What is the sufficient condition of integration by parts?
$begingroup$
From the product rule of differentiation, we can derive $duv = udv + vdu$. By integrating the both side, we can get $uv = int{udv} + int{vdu}$.
What is the sufficient condition that allows this equation, $uv = int{udv} + int{vdu}$?
I was confused by the following two equations.
If $u=x$, $v=y$, then $xy = int{xdy} + int{ydx}$. This is wrong.
If $u=costheta$, $v=sintheta$, then $costhetasintheta = int{cos{theta}dsintheta} + int{sintheta d costheta}$. This is correct.
Because both $left(x,yright)$ and $left(costheta,sinthetaright)$ are linear independent, it seems that the linear independence is not the sufficient condition of the integration by parts. Did I right? Or I was confused vectors with functions?
integration
asked Jan 10 at 18:50
Iven CJ7Iven CJ7
52
$endgroup$
add a comment |
$begingroup$
From the product rule of differentiation, we can derive $duv = udv + vdu$. By integrating the both side, we can get $uv = int{udv} + int{vdu}$.
What is the sufficient condition that allows this equation, $uv = int{udv} + int{vdu}$?
I was confused by the following two equations.
If $u=x$, $v=y$, then $xy = int{xdy} + int{ydx}$. This is wrong.
If $u=costheta$, $v=sintheta$, then $costhetasintheta = int{cos{theta}dsintheta} + int{sintheta d costheta}$. This is correct.
Because both $left(x,yright)$ and $left(costheta,sinthetaright)$ are linear independent, it seems that the linear independence is not the sufficient condition of the integration by parts. Did I right? Or I was confused vectors with functions?
integration
asked Jan 10 at 18:50
Iven CJ7Iven CJ7
52
$endgroup$
$begingroup$
What do you think that $int u,dv$ means?
$endgroup$
– Christian Blatter
Jan 10 at 19:20
$begingroup$
I think it means the sum of all $udv$ for all $u$.
$endgroup$
– Iven CJ7
Jan 11 at 2:13
add a comment |
$begingroup$
From the product rule of differentiation, we can derive $duv = udv + vdu$. By integrating the both side, we can get $uv = int{udv} + int{vdu}$.
What is the sufficient condition that allows this equation, $uv = int{udv} + int{vdu}$?
I was confused by the following two equations.
If $u=x$, $v=y$, then $xy = int{xdy} + int{ydx}$. This is wrong.
If $u=costheta$, $v=sintheta$, then $costhetasintheta = int{cos{theta}dsintheta} + int{sintheta d costheta}$. This is correct.
Because both $left(x,yright)$ and $left(costheta,sinthetaright)$ are linear independent, it seems that the linear independence is not the sufficient condition of the integration by parts. Did I right? Or I was confused vectors with functions?
integration
asked Jan 10 at 18:50
Iven CJ7Iven CJ7
52
$endgroup$
From the product rule of differentiation, we can derive $duv = udv + vdu$. By integrating the both side, we can get $uv = int{udv} + int{vdu}$.
What is the sufficient condition that allows this equation, $uv = int{udv} + int{vdu}$?
I was confused by the following two equations.
If $u=x$, $v=y$, then $xy = int{xdy} + int{ydx}$. This is wrong.
If $u=costheta$, $v=sintheta$, then $costhetasintheta = int{cos{theta}dsintheta} + int{sintheta d costheta}$. This is correct.
Because both $left(x,yright)$ and $left(costheta,sinthetaright)$ are linear independent, it seems that the linear independence is not the sufficient condition of the integration by parts. Did I right? Or I was confused vectors with functions?
integration
integration
asked Jan 10 at 18:50
Iven CJ7Iven CJ7
52
asked Jan 10 at 18:50
Iven CJ7Iven CJ7
52
asked Jan 10 at 18:50
Iven CJ7Iven CJ7
52
asked Jan 10 at 18:50
Iven CJ7Iven CJ7
52
asked Jan 10 at 18:50
Iven CJ7Iven CJ7
52
52
$begingroup$
What do you think that $int u,dv$ means?
$endgroup$
– Christian Blatter
Jan 10 at 19:20
$begingroup$
I think it means the sum of all $udv$ for all $u$.
$endgroup$
– Iven CJ7
Jan 11 at 2:13
add a comment |
$begingroup$
What do you think that $int u,dv$ means?
$endgroup$
– Christian Blatter
Jan 10 at 19:20
$begingroup$
I think it means the sum of all $udv$ for all $u$.
