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Question about substitutions in the double integral $int_0^1int_0^1 frac-ln xy1-xy dx, dy = 2zeta(3)$
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beginalign
int_0^1int_0^1 frac-ln xy1-xy dx, dy = 2zeta(3) tag1
endalign
Since,
beginalign
int_0^1 frac11-az dz = frac-ln (1-a)a
endalign
and putting $a = 1-xy$, $(1)$ becomes:
beginalign
int_0^1int_0^1 int_0^1 frac11-z(1-xy) dx, dy, dz = 2zeta(3) tag2
endalign
with $t = 1-z(1-xy) implies dt = -(1-xy) , dz$. When $z=0$, $t=1$ and when $z=1$, $t=xy.$ Making these substitutions in $(2)$:
beginalign
-int_1^xyint_0^1 int_0^1 frac1t(1-xy) dx, dy, dt tag3
endalign
Is $(3)$ still equal to $2zeta(3)$? I think $(3)$ will be a function of $x$ and $y$, but after making these substitutions the integral should stay the same.
integration substitution
asked Jul 21 at 0:10
Pinteco
441110
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beginalign
int_0^1int_0^1 frac-ln xy1-xy dx, dy = 2zeta(3) tag1
endalign
Since,
beginalign
int_0^1 frac11-az dz = frac-ln (1-a)a
endalign
and putting $a = 1-xy$, $(1)$ becomes:
beginalign
int_0^1int_0^1 int_0^1 frac11-z(1-xy) dx, dy, dz = 2zeta(3) tag2
endalign
with $t = 1-z(1-xy) implies dt = -(1-xy) , dz$. When $z=0$, $t=1$ and when $z=1$, $t=xy.$ Making these substitutions in $(2)$:
beginalign
-int_1^xyint_0^1 int_0^1 frac1t(1-xy) dx, dy, dt tag3
endalign
Is $(3)$ still equal to $2zeta(3)$? I think $(3)$ will be a function of $x$ and $y$, but after making these substitutions the integral should stay the same.
integration substitution
asked Jul 21 at 0:10
Pinteco
441110
I am pretty sure, your outer integral should actually be the inner integral, i.e., you should have $$ - int_0^1int_0^1 int_1^xy frac1t(1-xy),mathrmd x ,mathrmd y,mathrmd t$$ instead.
– Stan Tendijck
Jul 21 at 0:17
If $dz$ belongs to the outer most integral, why we change the limits of the innner most integral?
– Pinteco
Jul 21 at 0:30
First of all, I made a mistake, in my integral you should have $mathrmd t,mathrmd x,mathrmd y$. Second of all, you can bring change the order of integration (Fubini's theorem) and third of all, if you want to have $z$ as the outer one. You should have $z$ from $0$ to $1$. Then the limit for $y$ is a function of $z$ and the limit for $x$ is a function of $y$ and $z$.
– Stan Tendijck
Jul 21 at 12:47
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
beginalign
int_0^1int_0^1 frac-ln xy1-xy dx, dy = 2zeta(3) tag1
endalign
Since,
beginalign
int_0^1 frac11-az dz = frac-ln (1-a)a
endalign
and putting $a = 1-xy$, $(1)$ becomes:
beginalign
int_0^1int_0^1 int_0^1 frac11-z(1-xy) dx, dy, dz = 2zeta(3) tag2
endalign
with $t = 1-z(1-xy) implies dt = -(1-xy) , dz$. When $z=0$, $t=1$ and when $z=1$, $t=xy.$ Making these substitutions in $(2)$:
beginalign
-int_1^xyint_0^1 int_0^1 frac1t(1-xy) dx, dy, dt tag3
endalign
Is $(3)$ still equal to $2zeta(3)$? I think $(3)$ will be a function of $x$ and $y$, but after making these substitutions the integral should stay the same.
integration substitution
asked Jul 21 at 0:10
Pinteco
441110
beginalign
int_0^1int_0^1 frac-ln xy1-xy dx, dy = 2zeta(3) tag1
endalign
Since,
beginalign
int_0^1 frac11-az dz = frac-ln (1-a)a
endalign
and putting $a = 1-xy$, $(1)$ becomes:
beginalign
int_0^1int_0^1 int_0^1 frac11-z(1-xy) dx, dy, dz = 2zeta(3) tag2
endalign
with $t = 1-z(1-xy) implies dt = -(1-xy) , dz$. When $z=0$, $t=1$ and when $z=1$, $t=xy.$ Making these substitutions in $(2)$:
beginalign
-int_1^xyint_0^1 int_0^1 frac1t(1-xy) dx, dy, dt tag3
endalign
Is $(3)$ still equal to $2zeta(3)$? I think $(3)$ will be a function of $x$ and $y$, but after making these substitutions the integral should stay the same.
integration substitution
asked Jul 21 at 0:10
Pinteco
441110
asked Jul 21 at 0:10
Pinteco
441110
asked Jul 21 at 0:10
Pinteco
441110
asked Jul 21 at 0:10
Pinteco
441110
441110
I am pretty sure, your outer integral should actually be the inner integral, i.e., you should have $$ - int_0^1int_0^1 int_1^xy frac1t(1-xy),mathrmd x ,mathrmd y,mathrmd t$$ instead.
