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How to know if this double integral integrates to 0
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I would like some help to find out if the value of this double integral is $0$ or not. I do not think it is:
$$int_0^1int_-1^1e^x^2+y^2sin(y),dx,dy$$
Separating the integrals with respect to $x$ and to $y$. I could see by using the power series expansion of $e^x^2$ that $int_-1^1e^x^2dx$ is positive.
What is an easy way to see that $int_0^1e^y^2sin(y),dy$ is not $0$?
integration definite-integrals
edited Aug 10 at 17:31
Robert Howard
1,329620
asked Jul 22 at 7:02
Arthur Carvalho Brito
165
 |Â
show 1 more comment
up vote
0
down vote
favorite
I would like some help to find out if the value of this double integral is $0$ or not. I do not think it is:
$$int_0^1int_-1^1e^x^2+y^2sin(y),dx,dy$$
Separating the integrals with respect to $x$ and to $y$. I could see by using the power series expansion of $e^x^2$ that $int_-1^1e^x^2dx$ is positive.
What is an easy way to see that $int_0^1e^y^2sin(y),dy$ is not $0$?
integration definite-integrals
edited Aug 10 at 17:31
Robert Howard
1,329620
asked Jul 22 at 7:02
Arthur Carvalho Brito
165
If you swap $mathrm dx$ and $mathrm dy$ then you get $0$ though.
– Kenny Lau
Jul 22 at 7:04
The integrand is positive.
– Lord Shark the Unknown
Jul 22 at 7:04
@Surb wolframalpha.com/input/…
– Kenny Lau
Jul 22 at 7:48
@Surb If you do swap $dx$ and $dy$ (without swapping the integral signs, so that it is $y$ which goes from $-1$ to $1$), then the substitution $ymapsto -y$ gives you the exact same integral with a minus sign in front. Thus the integral is equal to its own negative, and must be $0$. Alternatively, the integrand is odd in $y$, so integrating over an area symmetric across the $y$-axis must yield $0$. That's what Kenny Lau means.
– Arthur
Jul 22 at 8:14
"What is an easy way to see that $int_0^1 e^y^2sin(y) dy$ is not 0??" To check that the integrand is positive on $(0,1)$. Is this your question?
– Did
Jul 22 at 8:35
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I would like some help to find out if the value of this double integral is $0$ or not. I do not think it is:
$$int_0^1int_-1^1e^x^2+y^2sin(y),dx,dy$$
Separating the integrals with respect to $x$ and to $y$. I could see by using the power series expansion of $e^x^2$ that $int_-1^1e^x^2dx$ is positive.
What is an easy way to see that $int_0^1e^y^2sin(y),dy$ is not $0$?
integration definite-integrals
edited Aug 10 at 17:31
Robert Howard
1,329620
asked Jul 22 at 7:02
Arthur Carvalho Brito
165
I would like some help to find out if the value of this double integral is $0$ or not. I do not think it is:
$$int_0^1int_-1^1e^x^2+y^2sin(y),dx,dy$$
Separating the integrals with respect to $x$ and to $y$. I could see by using the power series expansion of $e^x^2$ that $int_-1^1e^x^2dx$ is positive.
What is an easy way to see that $int_0^1e^y^2sin(y),dy$ is not $0$?
integration definite-integrals
edited Aug 10 at 17:31
Robert Howard
1,329620
asked Jul 22 at 7:02
Arthur Carvalho Brito
165
edited Aug 10 at 17:31
Robert Howard
1,329620
edited Aug 10 at 17:31
Robert Howard
1,329620
edited Aug 10 at 17:31
Robert Howard
1,329620
1,329620
asked Jul 22 at 7:02
Arthur Carvalho Brito
165
asked Jul 22 at 7:02
Arthur Carvalho Brito
165
asked Jul 22 at 7:02
Arthur Carvalho Brito
165
165
If you swap $mathrm dx$ and $mathrm dy$ then you get $0$ though.
– Kenny Lau
Jul 22 at 7:04
The integrand is positive.
– Lord Shark the Unknown
Jul 22 at 7:04
@Surb wolframalpha.com/input/…
– Kenny Lau
Jul 22 at 7:48
@Surb If you do swap $dx$ and $dy$ (without swapping the integral signs, so that it is $y$ which goes from $-1$ to $1$), then the substitution $ymapsto -y$ gives you the exact same integral with a minus sign in front. Thus the integral is equal to its own negative, and must be $0$. Alternatively, the integrand is odd in $y$, so integrating over an area symmetric across the $y$-axis must yield $0$. That's what Kenny Lau means.
– Arthur
Jul 22 at 8:14
"What is an easy way to see that $int_0^1 e^y^2sin(y) dy$ is not 0??" To check that the integrand is positive on $(0,1)$. Is this your question?
– Did
Jul 22 at 8:35
 |Â
show 1 more comment
If you swap $mathrm dx$ and $mathrm dy$ then you get $0$ though.
– Kenny Lau
Jul 22 at 7:04
The integrand is positive.
