- #1
- 1,995
- 7
Physicists do it all the time: interchanging limits, assuming uniform convergence, differentiation a delta function..
Usually it's all valid, but I'd like to see when we can interchange differentiation with respect to one variabele with integration over another.
It seems the following theorem exists:
[tex]\frac{d}{dy}\int_{x_1(y)}^{x_2(y)}f(x,y)dx=f(x_2,y)\frac{dx_2}{dy}-f(x_1,y)\frac{dx_1}{dy}+\int_{x_1(y)}^{x_2(y)}\frac{\partial f}{\partial y}dx[/tex]
So when the boundaries do not depend on [tex]y[/tex] we may simply bring the derivative under the integral sign.
I couldn't find a proof of this so I set out to prove it myself.
I'd like to know if I made any mistakes.
Let [tex]G(x,y)=\int^xf(x',y)dx'[/tex].
So that [tex]\frac{\partial G}{\partial x}=f(x,y)[/tex]
Using this we have:
(1) [tex]\int_{x_1(y)}^{x_2(y)}\frac{\partial f}{\partial y}dx= \int_{x_1(y)}^{x_2(y)}\frac{\partial^2 G}{\partial y \partial x}dx= \int_{x_1(y)}^{x_2(y)}\frac{\partial^2 G}{\partial x \partial y}dx=[/tex]
[tex]\int_{x_1(y)}^{x_2(y)}\frac{\partial}{\partial x}\left(\frac{\partial G}{\partial y}\right)dx= \frac{\partial G}{\partial y}(x_2(y),y)-\frac{\partial G}{\partial y}(x_1(y),y)=[/tex]
[tex] \frac{\partial}{\partial y}\left(\int^{x_2(y)}f(x,y)dx-\int^{x_1(y)}f(x,y)dx\right)=\frac{\partial}{\partial y}\int_{x_1(y)}^{x_2(y)}f(x,y)dx[/tex]
Which is pretty close. Actually, I realized I need total differentiation with respect to y, not partial differentiation. So I can use:
[tex]\frac{d}{dy}(G(x_2(y),y)-G(x_1(y),y))=\frac{\partial G}{\partial x_2}\frac{dx_2}{dy}+\frac{\partial G}{\partial y}(x_2,y)-\frac{\partial G}{\partial x_1}\frac{dx_1}{dy}-\frac{\partial G}{\partial y}(x_1,y)[/tex]
Which is equal to
[tex]\left(\frac{\partial}{\partial x_2}\int^{x_2(y)}f(x,y)dx\right)\frac{dx_2}{dy}-\left(\frac{\partial}{\partial x_1}\int^{x_1(y)}f(x,y)dx\right)\frac{dx_1}{dy}+[/tex]
[tex]\frac{\partial}{\partial y}\int_{x_1(y)}^{x_2(y)}f(x,y)dx[/tex]
Which is the right answer if
[tex]\frac{\partial}{\partial x_2}\int^{x_2(y)}f(x,y)dx=f(x_2(y),y)[/tex]
But I`m not sure if this is true. I know that
[tex]\frac{d}{dx}\int^xf(x)dx=f(x)[/tex], but I`m not sure if I can use that here. Furthermore, we need:
[tex]\int_{x_1(y)}^{x_2(y)}\frac{\partial}{\partial x}\left(\frac{\partial G}{\partial y}\right)dx= \frac{d}{dy}(G(x_2(y),y)-G(x_1(y),y))[/tex]
instead of partial differentiation.
Can anyone give me pointers please?
Usually it's all valid, but I'd like to see when we can interchange differentiation with respect to one variabele with integration over another.
It seems the following theorem exists:
[tex]\frac{d}{dy}\int_{x_1(y)}^{x_2(y)}f(x,y)dx=f(x_2,y)\frac{dx_2}{dy}-f(x_1,y)\frac{dx_1}{dy}+\int_{x_1(y)}^{x_2(y)}\frac{\partial f}{\partial y}dx[/tex]
So when the boundaries do not depend on [tex]y[/tex] we may simply bring the derivative under the integral sign.
I couldn't find a proof of this so I set out to prove it myself.
I'd like to know if I made any mistakes.
Let [tex]G(x,y)=\int^xf(x',y)dx'[/tex].
So that [tex]\frac{\partial G}{\partial x}=f(x,y)[/tex]
Using this we have:
(1) [tex]\int_{x_1(y)}^{x_2(y)}\frac{\partial f}{\partial y}dx= \int_{x_1(y)}^{x_2(y)}\frac{\partial^2 G}{\partial y \partial x}dx= \int_{x_1(y)}^{x_2(y)}\frac{\partial^2 G}{\partial x \partial y}dx=[/tex]
[tex]\int_{x_1(y)}^{x_2(y)}\frac{\partial}{\partial x}\left(\frac{\partial G}{\partial y}\right)dx= \frac{\partial G}{\partial y}(x_2(y),y)-\frac{\partial G}{\partial y}(x_1(y),y)=[/tex]
[tex] \frac{\partial}{\partial y}\left(\int^{x_2(y)}f(x,y)dx-\int^{x_1(y)}f(x,y)dx\right)=\frac{\partial}{\partial y}\int_{x_1(y)}^{x_2(y)}f(x,y)dx[/tex]
Which is pretty close. Actually, I realized I need total differentiation with respect to y, not partial differentiation. So I can use:
[tex]\frac{d}{dy}(G(x_2(y),y)-G(x_1(y),y))=\frac{\partial G}{\partial x_2}\frac{dx_2}{dy}+\frac{\partial G}{\partial y}(x_2,y)-\frac{\partial G}{\partial x_1}\frac{dx_1}{dy}-\frac{\partial G}{\partial y}(x_1,y)[/tex]
Which is equal to
[tex]\left(\frac{\partial}{\partial x_2}\int^{x_2(y)}f(x,y)dx\right)\frac{dx_2}{dy}-\left(\frac{\partial}{\partial x_1}\int^{x_1(y)}f(x,y)dx\right)\frac{dx_1}{dy}+[/tex]
[tex]\frac{\partial}{\partial y}\int_{x_1(y)}^{x_2(y)}f(x,y)dx[/tex]
Which is the right answer if
[tex]\frac{\partial}{\partial x_2}\int^{x_2(y)}f(x,y)dx=f(x_2(y),y)[/tex]
But I`m not sure if this is true. I know that
[tex]\frac{d}{dx}\int^xf(x)dx=f(x)[/tex], but I`m not sure if I can use that here. Furthermore, we need:
[tex]\int_{x_1(y)}^{x_2(y)}\frac{\partial}{\partial x}\left(\frac{\partial G}{\partial y}\right)dx= \frac{d}{dy}(G(x_2(y),y)-G(x_1(y),y))[/tex]
instead of partial differentiation.
Can anyone give me pointers please?