– Stan Tendijck
Jul 21 at 0:17
If $dz$ belongs to the outer most integral, why we change the limits of the innner most integral?
– Pinteco
Jul 21 at 0:30
First of all, I made a mistake, in my integral you should have $mathrmd t,mathrmd x,mathrmd y$. Second of all, you can bring change the order of integration (Fubini's theorem) and third of all, if you want to have $z$ as the outer one. You should have $z$ from $0$ to $1$. Then the limit for $y$ is a function of $z$ and the limit for $x$ is a function of $y$ and $z$.
– Stan Tendijck
Jul 21 at 12:47
add a comment |Â
I am pretty sure, your outer integral should actually be the inner integral, i.e., you should have $$ - int_0^1int_0^1 int_1^xy frac1t(1-xy),mathrmd x ,mathrmd y,mathrmd t$$ instead.
– Stan Tendijck
Jul 21 at 0:17
If $dz$ belongs to the outer most integral, why we change the limits of the innner most integral?
– Pinteco
Jul 21 at 0:30
First of all, I made a mistake, in my integral you should have $mathrmd t,mathrmd x,mathrmd y$. Second of all, you can bring change the order of integration (Fubini's theorem) and third of all, if you want to have $z$ as the outer one. You should have $z$ from $0$ to $1$. Then the limit for $y$ is a function of $z$ and the limit for $x$ is a function of $y$ and $z$.
– Stan Tendijck
Jul 21 at 12:47
I am pretty sure, your outer integral should actually be the inner integral, i.e., you should have $$ - int_0^1int_0^1 int_1^xy frac1t(1-xy),mathrmd x ,mathrmd y,mathrmd t$$ instead.
– Stan Tendijck
Jul 21 at 0:17
I am pretty sure, your outer integral should actually be the inner integral, i.e., you should have $$ - int_0^1int_0^1 int_1^xy frac1t(1-xy),mathrmd x ,mathrmd y,mathrmd t$$ instead.
– Stan Tendijck
Jul 21 at 0:17
If $dz$ belongs to the outer most integral, why we change the limits of the innner most integral?
– Pinteco
Jul 21 at 0:30
If $dz$ belongs to the outer most integral, why we change the limits of the innner most integral?
– Pinteco
Jul 21 at 0:30
First of all, I made a mistake, in my integral you should have $mathrmd t,mathrmd x,mathrmd y$. Second of all, you can bring change the order of integration (Fubini's theorem) and third of all, if you want to have $z$ as the outer one. You should have $z$ from $0$ to $1$. Then the limit for $y$ is a function of $z$ and the limit for $x$ is a function of $y$ and $z$.
– Stan Tendijck
Jul 21 at 12:47
First of all, I made a mistake, in my integral you should have $mathrmd t,mathrmd x,mathrmd y$. Second of all, you can bring change the order of integration (Fubini's theorem) and third of all, if you want to have $z$ as the outer one. You should have $z$ from $0$ to $1$. Then the limit for $y$ is a function of $z$ and the limit for $x$ is a function of $y$ and $z$.
– Stan Tendijck
Jul 21 at 12:47
add a comment |Â
1 Answer
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By geometric series, your integral equals
$$-int^1_0int^1_0(ln x+ln y)sum^infty_k=0(xy)^kdydx$$
Due to the symmetry in $x$ and $y$, this equals
$$-2int^1_0int^1_0(ln x)sum^infty_k=0(xy)^kdydx$$
Since the series converges uniformly, we can integrate it termwise to obtain
$$-2sum^infty_k=0int^1_0 x^kln xcdotfrac11+k dx$$
Since $int^1_0 x^kln xdx=-frac1(1+k)^2$, this simplifies to
$$2sum^infty_k=0frac1(1+k)^3 dx=2zeta(3)$$ by definition.
Prove $int^1_0 x^kln xdx=-frac1(1+k)^2$:
By the substitution $t=-(k+1)ln x$,
$$-int_infty^0 e^frac-ktk+1fractk+1cdotfrac-11+ke^frac-tk+1dt=-frac1(k+1)^2int^infty_0te^-tdt=-frac1(k+1)^2(1!)=colorred-frac1(k+1)^2$$
answered Jul 21 at 15:20
Szeto
4,1631521
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
By geometric series, your integral equals
$$-int^1_0int^1_0(ln x+ln y)sum^infty_k=0(xy)^kdydx$$
Due to the symmetry in $x$ and $y$, this equals
$$-2int^1_0int^1_0(ln x)sum^infty_k=0(xy)^kdydx$$
Since the series converges uniformly, we can integrate it termwise to obtain
$$-2sum^infty_k=0int^1_0 x^kln xcdotfrac11+k dx$$
Since $int^1_0 x^kln xdx=-frac1(1+k)^2$, this simplifies to
$$2sum^infty_k=0frac1(1+k)^3 dx=2zeta(3)$$ by definition.
