Fubini's theorem
Proof

Let V (x) {\displaystyle V(x)} and W (x) {\displaystyle W(x)} are primitive functions of the functions v (x) {\displaystyle v(x)} and w (x) {\displaystyle w(x)} respectively, which pass through the origin:

Therefore, we have

= V (u) W (u) {\displaystyle {\biggl }{\biggl }=V(u)W(u)}

By the product rule, the derivative of the right-hand side is

d d x = V (x) w (x) + v (x) W (x) {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}{\bigl }=V(x)w(x)+v(x)W(x)}

and by integrating we have:

∫ 0 u V (x) w (x) + v (x) W (x) d x = V (u) W (u) {\displaystyle \int _{0}^{u}V(x)w(x)+v(x)W(x)\,\mathrm {d} x=V(u)W(u)}

Thus, the equation from the beginning we get:

= ∫ 0 u V (x) w (x) + v (x) W (x) d x {\displaystyle {\biggl }{\biggl }=\int _{0}^{u}V(x)w(x)+v(x)W(x)\,\mathrm {d} x}

Now, we introduce a second integration parameter y {\displaystyle y} for the description of the antiderivatives V (x) {\displaystyle V(x)} and W (x) {\displaystyle W(x)}:

By insertion, a double integral appears:

= ∫ 0 u w (x) + v (x) d x {\displaystyle {\biggl }{\biggl }=\int _{0}^{u}{\biggl }w(x)+v(x){\biggl }\,\mathrm {d} x}

Functions that are foreign to the concerned integration parameter can be imported into the inner function as a factor:

= ∫ 0 u + d x {\displaystyle {\biggl }{\biggl }=\int _{0}^{u}{\biggl }+{\biggl }\,\mathrm {d} x}

In the next step, the sum rule is applied to the integrals:

= ∫ 0 u ∫ 0 1 x v (x y) w (x) + x v (x) w (x y) d y d x {\displaystyle {\biggl }{\biggl }=\int _{0}^{u}\int _{0}^{1}x\,v(xy)\,w(x)+x\,v(x)\,w(xy)\,\mathrm {d} y\,\mathrm {d} x}

And finally, we use the Fubini theorem

= ∫ 0 1 ∫ 0 u x v (x y) w (x) + x v (x) w (x y) d x d y {\displaystyle {\biggl }{\biggl }=\int _{0}^{1}\int _{0}^{u}x\,v(xy)\,w(x)+x\,v(x)\,w(xy)\,\mathrm {d} x\,\mathrm {d} y}
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