Solve the integral $int_0^1int^1_xy^4e^{xy^2}dydx$.
$begingroup$
Solve the integral $int_0^1int^1_xy^4e^{xy^2}dydx$.
I think that variables substituation is neede here. I've substitute
$$
\ left{begin{matrix}
u=xy^2\
v=y
end{matrix}right.
$$
and calculated
$$\J=begin{vmatrix}
y^2 & 2xy\
0 & 1
end{vmatrix}=y^2
$$
Then, the new integrand is $v^2e^u$. But what is the new domain? Thanks.
integration multivariable-calculus substitution jacobian
asked Jan 6 at 17:43
J. DoeJ. Doe
1436
$endgroup$
|
show 1 more comment
$begingroup$
Solve the integral $int_0^1int^1_xy^4e^{xy^2}dydx$.
I think that variables substituation is neede here. I've substitute
$$
\ left{begin{matrix}
u=xy^2\
v=y
end{matrix}right.
$$
and calculated
$$\J=begin{vmatrix}
y^2 & 2xy\
0 & 1
end{vmatrix}=y^2
$$
Then, the new integrand is $v^2e^u$. But what is the new domain? Thanks.
integration multivariable-calculus substitution jacobian
asked Jan 6 at 17:43
J. DoeJ. Doe
1436
$endgroup$
1
$begingroup$
Maybe try reversing order of integration?
$endgroup$
– Zachary Selk
Jan 6 at 17:44
2
$begingroup$
Obviously, $$x=u/v^2qquad v=y$$ hence the domain $$0<x<y<1$$ translates as $$0<u/v^2<v<1$$ that is, $$0<u<v^3<1$$
$endgroup$
– Did
Jan 6 at 17:46
$begingroup$
@ZacharySelk Ok, but then, what meaning does the $x$ in the lower bound of the external integral have: $int_x^1int_0^1 A dxdy$ (if $A$ is the integrand)?
$endgroup$
– J. Doe
Jan 6 at 17:52
1
$begingroup$
Draw a picture of the region.
$endgroup$
– Zachary Selk
Jan 6 at 17:53
1
$begingroup$
Yeah now it's totally doable.
$endgroup$
– Zachary Selk
Jan 6 at 18:12
|
show 1 more comment
$begingroup$
Solve the integral $int_0^1int^1_xy^4e^{xy^2}dydx$.
I think that variables substituation is neede here. I've substitute
$$
\ left{begin{matrix}
u=xy^2\
v=y
end{matrix}right.
$$
and calculated
$$\J=begin{vmatrix}
y^2 & 2xy\
0 & 1
end{vmatrix}=y^2
$$
Then, the new integrand is $v^2e^u$. But what is the new domain? Thanks.
integration multivariable-calculus substitution jacobian
asked Jan 6 at 17:43
J. DoeJ. Doe
1436
$endgroup$
Solve the integral $int_0^1int^1_xy^4e^{xy^2}dydx$.
I think that variables substituation is neede here. I've substitute
$$
\ left{begin{matrix}
u=xy^2\
v=y
end{matrix}right.
$$
and calculated
$$\J=begin{vmatrix}
y^2 & 2xy\
0 & 1
end{vmatrix}=y^2
$$
Then, the new integrand is $v^2e^u$. But what is the new domain? Thanks.
integration multivariable-calculus substitution jacobian
integration multivariable-calculus substitution jacobian
asked Jan 6 at 17:43
J. DoeJ. Doe
1436
asked Jan 6 at 17:43
J. DoeJ. Doe
1436
asked Jan 6 at 17:43
J. DoeJ. Doe
1436
asked Jan 6 at 17:43
J. DoeJ. Doe
1436
asked Jan 6 at 17:43
J. DoeJ. Doe
1436
1436
1
$begingroup$
Maybe try reversing order of integration?
$endgroup$
– Zachary Selk
Jan 6 at 17:44
2
$begingroup$
Obviously, $$x=u/v^2qquad v=y$$ hence the domain $$0<x<y<1$$ translates as $$0<u/v^2<v<1$$ that is, $$0<u<v^3<1$$
$endgroup$
– Did
Jan 6 at 17:46
$begingroup$
@ZacharySelk Ok, but then, what meaning does the $x$ in the lower bound of the external integral have: $int_x^1int_0^1 A dxdy$ (if $A$ is the integrand)?