Prove $int^1_0 x^kln xdx=-frac1(1+k)^2$:
By the substitution $t=-(k+1)ln x$,
$$-int_infty^0 e^frac-ktk+1fractk+1cdotfrac-11+ke^frac-tk+1dt=-frac1(k+1)^2int^infty_0te^-tdt=-frac1(k+1)^2(1!)=colorred-frac1(k+1)^2$$
answered Jul 21 at 15:20
Szeto
4,1631521
add a comment |Â
up vote
0
down vote
By geometric series, your integral equals
$$-int^1_0int^1_0(ln x+ln y)sum^infty_k=0(xy)^kdydx$$
Due to the symmetry in $x$ and $y$, this equals
$$-2int^1_0int^1_0(ln x)sum^infty_k=0(xy)^kdydx$$
Since the series converges uniformly, we can integrate it termwise to obtain
$$-2sum^infty_k=0int^1_0 x^kln xcdotfrac11+k dx$$
Since $int^1_0 x^kln xdx=-frac1(1+k)^2$, this simplifies to
$$2sum^infty_k=0frac1(1+k)^3 dx=2zeta(3)$$ by definition.
Prove $int^1_0 x^kln xdx=-frac1(1+k)^2$:
By the substitution $t=-(k+1)ln x$,
$$-int_infty^0 e^frac-ktk+1fractk+1cdotfrac-11+ke^frac-tk+1dt=-frac1(k+1)^2int^infty_0te^-tdt=-frac1(k+1)^2(1!)=colorred-frac1(k+1)^2$$
answered Jul 21 at 15:20
Szeto
4,1631521
add a comment |Â
up vote
0
down vote
up vote
0
down vote
By geometric series, your integral equals
$$-int^1_0int^1_0(ln x+ln y)sum^infty_k=0(xy)^kdydx$$
Due to the symmetry in $x$ and $y$, this equals
$$-2int^1_0int^1_0(ln x)sum^infty_k=0(xy)^kdydx$$
Since the series converges uniformly, we can integrate it termwise to obtain
$$-2sum^infty_k=0int^1_0 x^kln xcdotfrac11+k dx$$
Since $int^1_0 x^kln xdx=-frac1(1+k)^2$, this simplifies to
$$2sum^infty_k=0frac1(1+k)^3 dx=2zeta(3)$$ by definition.
Prove $int^1_0 x^kln xdx=-frac1(1+k)^2$:
By the substitution $t=-(k+1)ln x$,
$$-int_infty^0 e^frac-ktk+1fractk+1cdotfrac-11+ke^frac-tk+1dt=-frac1(k+1)^2int^infty_0te^-tdt=-frac1(k+1)^2(1!)=colorred-frac1(k+1)^2$$
answered Jul 21 at 15:20
Szeto
4,1631521
By geometric series, your integral equals
$$-int^1_0int^1_0(ln x+ln y)sum^infty_k=0(xy)^kdydx$$
Due to the symmetry in $x$ and $y$, this equals
$$-2int^1_0int^1_0(ln x)sum^infty_k=0(xy)^kdydx$$
Since the series converges uniformly, we can integrate it termwise to obtain
$$-2sum^infty_k=0int^1_0 x^kln xcdotfrac11+k dx$$
Since $int^1_0 x^kln xdx=-frac1(1+k)^2$, this simplifies to
$$2sum^infty_k=0frac1(1+k)^3 dx=2zeta(3)$$ by definition.
Prove $int^1_0 x^kln xdx=-frac1(1+k)^2$:
By the substitution $t=-(k+1)ln x$,
$$-int_infty^0 e^frac-ktk+1fractk+1cdotfrac-11+ke^frac-tk+1dt=-frac1(k+1)^2int^infty_0te^-tdt=-frac1(k+1)^2(1!)=colorred-frac1(k+1)^2$$
answered Jul 21 at 15:20
Szeto
4,1631521
edited Jul 21 at 15:30
answered Jul 21 at 15:20
Szeto
4,1631521
answered Jul 21 at 15:20
Szeto
4,1631521
answered Jul 21 at 15:20
Szeto
4,1631521
4,1631521
add a comment |Â
add a comment |Â
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I am pretty sure, your outer integral should actually be the inner integral, i.e., you should have $$ - int_0^1int_0^1 int_1^xy frac1t(1-xy),mathrmd x ,mathrmd y,mathrmd t$$ instead.
– Stan Tendijck
Jul 21 at 0:17
If $dz$ belongs to the outer most integral, why we change the limits of the innner most integral?
– Pinteco
Jul 21 at 0:30
First of all, I made a mistake, in my integral you should have $mathrmd t,mathrmd x,mathrmd y$. Second of all, you can bring change the order of integration (Fubini's theorem) and third of all, if you want to have $z$ as the outer one. You should have $z$ from $0$ to $1$. Then the limit for $y$ is a function of $z$ and the limit for $x$ is a function of $y$ and $z$.
– Stan Tendijck
Jul 21 at 12:47