Mixing
What is the meaning of this notation $int dy int f(x,y)dx$
$begingroup$
I was asked to determine whether the integrals $int_{0}^{infty}dyint_{0}^{infty}e^{-xy}sin(x)dx$ and $int_{0}^{infty}dxint_{0}^{infty}e^{-xy}sin(x)dy$ converge, and if they do, calculate them.
I'm not sure what this notation means. Is $int_{0}^{infty}dyint_{0}^{infty}e^{-xy}sin(x)dx$ supposed to mean $int_{0}^{infty}int_{0}^{infty}e^{-xy}sin(x)dxdy$? Or does this literally mean the product of the integrals: $int_{0}^{infty} dycdot int_{0}^{infty}e^{-xy}sin(x)dx$? where the $y$ in the second integral is just some unknown parameter now, not related to the first integral.
Anyone seen this notation before?
calculus integration notation improper-integrals
asked Jan 2 at 9:13
Oria GruberOria Gruber
6,50732360
$endgroup$
add a comment |
$begingroup$
I was asked to determine whether the integrals $int_{0}^{infty}dyint_{0}^{infty}e^{-xy}sin(x)dx$ and $int_{0}^{infty}dxint_{0}^{infty}e^{-xy}sin(x)dy$ converge, and if they do, calculate them.
I'm not sure what this notation means. Is $int_{0}^{infty}dyint_{0}^{infty}e^{-xy}sin(x)dx$ supposed to mean $int_{0}^{infty}int_{0}^{infty}e^{-xy}sin(x)dxdy$? Or does this literally mean the product of the integrals: $int_{0}^{infty} dycdot int_{0}^{infty}e^{-xy}sin(x)dx$? where the $y$ in the second integral is just some unknown parameter now, not related to the first integral.
Anyone seen this notation before?
calculus integration notation improper-integrals
asked Jan 2 at 9:13
Oria GruberOria Gruber
6,50732360
$endgroup$
$begingroup$
It means the former.
$endgroup$
– Song
Jan 2 at 9:15
add a comment |
$begingroup$
I was asked to determine whether the integrals $int_{0}^{infty}dyint_{0}^{infty}e^{-xy}sin(x)dx$ and $int_{0}^{infty}dxint_{0}^{infty}e^{-xy}sin(x)dy$ converge, and if they do, calculate them.
I'm not sure what this notation means. Is $int_{0}^{infty}dyint_{0}^{infty}e^{-xy}sin(x)dx$ supposed to mean $int_{0}^{infty}int_{0}^{infty}e^{-xy}sin(x)dxdy$? Or does this literally mean the product of the integrals: $int_{0}^{infty} dycdot int_{0}^{infty}e^{-xy}sin(x)dx$? where the $y$ in the second integral is just some unknown parameter now, not related to the first integral.
Anyone seen this notation before?
calculus integration notation improper-integrals
asked Jan 2 at 9:13
Oria GruberOria Gruber
6,50732360
$endgroup$
I was asked to determine whether the integrals $int_{0}^{infty}dyint_{0}^{infty}e^{-xy}sin(x)dx$ and $int_{0}^{infty}dxint_{0}^{infty}e^{-xy}sin(x)dy$ converge, and if they do, calculate them.
I'm not sure what this notation means. Is $int_{0}^{infty}dyint_{0}^{infty}e^{-xy}sin(x)dx$ supposed to mean $int_{0}^{infty}int_{0}^{infty}e^{-xy}sin(x)dxdy$? Or does this literally mean the product of the integrals: $int_{0}^{infty} dycdot int_{0}^{infty}e^{-xy}sin(x)dx$? where the $y$ in the second integral is just some unknown parameter now, not related to the first integral.