$endgroup$
– Iven CJ7
Jan 11 at 2:13
$begingroup$
What do you think that $int u,dv$ means?
$endgroup$
– Christian Blatter
Jan 10 at 19:20
$begingroup$
What do you think that $int u,dv$ means?
$endgroup$
– Christian Blatter
Jan 10 at 19:20
$begingroup$
I think it means the sum of all $udv$ for all $u$.
$endgroup$
– Iven CJ7
Jan 11 at 2:13
$begingroup$
I think it means the sum of all $udv$ for all $u$.
$endgroup$
– Iven CJ7
Jan 11 at 2:13
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In your first example ($u=x$ and $v=y$), the two functions are functions of different variables. But in your second example ($u=costheta$ and $v=sintheta$), although they are independent, they are both functions of $theta$, and so they are related in that way. In your first example, if it was instead $u=x$ and $v=y(x)$, then integration by parts would work.
answered Jan 10 at 18:53
Calvin GodfreyCalvin Godfrey
633311
$endgroup$
$begingroup$
Although $x$ and $y$ are not related, but $dxy = xdy+ydx$ is still meaningful. Is it? $dxy = xdy+ydx$ is meaningful but $int dxy = int xdy+ int ydx$ is not, so that $u$ and $v$ is related is the sufficient condition allows integrating both sides of $duv = udv + vdu$?
$endgroup$
– Iven CJ7
Jan 11 at 8:58
add a comment |
$begingroup$
Take $u=f(x), v=g(x)$
What IBP states is that:
$$f(x)g(x)=int{f(x)g'(x)+g(x)f'(x)space dx}$$
Differentiating, we get:
$$[f(x)g(x)]'=f(x)g'(x)+g(x)f'(x)$$
Which is just the product rule.
So as long as the two variables used are functions of the same variable, this holds.
answered Jan 10 at 18:57
Rhys HughesRhys Hughes
6,9571530
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
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active
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$begingroup$
In your first example ($u=x$ and $v=y$), the two functions are functions of different variables. But in your second example ($u=costheta$ and $v=sintheta$), although they are independent, they are both functions of $theta$, and so they are related in that way. In your first example, if it was instead $u=x$ and $v=y(x)$, then integration by parts would work.
answered Jan 10 at 18:53
Calvin GodfreyCalvin Godfrey
633311
$endgroup$
$begingroup$
Although $x$ and $y$ are not related, but $dxy = xdy+ydx$ is still meaningful. Is it? $dxy = xdy+ydx$ is meaningful but $int dxy = int xdy+ int ydx$ is not, so that $u$ and $v$ is related is the sufficient condition allows integrating both sides of $duv = udv + vdu$?
$endgroup$
– Iven CJ7
Jan 11 at 8:58
add a comment |
$begingroup$
In your first example ($u=x$ and $v=y$), the two functions are functions of different variables. But in your second example ($u=costheta$ and $v=sintheta$), although they are independent, they are both functions of $theta$, and so they are related in that way. In your first example, if it was instead $u=x$ and $v=y(x)$, then integration by parts would work.
answered Jan 10 at 18:53
Calvin GodfreyCalvin Godfrey
633311
$endgroup$
$begingroup$
Although $x$ and $y$ are not related, but $dxy = xdy+ydx$ is still meaningful. Is it? $dxy = xdy+ydx$ is meaningful but $int dxy = int xdy+ int ydx$ is not, so that $u$ and $v$ is related is the sufficient condition allows integrating both sides of $duv = udv + vdu$?
$endgroup$
– Iven CJ7
Jan 11 at 8:58
add a comment |
$begingroup$
In your first example ($u=x$ and $v=y$), the two functions are functions of different variables. But in your second example ($u=costheta$ and $v=sintheta$), although they are independent, they are both functions of $theta$, and so they are related in that way. In your first example, if it was instead $u=x$ and $v=y(x)$, then integration by parts would work.
answered Jan 10 at 18:53
Calvin GodfreyCalvin Godfrey
633311
$endgroup$
In your first example ($u=x$ and $v=y$), the two functions are functions of different variables. But in your second example ($u=costheta$ and $v=sintheta$), although they are independent, they are both functions of $theta$, and so they are related in that way. In your first example, if it was instead $u=x$ and $v=y(x)$, then integration by parts would work.
answered Jan 10 at 18:53
Calvin GodfreyCalvin Godfrey
633311
answered Jan 10 at 18:53
Calvin GodfreyCalvin Godfrey
633311
answered Jan 10 at 18:53
Calvin GodfreyCalvin Godfrey
633311
answered Jan 10 at 18:53
Calvin GodfreyCalvin Godfrey
633311
633311
$begingroup$
Although $x$ and $y$ are not related, but $dxy = xdy+ydx$ is still meaningful. Is it? $dxy = xdy+ydx$ is meaningful but $int dxy = int xdy+ int ydx$ is not, so that $u$ and $v$ is related is the sufficient condition allows integrating both sides of $duv = udv + vdu$?
