Proof of multiple integral equation

Question

I am trying to prove the following equation: $$ \int_a^x\int_a^{t_1}\cdots\int_a^{t_{n-1}}\,dt_n\cdots\,dt_2\,dt_1=\frac{(x-a)^n}{n!}$$ I am not really sure where to begin. I would appreciate any help.

Answer 1

Let $I_n(x)$ be the integral above. It is straightforward to verify the equation for $n=1$.

Note that we have $I_{n+1}'(x) = I_n(x)$.

Suppose the equation is true for $1,...,n$.

Then integrating the equation above gives $I_{n+1}(x) = \int_a^x I_{n+1}'(t)dt = \int_a^x I_{n}(t)dt= \int_a^x {(t-a)^n \over n! }dt $, and evaluating the integral yields the desired result.

Answer 2

@Zardo suggested mathematical induction on $n$. Let's try a crude form of that and see if you can take it from there. $$ \int_a^{t_{n-1}} \, dt_n = t_{n-1} - a. $$ \begin{align} \int_a^{t_{n-2}} \int_a^{t_{n-1}} \, dt_n\,dt_{n-1} & = \int_a^{t_{n-2}} (t_{n-1} - a)\,dt_{n-1} \\[8pt] & = \left.\vphantom{\frac11} \frac{(t_{n-1} - a)^2} 2 \right|_{t_{n-1}:=a}^{t_{n-1}:=t_{n-2}} = \frac{(t_{n-2}-a)^2} 2 \end{align}

\begin{align} \int_a^{t_{n-3}} \int_a^{t_{n-2}} \int_a^{t_{n-1}} \, dt_n\,dt_{n-1} \, dt_{n-2} & = \int_a^{t_{n-3}} \frac{(t_{n-2}-a)^2} 2\,dt_{n-1} \\[8pt] & = \left.\vphantom{\frac11} \frac{(t_{n-2} - a)^3} 6 \right|_{t_{n-2}:=a}^{t_{n-2}:=t_{n-3}} = \frac{(t_{n-3}-a)^3} 6 \end{align}

$\ldots$ and so on. That is how mathematical induction is done, but it's stated somewhat more abstractly.

Let's make it a little bit more abstract:

Induction hypothesis: $$ \int_a^{t_{n-(k-1)}} \int_a^{t_{n-(k-2)}} \cdots \int_1^{t_{n-1}} dt_n \, dt_{n-1} \cdots dt_{n-(k-2)} = \frac{(t_{n-(k-1)}-a)^{n-(k-1)}}{(n-(k-1))!}. $$

Induction step: \begin{align} & \int_a^{t_{n-k}} \int_a^{t_{n-(k-1)}} \int_a^{t_{n-(k-2)}} \cdots \int_a^{t_{n-1}} dt_n \, dt_{n-1} \cdots dt_{n-(k-2)} \, dt_{n-(k-1)} \\[10pt] = {} & \int_a^{t_{n-k}} \frac{(t_{n-(k-1)}-a)^{n-(k-1)}}{(n-(k-1))!} \, dt_{n-(k-1)} \\[10pt] = {} & \left. \frac{(t_{n-(k-1)} - a)^{n-k}}{(n-k)!} \right|_{t_{n-(k-1)}\,:=\,a}^{t_{n-(k-1)}\,:=\,t_{n-k}} \\[10pt] = {} & \frac{(t_{n-k} -a)^{n-k}}{(n-k)!}. \end{align} This is the same argument that was presented more concretely above.

Answer 3

HINT:

$$\begin{align} \int_a^{t_{n-1}}dt_n&=\frac{1}{1}(t_{n-1}-a)^1\\\\ \int_a^{t_{n-2}}(t_{n-1}-a)dt&=\frac{1}{2\cdot 1}(t_{n-2}-a)^2\\\\ \int_a^{t_{n-3}}\frac12 (t_{n-2}-a)^2dt&=\frac{1}{3\cdot 2\cdot 1}(t_{n-3}-a)^3 \end{align}$$

---

HINT 2:

We have established a benchmark that the result is true if $n=3$ ($x=t_{n-3}=t_0$). Now assume that the relationship is true for some number $k$. Show that it is true for $k+1$ to complete the proof by induction.