Dgdrxrt

IC on Digikey is 5x more expensive than board containing same IC on Alibaba: How?

Is there a spell that can create a permanent fire?

How to get a flat-head nail out of a piece of wood?

Can a Knight grant Knighthood to another?

Fit odd number of triplets in a measure?

Marquee sign letters

What helicopter has the most rotor blades?

Changing order of draw operation in PGFPlots

Why complex landing gears are used instead of simple, reliable and light weight muscle wire or shape memory alloys?

The test team as an enemy of development? And how can this be avoided?

Is honorific speech ever used in the first person?

Can stored/leased 737s be used to substitute for grounded MAXs?

How can I prevent/balance waiting and turtling as a response to cooldown mechanics

"Destructive power" carried by a B-52?

By what mechanism was the 2017 UK General Election called?

Shimano 105 brifters (5800) and Avid BB5 compatibility

calculator's angle answer for trig ratios that can work in more than 1 quadrant on the unit circle

Why not use the yoke to control yaw, as well as pitch and roll?

Why is there so little support for joining EFTA in the British parliament?

Pointing to problems without suggesting solutions

Is it OK if I do not take the receipt in Germany?

Why did Bronn offer to be Tyrion Lannister's champion in trial by combat?

Table formatting with tabularx?

Short story about astronauts fertilizing soil with their own bodies

Multidimensional integral

2

$begingroup$

I am trying to understand a proof where the following equality, without any further details, appears:

$$int_B_qe^-ilangle x,trangledx = c(k)int_-q^q e^-i(q^2-y^2)^k/2dy,$$

where $B_q = leq q$, $tinmathbbR^k+1$, $|x| = (x_1^2+ldots + x_k+1^2)^1/2$, $langlecdot,cdotrangle$ is the standard scalar product and $c(k)$ is a constant depending only on $k$.

I am not able to see how can I pass from a $(k+1)$-dimensional integral to 1-dimensional integral. My intuition tells me that this constant $c(k)$ must be related with the surface area of $mathbbS^k=x$ and then the remaining integral will be related with the radius of $B_q$.

multivariable-calculus spherical-coordinates

share|cite|improve this question

edited Apr 2 at 17:43

Daniele Tampieri

2,79221023

asked Apr 2 at 17:26

djpoliridjpoliri

135

$endgroup$

add a comment | 

2

$begingroup$

I am trying to understand a proof where the following equality, without any further details, appears:

$$int_B_qe^-ilangle x,trangledx = c(k)int_-q^q e^-i(q^2-y^2)^k/2dy,$$

where $B_q = leq q$, $tinmathbbR^k+1$, $|x| = (x_1^2+ldots + x_k+1^2)^1/2$, $langlecdot,cdotrangle$ is the standard scalar product and $c(k)$ is a constant depending only on $k$.

I am not able to see how can I pass from a $(k+1)$-dimensional integral to 1-dimensional integral. My intuition tells me that this constant $c(k)$ must be related with the surface area of $mathbbS^k=x$ and then the remaining integral will be related with the radius of $B_q$.

multivariable-calculus spherical-coordinates

share|cite|improve this question

edited Apr 2 at 17:43

Daniele Tampieri

2,79221023

asked Apr 2 at 17:26

djpoliridjpoliri

135

$endgroup$

add a comment | 

$begingroup$

I am trying to understand a proof where the following equality, without any further details, appears:

$$int_B_qe^-ilangle x,trangledx = c(k)int_-q^q e^-i(q^2-y^2)^k/2dy,$$

where $B_q = leq q$, $tinmathbbR^k+1$, $|x| = (x_1^2+ldots + x_k+1^2)^1/2$, $langlecdot,cdotrangle$ is the standard scalar product and $c(k)$ is a constant depending only on $k$.

I am not able to see how can I pass from a $(k+1)$-dimensional integral to 1-dimensional integral. My intuition tells me that this constant $c(k)$ must be related with the surface area of $mathbbS^k=x$ and then the remaining integral will be related with the radius of $B_q$.

