# Homework Help: Prove that a function must have only x dependency

1. Jun 12, 2010

### matteo86bo

If I have this equation:
$$\int_0^{+\infty}{\frac{e^{-x}}{\int_{\Delta T}f(x,t)}f(x,t)xdx}=Ag(t)$$

I am interested to show that f(x,t) must be g(t). This condition must be necessary and sufficient. Is this possible to prove?

2. Jun 12, 2010

### pellman

First, what does $$\int_{\Delta T}f(x,t)$$ mean?

Second, is the problem to show that the only way for this statement to be true is for f(x,t) to be equal to g(t) (for all x)?

Last edited: Jun 12, 2010
3. Jun 13, 2010

### matteo86bo

$$\int_{\Delta T}f(x,t)=\int_0^{\Delta T}f(x,t)dt$$
sorry, I skip the "dt"
yes, for all positive x.

4. Jun 13, 2010

### Tedjn

What is ΔT?

5. Jun 13, 2010

### matteo86bo

a positive real number, sorry

6. Jun 13, 2010

### Tedjn

Once you do the outer integration, all instances of x disappear, so you are left with a function of t. Of course, it's completely possible this may equal ±∞ for some or all values of t.

7. Jun 13, 2010

### matteo86bo

if the integrandus if choose to be positive for all x and t, can't you say nothing?

furthermore,if $$f(x,t)=g(t)+h(x)$$
is now possibile to prove that h(x)=0 for all x??

8. Jun 13, 2010

### Tedjn

I don't understand your line of questioning. What does f(x,t) being positive have to do with anything? Moreover, you can not say that f(x,t) = g(t) + h(x) without further constraints on f(x,t); who knows how mixed up x and t are. The process of integration gets rid of the x.

9. Jun 13, 2010

### matteo86bo

of course it does.
sorry, I wasn't as clear as possibile.
If now I fix the x dependency by means of f(x,t)=g(t)+h(x), it can be proved that h(x)=0 or I'm fancy about it?

10. Jun 13, 2010

### HallsofIvy

Why?

$$\int_a^b f(t)dt= \int_a^b f(u)du= \int_a^b f(x)dx= \int_a^b f(y)dy= F(b)- F(a)$$.

Whatever variable the integral is "with respect to" is a "dummy" variable and does not appear in the result. "t" is the only "true" variable in that equation.