Evaluate a Nested Integral

Question

According to two questions [1] and [2] asked on this site earlier there exists a nice relation:

$$\frac1{n!} \left(\int_{0}^t\mathrm dt \; f(t)\right)^n = \int_{0}^t\mathrm dt_1 \int_{0}^{t_1}\mathrm dt_2 \cdots \int_{0}^{t_{n-1}}\mathrm dt_n\; f(t_1)\,f(t_2) \ldots f(t_n). $$

Unfrotunatly I am more interested in integrals of the form

$$\int_{0}^t\mathrm dt_1 \int_{0}^{t_1}\mathrm dt_2 \cdots \int_{0}^{t_{n-1}}\mathrm dt_n\; f(t_n). $$

Can it be somehow brought to an unary integral?

All I have figured out by now is that

$$\int_{0}^t\mathrm dt_1 \int_{0}^{t_1}\mathrm dt_2 \cdots \int_{0}^{t_{n-1}}\mathrm dt_n\; t_n^m = \frac{m!}{(m+n)!} t^{m+n} .$$

Answer 1

Assume $f$ is integrable on $[0,t]$, we have

$$\int_{0}^t\mathrm dt_1 \int_{0}^{t_1}\mathrm dt_2 \cdots \int_{0}^{t_{n-1}}\mathrm dt_n\; f(t_n) = \frac{1}{(n-1)!}\int_0^t (t-t_n)^{n-1} f(t_n) \mathrm dt_n$$

We can prove this using induction. Assume above statement is true up to a particular $n$, we have

$$\int_0^t\mathrm dt_1 \int_0^{t_1}\mathrm dt_2 \cdots \int_0^{t_{n}}\mathrm dt_{n+1}\; f(t_{n+1}) = \int_0^t \mathrm dt_1 \int_0^{t_1} dt_{n+1} \frac{(t_1 - t_{n+1})^{n-1}}{(n-1)!}f(t_{n+1})\\ = \int_0^t \left(\int_{t_{n+1}}^t dt_1 \frac{(t_1 - t_{n+1})^{n-1}}{(n-1)!} \right) f(t_{n+1}) dt_{n+1} = \int_0^t \frac{(t-t_{n+1})^n}{n!} f(t_{n+1}) dt_{n+1} $$ i.e. the statement will also be true for $n+1$.

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}^{t}\dd t_{1}\int_{0}^{t_{1}}\dd t_{2}\cdots\int_{0}^{t_{n - 1}} \dd t_{n}\,\fermi\pars{t_{1}}\fermi\pars{t_{2}}\ldots\fermi\pars{t_{n}}} \\[3mm]&=\int_{0}^{t}\dd t_{1}\int_{0}^{t}\dd t_{2}\cdots\int_{0}^{t} \dd t_{n}\,\Theta\pars{t_{1} - t_{2}}\Theta\pars{t_{2} - t_{3}} \ldots\Theta\pars{t_{n - 1} - t_{n}} \fermi\pars{t_{1}}\fermi\pars{t_{2}}\ldots\fermi\pars{t_{n}} \\[3mm]&={1 \over n!}\sum_{P\braces{t_{i}}} \\[3mm]&\int_{0}^{t}\dd t_{1}\int_{0}^{t}\dd t_{2}\cdots\int_{0}^{t} \dd t_{n}\,\Theta\pars{t_{1} - t_{2}}\Theta\pars{t_{2} - t_{3}} \ldots\Theta\pars{t_{n - 1} - t_{n}} \fermi\pars{t_{1}}\fermi\pars{t_{2}}\ldots\fermi\pars{t_{n}} \end{align} where $\ds{\sum_{P\braces{t_{i}}}}$ means sum over all permutations of $\ds{\braces{t_{1},t_{2},\ldots,t_{n}}}$.

Then, \begin{align} &\color{#66f}{\large\int_{0}^{t}\dd t_{1}\int_{0}^{t_{1}}\dd t_{2}\cdots\int_{0}^{t_{n - 1}} \dd t_{n}\,\fermi\pars{t_{1}}\fermi\pars{t_{2}}\ldots\fermi\pars{t_{n}}} \\[3mm]&={1 \over n!}\int_{0}^{t}\dd t_{1}\int_{0}^{t}\dd t_{2}\cdots\int_{0}^{t} \dd t_{n}\, \\[3mm]&\underbrace{\ \bracks{% \sum_{P\braces{t_{i}}}\Theta\pars{t_{1} - t_{2}}\Theta\pars{t_{2} - t_{3}} \ldots\Theta\pars{t_{n - 1} - t_{n}}}\ }_{\ds{=\ 1}}\ \fermi\pars{t_{1}}\fermi\pars{t_{2}}\ldots\fermi\pars{t_{n}} \\[5mm]&={1 \over n!}\bracks{\int_{0}^{t}\fermi\pars{t}\,\dd t}^{n} \quad\mbox{because}\quad \fermi\pars{t_{i}}\fermi\pars{t_{j}}=\fermi\pars{t_{j}}\fermi\pars{t_{i}}\,,\ \forall\ i,j = 1,2,\ldots,n \end{align}