Solve the integral $int_0^1int^1_xy^4e^{xy^2}dydx$.
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
1286
|
show 1 more comment
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
1286
1
Maybe try reversing order of integration?
– Zachary Selk
Jan 6 at 17:44
2
Obviously, $$x=u/v^2qquad v=y$$ hence the domain $$0<x<y<1$$ translates as $$0<u/v^2<v<1$$ that is, $$0<u<v^3<1$$
– Did
Jan 6 at 17:46
@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)?
– J. Doe
Jan 6 at 17:52
1
Draw a picture of the region.
– Zachary Selk
Jan 6 at 17:53
1
Yeah now it's totally doable.
– Zachary Selk
Jan 6 at 18:12
|
show 1 more comment
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
1286
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
1286
asked Jan 6 at 17:43
J. DoeJ. Doe
1286
asked Jan 6 at 17:43
J. DoeJ. Doe
1286
asked Jan 6 at 17:43
J. DoeJ. Doe
1286
asked Jan 6 at 17:43
J. DoeJ. Doe
1286
1286
1
Maybe try reversing order of integration?
– Zachary Selk
Jan 6 at 17:44
2
Obviously, $$x=u/v^2qquad v=y$$ hence the domain $$0<x<y<1$$ translates as $$0<u/v^2<v<1$$ that is, $$0<u<v^3<1$$
– Did
Jan 6 at 17:46
@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)?
– J. Doe
Jan 6 at 17:52
1
Draw a picture of the region.
– Zachary Selk
Jan 6 at 17:53
1
Yeah now it's totally doable.
– Zachary Selk
Jan 6 at 18:12
|
show 1 more comment
1
Maybe try reversing order of integration?
– Zachary Selk
Jan 6 at 17:44
2
Obviously, $$x=u/v^2qquad v=y$$ hence the domain $$0<x<y<1$$ translates as $$0<u/v^2<v<1$$ that is, $$0<u<v^3<1$$
– Did
Jan 6 at 17:46
@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)?
– J. Doe
Jan 6 at 17:52
1
Draw a picture of the region.
– Zachary Selk
Jan 6 at 17:53
1
Yeah now it's totally doable.
– Zachary Selk
Jan 6 at 18:12
1
1
Maybe try reversing order of integration?
– Zachary Selk
Jan 6 at 17:44
Maybe try reversing order of integration?
– Zachary Selk
Jan 6 at 17:44
2
2
Obviously, $$x=u/v^2qquad v=y$$ hence the domain $$0<x<y<1$$ translates as $$0<u/v^2<v<1$$ that is, $$0<u<v^3<1$$
– Did
Jan 6 at 17:46
Obviously, $$x=u/v^2qquad v=y$$ hence the domain $$0<x<y<1$$ translates as $$0<u/v^2<v<1$$ that is, $$0<u<v^3<1$$
– Did
Jan 6 at 17:46
@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)?
– J. Doe
Jan 6 at 17:52
@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)?
– J. Doe
Jan 6 at 17:52
1
1
Draw a picture of the region.
– Zachary Selk
Jan 6 at 17:53
Draw a picture of the region.
– Zachary Selk
Jan 6 at 17:53
1
1
Yeah now it's totally doable.
– Zachary Selk
Jan 6 at 18:12
Yeah now it's totally doable.
– Zachary Selk
Jan 6 at 18:12
|
show 1 more comment
2 Answers
2
active
oldest
votes
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
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newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
$ds{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x:
{LARGE ?}}$.
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
67.2k7107141
add a comment |
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
21.7k31240
add a comment |
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2 Answers
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2 Answers
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$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
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$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
67.2k7107141
add a comment |
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
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newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
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newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
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newcommand{verts}[1]{leftvert,{#1},rightvert}$
$ds{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x:
{LARGE ?}}$.
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
67.2k7107141
add a comment |
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
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newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
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newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
$ds{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x:
{LARGE ?}}$.
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
67.2k7107141
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
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newcommand{dd}{mathrm{d}}
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newcommand{verts}[1]{leftvert,{#1},rightvert}$
$ds{int_{0}^{1}int_{x}^{1}y^{4}expo{xy^{2}}dd y,dd x:
{LARGE ?}}$.
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
67.2k7107141
answered Jan 6 at 21:48
Felix MarinFelix Marin
67.2k7107141
answered Jan 6 at 21:48
Felix MarinFelix Marin
67.2k7107141
answered Jan 6 at 21:48
Felix MarinFelix Marin
67.2k7107141
67.2k7107141
add a comment |
add a comment |
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
21.7k31240
add a comment |
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
21.7k31240
add a comment |
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
21.7k31240
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
21.7k31240
edited Jan 7 at 20:05
answered Jan 6 at 20:52
JanJan
21.7k31240
answered Jan 6 at 20:52
JanJan
21.7k31240
answered Jan 6 at 20:52
JanJan
21.7k31240
21.7k31240
add a comment |
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1
Maybe try reversing order of integration?
– Zachary Selk
Jan 6 at 17:44
2
Obviously, $$x=u/v^2qquad v=y$$ hence the domain $$0<x<y<1$$ translates as $$0<u/v^2<v<1$$ that is, $$0<u<v^3<1$$
– Did
Jan 6 at 17:46
@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)?
– J. Doe
Jan 6 at 17:52
1
Draw a picture of the region.
– Zachary Selk
Jan 6 at 17:53
1
Yeah now it's totally doable.
– Zachary Selk
Jan 6 at 18:12