# Pretty simple question (I hope)

Hey guys, neat forum! I've got a quick calculus question that I'd love to get some help on. Suppose that I have some function $f$ of two variables $x$ and $y$. I can write this function as $f=f(x,y)$. Now suppose that this function satisfies $f(x,y)|_{x=0}=f(0,y)=0$. This means that I can write the function as

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

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

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

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

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

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

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

but this isn't quite what I want because the derivative of $f$ inside the integral is with respect to $tx$, and not with respect to $x$, 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.

Gib Z
Homework Helper
Hello couscous! Welcome to Physicsforums!

I hope we have more posters who set our their posts and ideas as clearly as you do, keep it up! For your first result, yes you are correct, you evaluating the integral correctly.

Your suspicion is also correct, though I'm not sure if how I did it is the easiest method. Basically I just recognized that in this problem, y can be treated as a constant and somewhat be ignored, so I'd write f(x) for f(x,y) here instead ; then I use the differentiation under the integral sign rule in reverse to take that partial derivative out. An anti derivative should not be too hard to find here and then your almost done ! Good luck!

I'd say your suspicion is wrong. Put $$u=tx$$ and use the chain rule

$$\frac{d}{d t}f( u,y) = \frac{df}{du}\frac{du}{dt}=x\frac{d}{d(xt)}f(xt,y)$$