multivariable-calculus spherical-coordinates

share|cite|improve this question

edited Apr 2 at 17:43

Daniele Tampieri

2,79221023

asked Apr 2 at 17:26

djpoliridjpoliri

135

$endgroup$

share|cite|improve this question

edited Apr 2 at 17:43

Daniele Tampieri

2,79221023

asked Apr 2 at 17:26

djpoliridjpoliri

135

share|cite|improve this question

edited Apr 2 at 17:43

Daniele Tampieri

2,79221023

asked Apr 2 at 17:26

djpoliridjpoliri

135

edited Apr 2 at 17:43

Daniele Tampieri

2,79221023

edited Apr 2 at 17:43

Daniele Tampieri

2,79221023

asked Apr 2 at 17:26

djpoliridjpoliri

135

asked Apr 2 at 17:26

djpoliridjpoliri

135

1 Answer
1

active

oldest

votes

0

$begingroup$

Let's make a linear variable change that is given by some orthogonal matrix that takes $t$ to $( 0, ldots, 0, |t|)$. Since the Jacobian determinant is 1, and the domain of integration is invariant, we get that the integral we are interested is equal to $int_B_q e^x_k+1 d x_1 ldots d x_k d x_k+1$. Now let's call $x_k+1=y$ and integrate by iteration, with integration with respect with $d x_1 ldots d x_k$ as the inner integral. That becomes $$int_B^k_sqrtq^2-y^2 e^-i d x_1 ldots d x_k=e^-iint_B^k_sqrtq^2-y^2 1 d x_1 ldots d x_k=e^-i c(k)(q^2-y^2)^k/2,$$ where $c(k)$ is the volume of the unit ball in $mathbbR^k$. Now the outer integral is exactly as stated.

share|cite|improve this answer

edited Apr 2 at 19:28

answered Apr 2 at 19:20

MaxMax

4,7071326

$endgroup$

add a comment | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
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: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3172148%2fmultidimensional-integral%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

0

$begingroup$

Let's make a linear variable change that is given by some orthogonal matrix that takes $t$ to $( 0, ldots, 0, |t|)$. Since the Jacobian determinant is 1, and the domain of integration is invariant, we get that the integral we are interested is equal to $int_B_q e^x_k+1 d x_1 ldots d x_k d x_k+1$. Now let's call $x_k+1=y$ and integrate by iteration, with integration with respect with $d x_1 ldots d x_k$ as the inner integral. That becomes $$int_B^k_sqrtq^2-y^2 e^-i d x_1 ldots d x_k=e^-iint_B^k_sqrtq^2-y^2 1 d x_1 ldots d x_k=e^-i c(k)(q^2-y^2)^k/2,$$ where $c(k)$ is the volume of the unit ball in $mathbbR^k$. Now the outer integral is exactly as stated.

share|cite|improve this answer

edited Apr 2 at 19:28

answered Apr 2 at 19:20

MaxMax

4,7071326

$endgroup$

add a comment | 

0

$begingroup$

Let's make a linear variable change that is given by some orthogonal matrix that takes $t$ to $( 0, ldots, 0, |t|)$. Since the Jacobian determinant is 1, and the domain of integration is invariant, we get that the integral we are interested is equal to $int_B_q e^x_k+1 d x_1 ldots d x_k d x_k+1$. Now let's call $x_k+1=y$ and integrate by iteration, with integration with respect with $d x_1 ldots d x_k$ as the inner integral. That becomes $$int_B^k_sqrtq^2-y^2 e^-i d x_1 ldots d x_k=e^-iint_B^k_sqrtq^2-y^2 1 d x_1 ldots d x_k=e^-i c(k)(q^2-y^2)^k/2,$$ where $c(k)$ is the volume of the unit ball in $mathbbR^k$. Now the outer integral is exactly as stated.

share|cite|improve this answer

edited Apr 2 at 19:28

answered Apr 2 at 19:20

MaxMax

4,7071326

$endgroup$

add a comment | 

$begingroup$

Let's make a linear variable change that is given by some orthogonal matrix that takes $t$ to $( 0, ldots, 0, |t|)$. Since the Jacobian determinant is 1, and the domain of integration is invariant, we get that the integral we are interested is equal to $int_B_q e^x_k+1 d x_1 ldots d x_k d x_k+1$. Now let's call $x_k+1=y$ and integrate by iteration, with integration with respect with $d x_1 ldots d x_k$ as the inner integral. That becomes $$int_B^k_sqrtq^2-y^2 e^-i d x_1 ldots d x_k=e^-iint_B^k_sqrtq^2-y^2 1 d x_1 ldots d x_k=e^-i c(k)(q^2-y^2)^k/2,$$ where $c(k)$ is the volume of the unit ball in $mathbbR^k$. Now the outer integral is exactly as stated.

share|cite|improve this answer

edited Apr 2 at 19:28

answered Apr 2 at 19:20

MaxMax

4,7071326

$endgroup$

share|cite|improve this answer

edited Apr 2 at 19:28

answered Apr 2 at 19:20

MaxMax

4,7071326

answered Apr 2 at 19:20

MaxMax

4,7071326

answered Apr 2 at 19:20

MaxMax

4,7071326