# Prove that a function must have only x dependency

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?

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)?

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

What is ΔT?

What is ΔT?

a positive real number, sorry

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.

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.

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??

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.

The process of integration gets rid of the x.
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?

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