Anyone seen this notation before?
calculus integration notation improper-integrals
calculus integration notation improper-integrals
asked Jan 2 at 9:13
Oria GruberOria Gruber
6,50732360
asked Jan 2 at 9:13
Oria GruberOria Gruber
6,50732360
asked Jan 2 at 9:13
Oria GruberOria Gruber
6,50732360
asked Jan 2 at 9:13
Oria GruberOria Gruber
6,50732360
asked Jan 2 at 9:13
Oria GruberOria Gruber
6,50732360
6,50732360
$begingroup$
It means the former.
$endgroup$
– Song
Jan 2 at 9:15
add a comment |
$begingroup$
It means the former.
$endgroup$
– Song
Jan 2 at 9:15
$begingroup$
It means the former.
$endgroup$
– Song
Jan 2 at 9:15
$begingroup$
It means the former.
$endgroup$
– Song
Jan 2 at 9:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$int_{0}^{infty}dyint_{0}^{infty}e^{-xy}sin(x)dx$ means $int_{0}^{infty}left(int_{0}^{infty}e^{-xy}sin(x)dxright)dy$. It is very common to put the $dx$ (or $dy$) at the end of an integral (thus signifying the end of the integrand). While still common, it is less common to do so in repeated integrals.
Probably, the purpose of this is to make it clearer which integral sign (with bounds) belongs to which variable. In this case both $x$ and $y$ go from $0$ to $infty$, but that will not always be so, and then it is important to knwo which variable has which bounds. Sticking the $d$ right next to the $int$ it belongs to makes that easier.
answered Jan 2 at 9:17
ArthurArthur
112k7107191
$endgroup$
$begingroup$
and $int_{0}^{infty}(int_{0}^{infty}e^{-xy}sin(x)dx)dy = lim_{a to infty} int_{0}^{a}(lim_{b to infty}int_{0}^{b}e^{-xy}sin(x)dx)dy$ right? It's the iterated limits, not the multivariable one $lim_{(a,b) to (infty,infty)}$
$endgroup$
– Oria Gruber
Jan 2 at 9:26
$begingroup$
@OriaGruber Exactly. It's one limit inside another, not a single multivariable limit. The single multivariable limit would correspond to $iint_X f(x, y)dA$, where $X$ is the relevant region of the plane (in this case the first quadrant).
$endgroup$
– Arthur
Jan 2 at 9:31
$begingroup$
there's a small problem though: $int_{0}^{infty}e^{-xy}sin(x)dy =lim_{a to infty}frac{sin(x)}{x}-frac{sin(x)}{x}e^{-ax}$ according to my calculation. Without knowing $x$ what can we do?
$endgroup$
– Oria Gruber
Jan 2 at 9:36
1
$begingroup$
@OriaGruber There shouldn't be any $x$ left over from $int_{0}^{infty}e^{-xy}sin(x)dx$. They ought to disappear, and give you a final integral of the form $int_0^infty g(y)dy$ without any $x$ whatsoever.
$endgroup$
– Arthur
Jan 2 at 9:37
$begingroup$
Im doing the $dy$ first because I think it's easier to integrate. Perhaps I'm wrong.
$endgroup$
– Oria Gruber
Jan 2 at 9:38
|
show 11 more comments
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1 Answer
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$begingroup$
$int_{0}^{infty}dyint_{0}^{infty}e^{-xy}sin(x)dx$ means $int_{0}^{infty}left(int_{0}^{infty}e^{-xy}sin(x)dxright)dy$. It is very common to put the $dx$ (or $dy$) at the end of an integral (thus signifying the end of the integrand). While still common, it is less common to do so in repeated integrals.
Probably, the purpose of this is to make it clearer which integral sign (with bounds) belongs to which variable. In this case both $x$ and $y$ go from $0$ to $infty$, but that will not always be so, and then it is important to knwo which variable has which bounds. Sticking the $d$ right next to the $int$ it belongs to makes that easier.
answered Jan 2 at 9:17
ArthurArthur
112k7107191
$endgroup$
$begingroup$
and $int_{0}^{infty}(int_{0}^{infty}e^{-xy}sin(x)dx)dy = lim_{a to infty} int_{0}^{a}(lim_{b to infty}int_{0}^{b}e^{-xy}sin(x)dx)dy$ right? It's the iterated limits, not the multivariable one $lim_{(a,b) to (infty,infty)}$
$endgroup$
– Oria Gruber
Jan 2 at 9:26
$begingroup$
@OriaGruber Exactly. It's one limit inside another, not a single multivariable limit. The single multivariable limit would correspond to $iint_X f(x, y)dA$, where $X$ is the relevant region of the plane (in this case the first quadrant).