– Lord Shark the Unknown
Jul 22 at 7:04
@Surb wolframalpha.com/input/…
– Kenny Lau
Jul 22 at 7:48
@Surb If you do swap $dx$ and $dy$ (without swapping the integral signs, so that it is $y$ which goes from $-1$ to $1$), then the substitution $ymapsto -y$ gives you the exact same integral with a minus sign in front. Thus the integral is equal to its own negative, and must be $0$. Alternatively, the integrand is odd in $y$, so integrating over an area symmetric across the $y$-axis must yield $0$. That's what Kenny Lau means.
– Arthur
Jul 22 at 8:14
"What is an easy way to see that $int_0^1 e^y^2sin(y) dy$ is not 0??" To check that the integrand is positive on $(0,1)$. Is this your question?
– Did
Jul 22 at 8:35
If you swap $mathrm dx$ and $mathrm dy$ then you get $0$ though.
– Kenny Lau
Jul 22 at 7:04
If you swap $mathrm dx$ and $mathrm dy$ then you get $0$ though.
– Kenny Lau
Jul 22 at 7:04
The integrand is positive.
– Lord Shark the Unknown
Jul 22 at 7:04
The integrand is positive.
– Lord Shark the Unknown
Jul 22 at 7:04
@Surb wolframalpha.com/input/…
– Kenny Lau
Jul 22 at 7:48
@Surb wolframalpha.com/input/…
– Kenny Lau
Jul 22 at 7:48
@Surb If you do swap $dx$ and $dy$ (without swapping the integral signs, so that it is $y$ which goes from $-1$ to $1$), then the substitution $ymapsto -y$ gives you the exact same integral with a minus sign in front. Thus the integral is equal to its own negative, and must be $0$. Alternatively, the integrand is odd in $y$, so integrating over an area symmetric across the $y$-axis must yield $0$. That's what Kenny Lau means.
– Arthur
Jul 22 at 8:14
@Surb If you do swap $dx$ and $dy$ (without swapping the integral signs, so that it is $y$ which goes from $-1$ to $1$), then the substitution $ymapsto -y$ gives you the exact same integral with a minus sign in front. Thus the integral is equal to its own negative, and must be $0$. Alternatively, the integrand is odd in $y$, so integrating over an area symmetric across the $y$-axis must yield $0$. That's what Kenny Lau means.
– Arthur
Jul 22 at 8:14
"What is an easy way to see that $int_0^1 e^y^2sin(y) dy$ is not 0??" To check that the integrand is positive on $(0,1)$. Is this your question?
– Did
Jul 22 at 8:35
"What is an easy way to see that $int_0^1 e^y^2sin(y) dy$ is not 0??" To check that the integrand is positive on $(0,1)$. Is this your question?
– Did
Jul 22 at 8:35
 |Â
show 1 more comment
2 Answers
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0
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$e^y^2 sin(y) > 0$ when $y in (0,1]$, and $mu((0,1]) = 1 > 0$, where $mu$ is Lebesgue measure.
answered Jul 22 at 7:03
Kenny Lau
18.7k2157
add a comment |Â
up vote
0
down vote
An integral of the form
$$int_a^b f(x) textrm dx $$
Is zero if
$f(x)=0 forall xin [a,b]$, or
if the surface of all areas with $f(x)>0$ equals the surface of all areas with $f(x)<0$ in the respective integral domain $[a,b]$.
The integral you wonder about is:
$$int_0^1e^y^2 sin(y)textrm dy, $$
And the function here is always positive in the domain $[0,1]$. Hence, the integral will not be zero.
Be aware, however, that it is unclear which limits belong to which variable. It might be better to write
$$int_-1^1textrm dxint_0^1textrm dy, e^x^2+y^2 sin(y), $$
which is bigger than zero, or
$$int_-1^1textrm dyint_0^1textrm dx, e^x^2+y^2 sin(y), $$
which equals zero
answered Aug 10 at 18:05
kvantour
22618
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
$e^y^2 sin(y) > 0$ when $y in (0,1]$, and $mu((0,1]) = 1 > 0$, where $mu$ is Lebesgue measure.
answered Jul 22 at 7:03
Kenny Lau
18.7k2157
add a comment |Â
up vote
0
down vote
$e^y^2 sin(y) > 0$ when $y in (0,1]$, and $mu((0,1]) = 1 > 0$, where $mu$ is Lebesgue measure.
answered Jul 22 at 7:03
Kenny Lau
18.7k2157
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$e^y^2 sin(y) > 0$ when $y in (0,1]$, and $mu((0,1]) = 1 > 0$, where $mu$ is Lebesgue measure.
answered Jul 22 at 7:03
Kenny Lau
18.7k2157
$e^y^2 sin(y) > 0$ when $y in (0,1]$, and $mu((0,1]) = 1 > 0$, where $mu$ is Lebesgue measure.
answered Jul 22 at 7:03
Kenny Lau
18.7k2157
edited Jul 22 at 7:16
answered Jul 22 at 7:03
Kenny Lau
18.7k2157
answered Jul 22 at 7:03
Kenny Lau
18.7k2157
answered Jul 22 at 7:03
Kenny Lau
18.7k2157
18.7k2157
add a comment |Â
add a comment |Â
up vote
0
down vote
An integral of the form
$$int_a^b f(x) textrm dx $$
Is zero if
$f(x)=0 forall xin [a,b]$, or
if the surface of all areas with $f(x)>0$ equals the surface of all areas with $f(x)<0$ in the respective integral domain $[a,b]$.