$endgroup$
– J. Doe
Jan 6 at 17:52
1
$begingroup$
Draw a picture of the region.
$endgroup$
– Zachary Selk
Jan 6 at 17:53
1
$begingroup$
Yeah now it's totally doable.
$endgroup$
– Zachary Selk
Jan 6 at 18:12
|
show 1 more comment
1
$begingroup$
Maybe try reversing order of integration?
$endgroup$
– Zachary Selk
Jan 6 at 17:44
2
$begingroup$
Obviously, $$x=u/v^2qquad v=y$$ hence the domain $$0<x<y<1$$ translates as $$0<u/v^2<v<1$$ that is, $$0<u<v^3<1$$
$endgroup$
– Did
Jan 6 at 17:46
$begingroup$
@ZacharySelk Ok, but then, what meaning does the $x$ in the lower bound of the external integral have: $int_x^1int_0^1 A dxdy$ (if $A$ is the integrand)?
$endgroup$
– J. Doe
Jan 6 at 17:52
1
$begingroup$
Draw a picture of the region.
$endgroup$
– Zachary Selk
Jan 6 at 17:53
1
$begingroup$
Yeah now it's totally doable.
$endgroup$
– Zachary Selk
Jan 6 at 18:12
1
1
$begingroup$
Maybe try reversing order of integration?
$endgroup$
– Zachary Selk
Jan 6 at 17:44
$begingroup$
Maybe try reversing order of integration?
$endgroup$
– Zachary Selk
Jan 6 at 17:44
2
2
$begingroup$
Obviously, $$x=u/v^2qquad v=y$$ hence the domain $$0<x<y<1$$ translates as $$0<u/v^2<v<1$$ that is, $$0<u<v^3<1$$
$endgroup$
– Did
Jan 6 at 17:46
$begingroup$
Obviously, $$x=u/v^2qquad v=y$$ hence the domain $$0<x<y<1$$ translates as $$0<u/v^2<v<1$$ that is, $$0<u<v^3<1$$
$endgroup$
– Did
Jan 6 at 17:46
$begingroup$
@ZacharySelk Ok, but then, what meaning does the $x$ in the lower bound of the external integral have: $int_x^1int_0^1 A dxdy$ (if $A$ is the integrand)?
$endgroup$
– J. Doe
Jan 6 at 17:52
$begingroup$
@ZacharySelk Ok, but then, what meaning does the $x$ in the lower bound of the external integral have: $int_x^1int_0^1 A dxdy$ (if $A$ is the integrand)?
$endgroup$
– J. Doe
Jan 6 at 17:52
1
1
$begingroup$
Draw a picture of the region.
$endgroup$
– Zachary Selk
Jan 6 at 17:53
$begingroup$
Draw a picture of the region.
$endgroup$
– Zachary Selk
Jan 6 at 17:53
1
1
$begingroup$
Yeah now it's totally doable.
$endgroup$
– Zachary Selk
Jan 6 at 18:12
$begingroup$
Yeah now it's totally doable.
$endgroup$
– Zachary Selk
Jan 6 at 18:12
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
$ds{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x:
{LARGE ?}}$.
