- #1

couscous1010

- 1

- 0

[tex]

f(x,y)=\int_{0}^{1}d t\frac{d}{d t}f( t x,y).

[/tex]

At least I think this is true because I can evaluate the integral on the r.h.s. as

[tex]

\int_{0}^{1}d t\frac{d}{d t}f( t x,y)=f( t x,y)|_{ t=0}^{ t=1}=f(x,y)-f(0,y)=f(x,y).

[/tex]

My problem is that I have a suspicion that [itex]f(x,y)[/itex] can also be written in the form

[tex]

f(x,y)=x\int_{0}^{1}d t\frac{\partial}{\partial x}f( t x,y),

[/tex]

but I can't quite see how one does this. For instance, an obvious approach is to try

[tex]

f(x,y) = \int_{0}^{1}d t\frac{d}{d t}f( t x,y)[/tex]

[tex]= \int_{0}^{1}d t\left[\frac{\partial( t x)}{\partial t}\frac{d}{d( t x)}f( t x,y)+\frac{\partial y}{\partial t}\frac{\partial}{\partial y}f( t x,y)\right][/tex]

[tex] = x\int_{0}^{1}d t\frac{d}{d( t x)}f( t x,y),[/tex]

but this isn't quite what I want because the derivative of [itex]f[/itex] inside the integral is with respect to [itex]tx[/itex], and

*not*with respect to [itex]x[/itex], which is what I really want.

I'm probably missing something blindingly obvious. Any help with this (even a simple hint or two) would be cool.