ({#,-Infinity,Infinity}&/@t)

]

Integrate[Exp[-p.p], p ∈ FullRegion[4]]

share|improve this answer

answered 8 hours ago

Carl WollCarl Woll

85.9k33 gold badges110110 silver badges220220 bronze badges

answered 8 hours ago

Carl WollCarl Woll

85.9k33 gold badges110110 silver badges220220 bronze badges

answered 8 hours ago

Carl WollCarl Woll

85.9k33 gold badges110110 silver badges220220 bronze badges