begin{align}
&bbox[10px,#ffd]{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x} =
int_{0}^{1}y^{4}int_{0}^{y}expo{xy^{2}}dd x,dd y =
int_{0}^{1}y^{4}{expo{y^{3}} - 1 over y^{2}},dd y
\[5mm] = &
int_{0}^{1}pars{y^{2}expo{y^{3}} - y^{2}}dd y =
left.{expo{y^{3}} - y^{3} over 3},rightvert_{0}^{1} =
bbx{expo{} - 2 over 3} approx 0.2394
end{align}
answered Jan 6 at 21:48
Felix MarinFelix Marin
67.4k7107141
$endgroup$
add a comment |
$begingroup$
Well, solving a much more general problem:
$$mathcal{I}_text{n}left(alpharight):=int_0^alphaint_x^alphatext{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}xtag1$$
Using that (for all $x$):
$$expleft(xright)=sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}tag2$$
We can write:
$$mathcal{I}_text{n}left(alpharight)=int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{left(xcdottext{y}^{text{n}-2}right)^text{k}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
$$int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{x^text{k}cdottext{y}^{text{k}left(text{n}-2right)}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
$$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^text{n}cdottext{y}^{text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
$$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^{text{n}+text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
$$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft[frac{text{y}^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right]_x^alpharight}spacetext{d}x=$$
$$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft(frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}-frac{x^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right)right}spacetext{d}x=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^text{k}spacetext{d}x-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^{1+text{n}+text{k}left(text{n}-1right)}spacetext{d}xright}=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+text{k}}}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+1+text{n}+text{k}left(text{n}-1right)}}{1+1+text{n}+text{k}left(text{n}-1right)}right}=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{1}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{k}}-frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}tag3$$
When $alpha=1$, we get:
$$mathcal{I}_text{n}left(alpharight):=int_0^1int_x^1text{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}x=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{1}{1+text{k}}-frac{1}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{k}}cdotfrac{1}{2+text{n}+text{k}left(text{n}-1right)}tag4$$
answered Jan 6 at 20:52
JanJan
21.8k31240
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3064163%2fsolve-the-integral-int-01-int1-xy4exy2dydx%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
$ds{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x:
{LARGE ?}}$.
begin{align}
&bbox[10px,#ffd]{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x} =
int_{0}^{1}y^{4}int_{0}^{y}expo{xy^{2}}dd x,dd y =
int_{0}^{1}y^{4}{expo{y^{3}} - 1 over y^{2}},dd y
\[5mm] = &
int_{0}^{1}pars{y^{2}expo{y^{3}} - y^{2}}dd y =
left.{expo{y^{3}} - y^{3} over 3},rightvert_{0}^{1} =
bbx{expo{} - 2 over 3} approx 0.2394
end{align}
answered Jan 6 at 21:48
Felix MarinFelix Marin
67.4k7107141
$endgroup$
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
$ds{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x:
{LARGE ?}}$.
begin{align}
&bbox[10px,#ffd]{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x} =
int_{0}^{1}y^{4}int_{0}^{y}expo{xy^{2}}dd x,dd y =
int_{0}^{1}y^{4}{expo{y^{3}} - 1 over y^{2}},dd y
\[5mm] = &
int_{0}^{1}pars{y^{2}expo{y^{3}} - y^{2}}dd y =
left.{expo{y^{3}} - y^{3} over 3},rightvert_{0}^{1} =
bbx{expo{} - 2 over 3} approx 0.2394
end{align}
answered Jan 6 at 21:48
Felix MarinFelix Marin
67.4k7107141
$endgroup$
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
$ds{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x:
{LARGE ?}}$.
begin{align}
&bbox[10px,#ffd]{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x} =
int_{0}^{1}y^{4}int_{0}^{y}expo{xy^{2}}dd x,dd y =
int_{0}^{1}y^{4}{expo{y^{3}} - 1 over y^{2}},dd y
\[5mm] = &
int_{0}^{1}pars{y^{2}expo{y^{3}} - y^{2}}dd y =
left.{expo{y^{3}} - y^{3} over 3},rightvert_{0}^{1} =
bbx{expo{} - 2 over 3} approx 0.2394
end{align}
answered Jan 6 at 21:48
Felix MarinFelix Marin
67.4k7107141
$endgroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
$ds{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x:
{LARGE ?}}$.
