- #1

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[tex]

\int_0^{+\infty}{\frac{e^{-x}}{\int_{\Delta T}f(x,t)}f(x,t)xdx}=Ag(t)

[/tex]

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?

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- Thread starter matteo86bo
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- #1

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[tex]

\int_0^{+\infty}{\frac{e^{-x}}{\int_{\Delta T}f(x,t)}f(x,t)xdx}=Ag(t)

[/tex]

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

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First, what does [tex]\int_{\Delta T}f(x,t)[/tex] 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)?

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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- #3

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[tex]\int_{\Delta T}f(x,t)=\int_0^{\Delta T}f(x,t)dt[/tex]First, what does [tex]\int_{\Delta T}f(x,t)[/tex] mean?

sorry, I skip the "dt"

yes, for all positive x.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)?

- #4

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What is ΔT?

- #5

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What is ΔT?

a positive real number, sorry

- #6

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- #7

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

furthermore,if [tex]f(x,t)=g(t)+h(x)[/tex]

is now possibile to prove that h(x)=0 for all x??

- #8

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- #9

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of course it does.The process of integration gets rid of the x.

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

HallsofIvy

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[tex]\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)[/tex].

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.

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