$\int_a^x \int_a^t f(t_1) dt_1 dt = \int_a^x (x-t) f(t) dt$?

Question

A quick question on confusion on some iterated integral in the text I am studying :

Defined in Special Functions by Roy :

How $\int_a^x \int_a^t f(t_1) dt_1 dt = \int_a^x (x-t) f(t) dt$ is true? Definition is only for $\int_a^x \int_a^t f(t_1) dt_1 dt$ and the equality is derived, I think?

Answer 1

For this small case, use more convenient notation: $\int_a^x \int_a^y f(z) dz dy$. There are now two ways to proceed to get this result. One is to integrate by parts in the outer integral. The other is to change the order of integration. To do the latter, note that you must have $y \geq z$, so the range of integration in the $dy$ integral for fixed $z$ is from $z$ to $x$. Meanwhile the largest possible value of $z$ will be $x$. So you have $\int_a^x \int_z^x f(z) dy dz$. So you can just do the inner integral now, obtaining $\int_a^x (x-z) f(z) dz$.

Related: https://en.wikipedia.org/wiki/Order_of_integration_(calculus)#Relation_to_integration_by_parts

Answer 2

Here's an alternative to Ian's approach:

Consider the functions

$$h_1(x)=\int_a^x \int_a^y f(z)dzdy$$ $$h_2(x)=\int_a^x (x-t)f(t)dt$$

We will show that these two functions have the same derivative, which will allow us to deduce that they differ by a constant in their common domain, namely $\text{dom}[f]$.

Since $f$ is (presumably) continuous, $\int_a^y f(z)dz$ is continuous with respect to $y$ over $\text{dom}[f]$. It follows from the first fundamental theorem of calculus that $h_1$ is differentiable over $\text{dom}[f]$, with derivative

$$h'(x)=\int_a^x f(z)dz\text{ for every }x\in\text{dom}[f]$$

Calculating the derivative of $h_2$ is also straightforward:

\begin{align} h_2(x) &= \int_a^x (x-t)f(t)dt\\ &= \int_a^x \left[xf(t)-tf(t)\right]dt\\ &= x\int_a^x f(t)dt-\int_a^x tf(t)dt\\ \implies h_2'(x) &= \int_a^x f(t)dt+xf(x)-xf(x)\\ &= \int_a^x f(t)dt \end{align}

It follows that $h_1$ and $h_2$ have the same derivative, so they must differ by a constant. Thus, for some $C\in\mathbb{R}$,

$$h_1(x)=h_2(x)+C\text{ for every }x\in\text{dom}[f]$$

Notice that $h_1(a)=h_2(a)=0$, so $C$ must be zero. We conclude that

$$h_1(x)=h_2(x)\text{ for every }x\in\text{dom}[f]$$

Answer 3

In my opnion, the easiest way to prove this result (assuming $f$ is continuous) is by integrating by parts:

Consider $F(t) = \int_a^t f(t_1) dt_1$ and $G(t)=t$. Then

\begin{align*} \int_a^x \int_a^t f(t_1) dt_1 dt &= \int_a^x F(t)G'(t) dt= \left [ F(t) G(t)\right ]_a^x - \int _a^x F'(t)G(t) dt = \\ & = \left [ t \int_a^t f(t_1) dt_1\right ]_a^x - \int_a^x t~f(t)~dt =\\ &= x \int_a^x f(t_1) dt_1 - \int_a^x t~f(t)~dt =\\ &= \int_a^x x ~f(t)~dt - \int_a^x t~f(t)~dt =\\ &= \int_a^x (x-t) f(t) dt \end{align*}