begin{align}
&bbox[10px,#ffd]{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x} =
int_{0}^{1}y^{4}int_{0}^{y}expo{xy^{2}}dd x,dd y =
int_{0}^{1}y^{4}{expo{y^{3}} - 1 over y^{2}},dd y
\[5mm] = &
int_{0}^{1}pars{y^{2}expo{y^{3}} - y^{2}}dd y =
left.{expo{y^{3}} - y^{3} over 3},rightvert_{0}^{1} =
bbx{expo{} - 2 over 3} approx 0.2394
end{align}
answered Jan 6 at 21:48
Felix MarinFelix Marin
67.4k7107141
answered Jan 6 at 21:48
Felix MarinFelix Marin
67.4k7107141
answered Jan 6 at 21:48
Felix MarinFelix Marin
67.4k7107141
answered Jan 6 at 21:48
Felix MarinFelix Marin
67.4k7107141
67.4k7107141
add a comment |
add a comment |
$begingroup$
Well, solving a much more general problem:
$$mathcal{I}_text{n}left(alpharight):=int_0^alphaint_x^alphatext{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}xtag1$$
Using that (for all $x$):
$$expleft(xright)=sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}tag2$$
We can write:
$$mathcal{I}_text{n}left(alpharight)=int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{left(xcdottext{y}^{text{n}-2}right)^text{k}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
$$int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{x^text{k}cdottext{y}^{text{k}left(text{n}-2right)}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
$$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^text{n}cdottext{y}^{text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
$$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^{text{n}+text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
$$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft[frac{text{y}^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right]_x^alpharight}spacetext{d}x=$$
$$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft(frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}-frac{x^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right)right}spacetext{d}x=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^text{k}spacetext{d}x-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^{1+text{n}+text{k}left(text{n}-1right)}spacetext{d}xright}=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+text{k}}}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+1+text{n}+text{k}left(text{n}-1right)}}{1+1+text{n}+text{k}left(text{n}-1right)}right}=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{1}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{k}}-frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}tag3$$
When $alpha=1$, we get:
$$mathcal{I}_text{n}left(alpharight):=int_0^1int_x^1text{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}x=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{1}{1+text{k}}-frac{1}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{k}}cdotfrac{1}{2+text{n}+text{k}left(text{n}-1right)}tag4$$
answered Jan 6 at 20:52
JanJan
21.8k31240
$endgroup$
add a comment |
$begingroup$
Well, solving a much more general problem:
$$mathcal{I}_text{n}left(alpharight):=int_0^alphaint_x^alphatext{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}xtag1$$
Using that (for all $x$):
$$expleft(xright)=sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}tag2$$
We can write:
$$mathcal{I}_text{n}left(alpharight)=int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{left(xcdottext{y}^{text{n}-2}right)^text{k}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
$$int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{x^text{k}cdottext{y}^{text{k}left(text{n}-2right)}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
$$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^text{n}cdottext{y}^{text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
$$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^{text{n}+text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
$$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft[frac{text{y}^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right]_x^alpharight}spacetext{d}x=$$
$$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft(frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}-frac{x^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right)right}spacetext{d}x=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^text{k}spacetext{d}x-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^{1+text{n}+text{k}left(text{n}-1right)}spacetext{d}xright}=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+text{k}}}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+1+text{n}+text{k}left(text{n}-1right)}}{1+1+text{n}+text{k}left(text{n}-1right)}right}=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{1}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{k}}-frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}tag3$$
When $alpha=1$, we get:
$$mathcal{I}_text{n}left(alpharight):=int_0^1int_x^1text{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}x=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{1}{1+text{k}}-frac{1}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{k}}cdotfrac{1}{2+text{n}+text{k}left(text{n}-1right)}tag4$$
answered Jan 6 at 20:52
JanJan
21.8k31240
$endgroup$
add a comment |
$begingroup$
Well, solving a much more general problem:
$$mathcal{I}_text{n}left(alpharight):=int_0^alphaint_x^alphatext{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}xtag1$$
Using that (for all $x$):
$$expleft(xright)=sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}tag2$$
We can write:
