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Can anyone provide a (non-geometrical) proof of the following change of variable trick? $$ \int_a^b\int_a^t f(s)g(t)\,ds\,dt=\int_a^b\int_s^b f(s)g(t)\,dt\,ds$$

For $f$ and $g$ complex-valued functions of the real variable sufficiently integrable for these expressions to make sense.

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2 Answers 2

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Let $k(s,t) = \begin{cases} 1 & s \leq t \\ 0 & s>t \end{cases}$. Then you can write the integral as $\int_I \int_I k(s,t) f(s)g(t) ds dt$. Then, under suitable integrability conditions, Fubini's theorem gives,

$$\int_I ( \int_I k(s,t) f(s)g(t) ds )dt = \int_{I\times I} k(s,t) f(s)g(t) d(s \times t) = \int_I ( \int_I k(s,t) f(s)g(t) dt )ds$$

Since $\int_I ( \int_I k(s,t) f(s)g(t) ds )dt = \int_a^b\int_a^t f(s)g(t)\,ds\,dt$ and $\int_I ( \int_I k(s,t) f(s)g(t) dt )ds = \int_a^b\int_s^b f(s)g(t)\,dt\,ds$, we have the desired result.

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I'm assuming you are comfortable with the switching of integrals on its own, which needs some theorem I can't recall. For the bounds, note that the bounds of integration in the first case are $a<t<b$ and $a<s<t$, so $a<s<t<b$ and extracting independent bounds for $s$ we get $a<s<b$, and for $t$ we have $s<t<b$ (just rearranging that inequality chain). Thus

$$\int_a^b\int_a^t f(s)g(t)\,ds\,dt=\int_a^b\int_s^b f(s)g(t)\,dt\,ds.$$

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