$endgroup$
– Arthur
Jan 2 at 9:31
$begingroup$
there's a small problem though: $int_{0}^{infty}e^{-xy}sin(x)dy =lim_{a to infty}frac{sin(x)}{x}-frac{sin(x)}{x}e^{-ax}$ according to my calculation. Without knowing $x$ what can we do?
$endgroup$
– Oria Gruber
Jan 2 at 9:36
1
$begingroup$
@OriaGruber There shouldn't be any $x$ left over from $int_{0}^{infty}e^{-xy}sin(x)dx$. They ought to disappear, and give you a final integral of the form $int_0^infty g(y)dy$ without any $x$ whatsoever.
$endgroup$
– Arthur
Jan 2 at 9:37
$begingroup$
Im doing the $dy$ first because I think it's easier to integrate. Perhaps I'm wrong.
$endgroup$
– Oria Gruber
Jan 2 at 9:38
|
show 11 more comments
$begingroup$
$int_{0}^{infty}dyint_{0}^{infty}e^{-xy}sin(x)dx$ means $int_{0}^{infty}left(int_{0}^{infty}e^{-xy}sin(x)dxright)dy$. It is very common to put the $dx$ (or $dy$) at the end of an integral (thus signifying the end of the integrand). While still common, it is less common to do so in repeated integrals.
Probably, the purpose of this is to make it clearer which integral sign (with bounds) belongs to which variable. In this case both $x$ and $y$ go from $0$ to $infty$, but that will not always be so, and then it is important to knwo which variable has which bounds. Sticking the $d$ right next to the $int$ it belongs to makes that easier.
answered Jan 2 at 9:17
ArthurArthur
112k7107191
$endgroup$
$begingroup$
and $int_{0}^{infty}(int_{0}^{infty}e^{-xy}sin(x)dx)dy = lim_{a to infty} int_{0}^{a}(lim_{b to infty}int_{0}^{b}e^{-xy}sin(x)dx)dy$ right? It's the iterated limits, not the multivariable one $lim_{(a,b) to (infty,infty)}$
$endgroup$
– Oria Gruber
Jan 2 at 9:26
$begingroup$
@OriaGruber Exactly. It's one limit inside another, not a single multivariable limit. The single multivariable limit would correspond to $iint_X f(x, y)dA$, where $X$ is the relevant region of the plane (in this case the first quadrant).
$endgroup$
– Arthur
Jan 2 at 9:31
$begingroup$
there's a small problem though: $int_{0}^{infty}e^{-xy}sin(x)dy =lim_{a to infty}frac{sin(x)}{x}-frac{sin(x)}{x}e^{-ax}$ according to my calculation. Without knowing $x$ what can we do?
$endgroup$
– Oria Gruber
Jan 2 at 9:36
1
$begingroup$
@OriaGruber There shouldn't be any $x$ left over from $int_{0}^{infty}e^{-xy}sin(x)dx$. They ought to disappear, and give you a final integral of the form $int_0^infty g(y)dy$ without any $x$ whatsoever.
$endgroup$
– Arthur
Jan 2 at 9:37
$begingroup$
Im doing the $dy$ first because I think it's easier to integrate. Perhaps I'm wrong.
$endgroup$
– Oria Gruber
Jan 2 at 9:38
|
show 11 more comments
$begingroup$
$int_{0}^{infty}dyint_{0}^{infty}e^{-xy}sin(x)dx$ means $int_{0}^{infty}left(int_{0}^{infty}e^{-xy}sin(x)dxright)dy$. It is very common to put the $dx$ (or $dy$) at the end of an integral (thus signifying the end of the integrand). While still common, it is less common to do so in repeated integrals.
