Changing order of integration (multiple integral)

Question

Prove $$ \int_0^a\left( \int_0^x \left( \int_0^y \left( \int_0^z f(u) \, du \right) dz \right) dy \right) dx = \int_0^a \frac {(a-t)^3}{3!} f(t) dt $$ where $a$ is constant.

So I began with two most inner integrals i.e. the double integral

$$ \int_0^y \left( \int_0^z f(u) du \right) dz $$

We are doing this over $0 \leq u \leq z \leq y \leq x \leq a$. So we want

$$ \int_0^y \left( \int_0^z f(u) du \right) dz = \int_?^? \left( \int_?^? f(u) dz \right) du $$

And immediately this problem got me stumped. How can one tell what the upper/lower bounds become? Keep in mind that drawing this region won't do much good as we are working in four dimensions.

EDIT: Forgot something crucial, edited now!

Answer 1

Related techniques: (I). Here is an approach. Start writing the integral as

$$ I = \int_{0}^{a}\int_{0}^{x}h(y)dydx. $$

then change the order of the integration which gives

$$ I = \int_{0}^{a}\int_{y}^{a}h(y)dxdy= \int_{0}^{a} (a-y) h(y)dy\,. $$

Repeat the same technique and tackle the integral

$$ I = \int_{0}^{a} (a-y) h(y)dy = \int_{0}^{a} (a-y) \int_{0}^{y} g(z)dz dy $$

and change the order of integration. Continue the process and you will get the answer.

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} &\color{#66f}{\large\int_{0}^{a}\int_{0}^{x}\int_{0}^{y}\int_{0}^{z} \fermi\pars{u}\,\dd u\,\dd z \,\dd y\,\dd x} \\[3mm]&=\int_{0}^{a}\int_{0}^{a}\int_{0}^{a}\int_{0}^{a} \fermi\pars{u}\Theta\pars{z - u}\Theta\pars{y - z}\Theta\pars{x - y} \,\dd u\,\dd z\,\dd y\,\dd x \\[5mm]&=\int_{0}^{a}\int_{0}^{a}\int_{0}^{a} \fermi\pars{u}\Theta\pars{z - u}\Theta\pars{y - z}\ \overbrace{\bracks{\int_{0}^{a}\Theta\pars{x - y}\,\dd x}} ^{\ds{=\ \color{#c00000}{a - y}}}\ \,\dd y\,\dd z\,\dd u \\[3mm]&=-\int_{0}^{a}\int_{0}^{a} \fermi\pars{u}\Theta\pars{z - u}\bracks{\int_{0}^{a}\pars{y - a} \Theta\pars{y - z}\,\dd y}\,\dd z\,\dd u \\[3mm]&=-\int_{0}^{a}\int_{0}^{a} \fermi\pars{u}\Theta\pars{z - u}\bracks{\int_{z}^{a}\pars{y - a} \,\dd y}\,\dd z\,\dd u \\[3mm]&=\int_{0}^{a}\int_{0}^{a} \fermi\pars{u}\Theta\pars{z - u}{\pars{z - a}^{2} \over 2}\,\dd z\,\dd u =\int_{0}^{a}\fermi\pars{u}\int_{u}^{a}{\pars{z - a}^{2} \over 2}\,\dd z\,\dd u \\[3mm]&=-\int_{0}^{a}\fermi\pars{u}{\pars{u - a}^{3} \over 3!}\,\dd u =\color{#66f}{\large\int_{0}^{a}{\pars{a - u}^{3} \over 3!}\fermi\pars{u}\,\dd u} \end{align}

$\ds{\Theta\pars{\xi}}$ is the Heaviside Step Function.

Answer 3

Suppose $f$ continuous. Put $\displaystyle F_1(z=\int_0^zf(u)du$. Then $F_1$ is the antiderivative of $f$ ( we have $F_1^{\prime}(z)=f(z)$)such that $F_1(0)=0$. Put $\displaystyle F_2(y)=\int_0^yF_1(z)dz$. Then by the same remark, we have $F_2^{\prime\prime}(y)=f(y)$, and $F_2(0)=F_2^{\prime}(0)=0$. We continue, and find that your integral $F(a)$ as function of $a$ is such that $F^{(4)}(a)=f(a)$ and $F^{(j)}(0)=0$ for $j=0,1,2,3$. Now use the Taylor formula: $$F(a)=F(0)+aF^{\prime}(0)+F^{\prime\prime}(0)\frac{a^2}{2!}+F^{(3)}(0)\frac{a^3}{3!}+\int_0^a \frac{(a-t)^3}{3!}F^{(4)}(t)dt$$ Hence $$F(a)=\int_0^a \frac{(a-t)^3}{3!}f(t)dt$$