The integral you wonder about is:
$$int_0^1e^y^2 sin(y)textrm dy, $$
And the function here is always positive in the domain $[0,1]$. Hence, the integral will not be zero.
Be aware, however, that it is unclear which limits belong to which variable. It might be better to write
$$int_-1^1textrm dxint_0^1textrm dy, e^x^2+y^2 sin(y), $$
which is bigger than zero, or
$$int_-1^1textrm dyint_0^1textrm dx, e^x^2+y^2 sin(y), $$
which equals zero
answered Aug 10 at 18:05
kvantour
22618
add a comment |Â
up vote
0
down vote
An integral of the form
$$int_a^b f(x) textrm dx $$
Is zero if
$f(x)=0 forall xin [a,b]$, or
if the surface of all areas with $f(x)>0$ equals the surface of all areas with $f(x)<0$ in the respective integral domain $[a,b]$.
The integral you wonder about is:
$$int_0^1e^y^2 sin(y)textrm dy, $$
And the function here is always positive in the domain $[0,1]$. Hence, the integral will not be zero.
Be aware, however, that it is unclear which limits belong to which variable. It might be better to write
$$int_-1^1textrm dxint_0^1textrm dy, e^x^2+y^2 sin(y), $$
which is bigger than zero, or
$$int_-1^1textrm dyint_0^1textrm dx, e^x^2+y^2 sin(y), $$
which equals zero
answered Aug 10 at 18:05
kvantour
22618
add a comment |Â
up vote
0
down vote
up vote
0
down vote
An integral of the form
$$int_a^b f(x) textrm dx $$
Is zero if
$f(x)=0 forall xin [a,b]$, or
if the surface of all areas with $f(x)>0$ equals the surface of all areas with $f(x)<0$ in the respective integral domain $[a,b]$.
The integral you wonder about is:
$$int_0^1e^y^2 sin(y)textrm dy, $$
And the function here is always positive in the domain $[0,1]$. Hence, the integral will not be zero.
Be aware, however, that it is unclear which limits belong to which variable. It might be better to write
$$int_-1^1textrm dxint_0^1textrm dy, e^x^2+y^2 sin(y), $$
which is bigger than zero, or
$$int_-1^1textrm dyint_0^1textrm dx, e^x^2+y^2 sin(y), $$
which equals zero
answered Aug 10 at 18:05
kvantour
22618
An integral of the form
$$int_a^b f(x) textrm dx $$
Is zero if
$f(x)=0 forall xin [a,b]$, or
if the surface of all areas with $f(x)>0$ equals the surface of all areas with $f(x)<0$ in the respective integral domain $[a,b]$.
The integral you wonder about is:
$$int_0^1e^y^2 sin(y)textrm dy, $$
And the function here is always positive in the domain $[0,1]$. Hence, the integral will not be zero.
Be aware, however, that it is unclear which limits belong to which variable. It might be better to write
$$int_-1^1textrm dxint_0^1textrm dy, e^x^2+y^2 sin(y), $$
which is bigger than zero, or
$$int_-1^1textrm dyint_0^1textrm dx, e^x^2+y^2 sin(y), $$
which equals zero
answered Aug 10 at 18:05
kvantour
22618
edited Aug 10 at 18:13
answered Aug 10 at 18:05
kvantour
22618
answered Aug 10 at 18:05
kvantour
22618
answered Aug 10 at 18:05
kvantour
22618
22618
add a comment |Â
add a comment |Â
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If you swap $mathrm dx$ and $mathrm dy$ then you get $0$ though.
– Kenny Lau
Jul 22 at 7:04
The integrand is positive.
– Lord Shark the Unknown
Jul 22 at 7:04
@Surb wolframalpha.com/input/…
– Kenny Lau
Jul 22 at 7:48
@Surb If you do swap $dx$ and $dy$ (without swapping the integral signs, so that it is $y$ which goes from $-1$ to $1$), then the substitution $ymapsto -y$ gives you the exact same integral with a minus sign in front. Thus the integral is equal to its own negative, and must be $0$. Alternatively, the integrand is odd in $y$, so integrating over an area symmetric across the $y$-axis must yield $0$. That's what Kenny Lau means.
– Arthur
Jul 22 at 8:14
"What is an easy way to see that $int_0^1 e^y^2sin(y) dy$ is not 0??" To check that the integrand is positive on $(0,1)$. Is this your question?
– Did
Jul 22 at 8:35