$$mathcal{I}_text{n}left(alpharight)=int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{left(xcdottext{y}^{text{n}-2}right)^text{k}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
$$int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{x^text{k}cdottext{y}^{text{k}left(text{n}-2right)}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
$$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^text{n}cdottext{y}^{text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
$$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^{text{n}+text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
$$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft[frac{text{y}^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right]_x^alpharight}spacetext{d}x=$$
$$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft(frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}-frac{x^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right)right}spacetext{d}x=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^text{k}spacetext{d}x-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^{1+text{n}+text{k}left(text{n}-1right)}spacetext{d}xright}=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+text{k}}}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+1+text{n}+text{k}left(text{n}-1right)}}{1+1+text{n}+text{k}left(text{n}-1right)}right}=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{1}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{k}}-frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}tag3$$
When $alpha=1$, we get:
$$mathcal{I}_text{n}left(alpharight):=int_0^1int_x^1text{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}x=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{1}{1+text{k}}-frac{1}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{k}}cdotfrac{1}{2+text{n}+text{k}left(text{n}-1right)}tag4$$
answered Jan 6 at 20:52
JanJan
21.8k31240
$endgroup$
Well, solving a much more general problem:
$$mathcal{I}_text{n}left(alpharight):=int_0^alphaint_x^alphatext{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}xtag1$$
Using that (for all $x$):
$$expleft(xright)=sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}tag2$$
We can write:
$$mathcal{I}_text{n}left(alpharight)=int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{left(xcdottext{y}^{text{n}-2}right)^text{k}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
$$int_0^alphaint_x^alphatext{y}^text{n}cdotsum_{text{k}=0}^inftyfrac{x^text{k}cdottext{y}^{text{k}left(text{n}-2right)}}{text{k}!}spacetext{d}text{y}spacetext{d}x=$$
$$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^text{n}cdottext{y}^{text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
$$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}int_x^alphatext{y}^{text{n}+text{k}left(text{n}-2right)}spacetext{d}text{y}right}spacetext{d}x=$$
$$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft[frac{text{y}^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right]_x^alpharight}spacetext{d}x=$$
$$int_0^alphaleft{sum_{text{k}=0}^inftyfrac{x^text{k}}{text{k}!}cdotleft(frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}-frac{x^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}right)right}spacetext{d}x=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^text{k}spacetext{d}x-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotint_0^alpha x^{1+text{n}+text{k}left(text{n}-1right)}spacetext{d}xright}=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{1+text{n}+text{k}left(text{n}-2right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+text{k}}}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{1+1+text{n}+text{k}left(text{n}-1right)}}{1+1+text{n}+text{k}left(text{n}-1right)}right}=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{1}{1+text{k}}-frac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotfrac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{1+text{k}}-frac{alpha^{2+text{n}+text{k}left(text{n}-1right)}}{2+text{n}+text{k}left(text{n}-1right)}right}tag3$$
When $alpha=1$, we get:
$$mathcal{I}_text{n}left(alpharight):=int_0^1int_x^1text{y}^text{n}cdotexpleft(xcdottext{y}^{text{n}-2}right)spacetext{d}text{y}spacetext{d}x=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{n}+text{k}left(text{n}-2right)}cdotleft{frac{1}{1+text{k}}-frac{1}{2+text{n}+text{k}left(text{n}-1right)}right}=$$
$$sum_{text{k}=0}^inftyfrac{1}{text{k}!}cdotfrac{1}{1+text{k}}cdotfrac{1}{2+text{n}+text{k}left(text{n}-1right)}tag4$$
answered Jan 6 at 20:52
JanJan
21.8k31240
edited Jan 7 at 20:05
answered Jan 6 at 20:52
JanJan
21.8k31240
answered Jan 6 at 20:52
JanJan
21.8k31240
answered Jan 6 at 20:52
JanJan
21.8k31240
21.8k31240
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3064163%2fsolve-the-integral-int-01-int1-xy4exy2dydx%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
Maybe try reversing order of integration?
$endgroup$
– Zachary Selk
Jan 6 at 17:44
2
$begingroup$
Obviously, $$x=u/v^2qquad v=y$$ hence the domain $$0<x<y<1$$ translates as $$0<u/v^2<v<1$$ that is, $$0<u<v^3<1$$
$endgroup$
– Did
Jan 6 at 17:46
$begingroup$
@ZacharySelk Ok, but then, what meaning does the $x$ in the lower bound of the external integral have: $int_x^1int_0^1 A dxdy$ (if $A$ is the integrand)?
$endgroup$
– J. Doe
Jan 6 at 17:52
1
$begingroup$
Draw a picture of the region.
$endgroup$
– Zachary Selk
Jan 6 at 17:53
1
$begingroup$
Yeah now it's totally doable.
$endgroup$
– Zachary Selk
Jan 6 at 18:12