Probably, the purpose of this is to make it clearer which integral sign (with bounds) belongs to which variable. In this case both $x$ and $y$ go from $0$ to $infty$, but that will not always be so, and then it is important to knwo which variable has which bounds. Sticking the $d$ right next to the $int$ it belongs to makes that easier.
answered Jan 2 at 9:17
ArthurArthur
112k7107191
$endgroup$
$int_{0}^{infty}dyint_{0}^{infty}e^{-xy}sin(x)dx$ means $int_{0}^{infty}left(int_{0}^{infty}e^{-xy}sin(x)dxright)dy$. It is very common to put the $dx$ (or $dy$) at the end of an integral (thus signifying the end of the integrand). While still common, it is less common to do so in repeated integrals.
Probably, the purpose of this is to make it clearer which integral sign (with bounds) belongs to which variable. In this case both $x$ and $y$ go from $0$ to $infty$, but that will not always be so, and then it is important to knwo which variable has which bounds. Sticking the $d$ right next to the $int$ it belongs to makes that easier.
answered Jan 2 at 9:17
ArthurArthur
112k7107191
edited Jan 2 at 9:23
answered Jan 2 at 9:17
ArthurArthur
112k7107191
answered Jan 2 at 9:17
ArthurArthur
112k7107191
answered Jan 2 at 9:17
ArthurArthur
112k7107191
112k7107191
$begingroup$
and $int_{0}^{infty}(int_{0}^{infty}e^{-xy}sin(x)dx)dy = lim_{a to infty} int_{0}^{a}(lim_{b to infty}int_{0}^{b}e^{-xy}sin(x)dx)dy$ right? It's the iterated limits, not the multivariable one $lim_{(a,b) to (infty,infty)}$
$endgroup$
– Oria Gruber
Jan 2 at 9:26
$begingroup$
@OriaGruber Exactly. It's one limit inside another, not a single multivariable limit. The single multivariable limit would correspond to $iint_X f(x, y)dA$, where $X$ is the relevant region of the plane (in this case the first quadrant).
$endgroup$
– Arthur
Jan 2 at 9:31
$begingroup$
there's a small problem though: $int_{0}^{infty}e^{-xy}sin(x)dy =lim_{a to infty}frac{sin(x)}{x}-frac{sin(x)}{x}e^{-ax}$ according to my calculation. Without knowing $x$ what can we do?
$endgroup$
– Oria Gruber
Jan 2 at 9:36
1
$begingroup$
@OriaGruber There shouldn't be any $x$ left over from $int_{0}^{infty}e^{-xy}sin(x)dx$. They ought to disappear, and give you a final integral of the form $int_0^infty g(y)dy$ without any $x$ whatsoever.
$endgroup$
– Arthur
Jan 2 at 9:37
$begingroup$
Im doing the $dy$ first because I think it's easier to integrate. Perhaps I'm wrong.
$endgroup$
– Oria Gruber
Jan 2 at 9:38
|
show 11 more comments
$begingroup$
and $int_{0}^{infty}(int_{0}^{infty}e^{-xy}sin(x)dx)dy = lim_{a to infty} int_{0}^{a}(lim_{b to infty}int_{0}^{b}e^{-xy}sin(x)dx)dy$ right? It's the iterated limits, not the multivariable one $lim_{(a,b) to (infty,infty)}$
$endgroup$
– Oria Gruber
Jan 2 at 9:26
$begingroup$
@OriaGruber Exactly. It's one limit inside another, not a single multivariable limit. The single multivariable limit would correspond to $iint_X f(x, y)dA$, where $X$ is the relevant region of the plane (in this case the first quadrant).
$endgroup$
– Arthur
Jan 2 at 9:31
$begingroup$
there's a small problem though: $int_{0}^{infty}e^{-xy}sin(x)dy =lim_{a to infty}frac{sin(x)}{x}-frac{sin(x)}{x}e^{-ax}$ according to my calculation. Without knowing $x$ what can we do?
