- #1
carlosbgois
- 68
- 0
Hi there. I think I have proved on little theorem on double integrals, showed below. Are my arguments 'correct' (I mean, rigorous enough)?
Let [itex]f[/itex] be a function of [itex]x, f(x)[/itex], [itex]g[/itex] be a function depending only on [itex]y, g(y)[/itex], and last, let [itex]A[/itex] be the set determined by [itex]a≤x≤b[/itex] and [itex]c≤y≤d[/itex]. By Fubini's theorem, [itex]\int\int_{A}f(x)g(y)dxdy=\int^{d}_{c}[\int^{b}_{a}f(x)g(y)dx]dy.[/itex] Knowing [itex]g[/itex] is a function only dependent on [itex]y[/itex], it is a constant when integrating over x, hence we have [itex]\int^{d}_{c}[g(y)\int^{b}_{a}f(x)dx]dy.[/itex] Still, as [itex]f[/itex] depends only on [itex]x[/itex], so does any integral over [itex]f[/itex], then finally [itex]\int^{d}_{c}g(y)dy \int^{b}_{a}f(x)dx.[/itex]
Many thanks.
Let [itex]f[/itex] be a function of [itex]x, f(x)[/itex], [itex]g[/itex] be a function depending only on [itex]y, g(y)[/itex], and last, let [itex]A[/itex] be the set determined by [itex]a≤x≤b[/itex] and [itex]c≤y≤d[/itex]. By Fubini's theorem, [itex]\int\int_{A}f(x)g(y)dxdy=\int^{d}_{c}[\int^{b}_{a}f(x)g(y)dx]dy.[/itex] Knowing [itex]g[/itex] is a function only dependent on [itex]y[/itex], it is a constant when integrating over x, hence we have [itex]\int^{d}_{c}[g(y)\int^{b}_{a}f(x)dx]dy.[/itex] Still, as [itex]f[/itex] depends only on [itex]x[/itex], so does any integral over [itex]f[/itex], then finally [itex]\int^{d}_{c}g(y)dy \int^{b}_{a}f(x)dx.[/itex]
Many thanks.