$endgroup$
– Iven CJ7
Jan 11 at 8:58
add a comment |
$begingroup$
Although $x$ and $y$ are not related, but $dxy = xdy+ydx$ is still meaningful. Is it? $dxy = xdy+ydx$ is meaningful but $int dxy = int xdy+ int ydx$ is not, so that $u$ and $v$ is related is the sufficient condition allows integrating both sides of $duv = udv + vdu$?
$endgroup$
– Iven CJ7
Jan 11 at 8:58
$begingroup$
Although $x$ and $y$ are not related, but $dxy = xdy+ydx$ is still meaningful. Is it? $dxy = xdy+ydx$ is meaningful but $int dxy = int xdy+ int ydx$ is not, so that $u$ and $v$ is related is the sufficient condition allows integrating both sides of $duv = udv + vdu$?
$endgroup$
– Iven CJ7
Jan 11 at 8:58
$begingroup$
Although $x$ and $y$ are not related, but $dxy = xdy+ydx$ is still meaningful. Is it? $dxy = xdy+ydx$ is meaningful but $int dxy = int xdy+ int ydx$ is not, so that $u$ and $v$ is related is the sufficient condition allows integrating both sides of $duv = udv + vdu$?
$endgroup$
– Iven CJ7
Jan 11 at 8:58
add a comment |
$begingroup$
Take $u=f(x), v=g(x)$
What IBP states is that:
$$f(x)g(x)=int{f(x)g'(x)+g(x)f'(x)space dx}$$
Differentiating, we get:
$$[f(x)g(x)]'=f(x)g'(x)+g(x)f'(x)$$
Which is just the product rule.
So as long as the two variables used are functions of the same variable, this holds.
answered Jan 10 at 18:57
Rhys HughesRhys Hughes
6,9571530
$endgroup$
add a comment |
$begingroup$
Take $u=f(x), v=g(x)$
What IBP states is that:
$$f(x)g(x)=int{f(x)g'(x)+g(x)f'(x)space dx}$$
Differentiating, we get:
$$[f(x)g(x)]'=f(x)g'(x)+g(x)f'(x)$$
Which is just the product rule.
So as long as the two variables used are functions of the same variable, this holds.
answered Jan 10 at 18:57
Rhys HughesRhys Hughes
6,9571530
$endgroup$
add a comment |
$begingroup$
Take $u=f(x), v=g(x)$
What IBP states is that:
$$f(x)g(x)=int{f(x)g'(x)+g(x)f'(x)space dx}$$
Differentiating, we get:
$$[f(x)g(x)]'=f(x)g'(x)+g(x)f'(x)$$
Which is just the product rule.
So as long as the two variables used are functions of the same variable, this holds.
answered Jan 10 at 18:57
Rhys HughesRhys Hughes
6,9571530
$endgroup$
Take $u=f(x), v=g(x)$
What IBP states is that:
$$f(x)g(x)=int{f(x)g'(x)+g(x)f'(x)space dx}$$
Differentiating, we get:
$$[f(x)g(x)]'=f(x)g'(x)+g(x)f'(x)$$
Which is just the product rule.
So as long as the two variables used are functions of the same variable, this holds.
answered Jan 10 at 18:57
Rhys HughesRhys Hughes
6,9571530
answered Jan 10 at 18:57
Rhys HughesRhys Hughes
6,9571530
answered Jan 10 at 18:57
Rhys HughesRhys Hughes
6,9571530
answered Jan 10 at 18:57
Rhys HughesRhys Hughes
6,9571530
6,9571530
add a comment |
add a comment |
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$begingroup$
What do you think that $int u,dv$ means?
$endgroup$
– Christian Blatter
Jan 10 at 19:20
$begingroup$
I think it means the sum of all $udv$ for all $u$.
$endgroup$
– Iven CJ7
Jan 11 at 2:13