$endgroup$
– Oria Gruber
Jan 2 at 9:36
1
$begingroup$
@OriaGruber There shouldn't be any $x$ left over from $int_{0}^{infty}e^{-xy}sin(x)dx$. They ought to disappear, and give you a final integral of the form $int_0^infty g(y)dy$ without any $x$ whatsoever.
$endgroup$
– Arthur
Jan 2 at 9:37
$begingroup$
Im doing the $dy$ first because I think it's easier to integrate. Perhaps I'm wrong.
$endgroup$
– Oria Gruber
Jan 2 at 9:38
$begingroup$
and $int_{0}^{infty}(int_{0}^{infty}e^{-xy}sin(x)dx)dy = lim_{a to infty} int_{0}^{a}(lim_{b to infty}int_{0}^{b}e^{-xy}sin(x)dx)dy$ right? It's the iterated limits, not the multivariable one $lim_{(a,b) to (infty,infty)}$
$endgroup$
– Oria Gruber
Jan 2 at 9:26
$begingroup$
and $int_{0}^{infty}(int_{0}^{infty}e^{-xy}sin(x)dx)dy = lim_{a to infty} int_{0}^{a}(lim_{b to infty}int_{0}^{b}e^{-xy}sin(x)dx)dy$ right? It's the iterated limits, not the multivariable one $lim_{(a,b) to (infty,infty)}$
$endgroup$
– Oria Gruber
Jan 2 at 9:26
$begingroup$
@OriaGruber Exactly. It's one limit inside another, not a single multivariable limit. The single multivariable limit would correspond to $iint_X f(x, y)dA$, where $X$ is the relevant region of the plane (in this case the first quadrant).
$endgroup$
– Arthur
Jan 2 at 9:31
$begingroup$
@OriaGruber Exactly. It's one limit inside another, not a single multivariable limit. The single multivariable limit would correspond to $iint_X f(x, y)dA$, where $X$ is the relevant region of the plane (in this case the first quadrant).
$endgroup$
– Arthur
Jan 2 at 9:31
$begingroup$
there's a small problem though: $int_{0}^{infty}e^{-xy}sin(x)dy =lim_{a to infty}frac{sin(x)}{x}-frac{sin(x)}{x}e^{-ax}$ according to my calculation. Without knowing $x$ what can we do?
$endgroup$
– Oria Gruber
Jan 2 at 9:36
$begingroup$
there's a small problem though: $int_{0}^{infty}e^{-xy}sin(x)dy =lim_{a to infty}frac{sin(x)}{x}-frac{sin(x)}{x}e^{-ax}$ according to my calculation. Without knowing $x$ what can we do?
$endgroup$
– Oria Gruber
Jan 2 at 9:36
1
1
$begingroup$
@OriaGruber There shouldn't be any $x$ left over from $int_{0}^{infty}e^{-xy}sin(x)dx$. They ought to disappear, and give you a final integral of the form $int_0^infty g(y)dy$ without any $x$ whatsoever.
$endgroup$
– Arthur
Jan 2 at 9:37
$begingroup$
@OriaGruber There shouldn't be any $x$ left over from $int_{0}^{infty}e^{-xy}sin(x)dx$. They ought to disappear, and give you a final integral of the form $int_0^infty g(y)dy$ without any $x$ whatsoever.
$endgroup$
– Arthur
Jan 2 at 9:37
$begingroup$
Im doing the $dy$ first because I think it's easier to integrate. Perhaps I'm wrong.
$endgroup$
– Oria Gruber
Jan 2 at 9:38
$begingroup$
Im doing the $dy$ first because I think it's easier to integrate. Perhaps I'm wrong.
$endgroup$
– Oria Gruber
Jan 2 at 9:38
|
show 11 more comments
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$begingroup$
It means the former.
$endgroup$
– Song
Jan 2 at 9:15