Let's consider the following integral:

$$C = \int_{a}^{\infty} h(x_2)dx_2$$

where $h(x_2)$ is an integral too:

$$h(x_2) = \int_{b}^{\infty} f(x_1)g(x_1 - x_2)dx_1$$

So $$C = \int_{a}^{\infty} \left[ \int_{b}^{\infty} f(x_1)g(x_1 - x_2)dx_1 \right] dx_2 = \int_{a}^{\infty} \int_{b}^{\infty} f(x_1)g(x_1 - x_2)dx_1dx_2$$

Let's substitute one variable as follows:

$$x_3 + x_1 = x_2$$ $$dx_3 = dx_2$$

so there will be $g(-x_3) = i(x_3)$ (note that $i$ is a function, not the imaginary unit) and the integral becomes separable:

$$C = \int_{c}^{d} \int_{b}^{\infty} f(x_1)i(x_3) dx_1 dx_3 = \int_{c}^{d} i(x_3) dx_3 \cdot \int_{b}^{\infty} f(x_1) dx_1 = A \cdot B$$

This problem is used in a physics proof. Anyway $f,g,h,i$ are real functions and $x_1, x_2, x_3$ are real variables.

First, I can't determine the new integration extrema $c$ and $d$. Looking above, $c$ is the value assumed by $x_3$ when $x_2 = a$; $d$ is the value of $x_3$ when $x_2 \to \infty$. But $x_3 + x_1 = x_2$, so in both cases $x_1$ is required too; anyway, $x_1$ is the variable in the integral $B$ and is not defined inside the integral $A$.

1) In presence of an integral with the product $f(x_1)g(x_1 - x_2)$ is it correct to introduce a new variable $x_3 = x_2 - x_1$ which appears as completely independent from $x_1$?

2) How to determine the values of the integration extrema $c,d$ if they depend on a variable, $x_1$, which is no more considered?

  • $\begingroup$ I modified the question trying to better explain what I am asking. Is it still unclear? $\endgroup$ – BowPark Jul 8 '16 at 8:12
  • $\begingroup$ I have written an answer, but I'm hesitating to post it. Is this substitution from an official proof? If so, do they state what $c$ and $d$ are? I get $c = -\infty$ and $d = \infty$ from my calculations. Also, do you have any further assumptions on the integrability of $f, g$ and $h$? $\endgroup$ – user331406 Jul 8 '16 at 17:20
  • $\begingroup$ I completely reformulated my answer. I hope that it answers your questions now more directly. $\endgroup$ – user331406 Jul 10 '16 at 13:59

Introduction: I think that the substitution $x_3 = x_2 - x_1$ works (and that a substitution is needed to split the integral), but as you pointed out, it's not very clear how to determine the new integration bounds. In fact, this substitution looks very similar to the "physicist's method" of substitution in the one dimensional case (no offense, we call it like that at our uni). Here's an example to illustrate what I mean: If we want to calculate

\begin{equation} \int_0^1 (2x + 1)^2 dx, \end{equation}

we can set $z = 2x + 1$, giving $\frac{dz}{dx} = 2 \Leftrightarrow \frac{1}{2}dz = dx$, and by plugging $0$ and $1$ into $2x + 1$ to get the new integration bounds our integral becomes to

\begin{equation} \int_0^1 (2x + 1)^2 dx = \int_1^3 \frac{1}{2}z^2 dz = \frac{13}{3} \end{equation}

While this method works quite well in the one-dimensional case, it's not very clear how to do it in the multi-dimensional case, and in fact this method hides the original statement of the integration by substitution rule (see below). This is unfortunate, because in order to avoid confusion it's very useful to write an integral exactly according to the rule when there's more than one variable around. I will show in the following section how to do this on your integral, using the substitutuion $x_3 = x_2 - x_1$.

Answer: For sake of completeness, I state the integration by substitution rule in the one- and multi-dimensional case.

In the one-dimensional case, the rule tells us that if $I \subset \mathbb{R}$ is an intervall, $f: I \rightarrow \mathbb{R}$ a continuous function and $\varphi: [a, b] \rightarrow I$ in $C^1$, we have

\begin{equation} \int_a^b f(\varphi(t))\varphi'(t)dt = \int_{\varphi(a)}^{\varphi(b)}f(x)dx, \end{equation}

and in the multi-dimensional case, the rule tells us that if $U \subseteq \mathbb{R}^n$ is open, $\varphi: U \rightarrow \mathbb{R}^n$ is a diffeomorphism and $D\varphi$ its Jacobian matrix, and $F$ is a continous function with existing integral on $\varphi(U)$, we have

\begin{equation} \int_{\varphi(U)}F(z)dz = \int_UF(\varphi(x))|\mathrm{det}(D\varphi)(x)|dx, \end{equation}

Now, we want to write your integral according to the rule in the multi-dimensional case, and we also want to use the substitution $x_3 = x_2 - x_1$. So we have to set

\begin{equation} U := (b, \infty) \times (a, \infty) \subset \mathbb{R}^2, \, \varphi(x) := (x_1, x_2 - x_1) = (x_1, x_3), \, F(m, n) := f(m)g(-n), \end{equation}

where $x = (x_1, x_2) \in U$, and look at your integral as (assuming that everything is integrable)

\begin{equation} \int_U F(\varphi(x))|\mathrm{det}(D\varphi)(x)|dx = \int_U F(\varphi(x))dx = \int_a^\infty \int_b^\infty f(x_1)g(x_1 - x_2)dx_1dx_2 \end{equation}

The first equality holds since $|\mathrm{det}(D\varphi)(u)| = 1$, and the second equality holds by Fubini and the defintion of $F$ and $\varphi$. Now, setting $z := \varphi(x) = (x_1, x_3)$ and using integration by substitution, we get

\begin{equation} \int_U F(\varphi(x))dx = \int_{\varphi(U)} F(z)dz \end{equation}

Since $\varphi(U) = (b, \infty) \times (-\infty, \infty) $, using Fubini and $g(-n) := i(n)$, we get

\begin{equation} \int_{\varphi(U)} F(z)dz = \int_{-\infty}^\infty \int_b^\infty f(x_1)g(-x_3)dx_1dx_3 = \int_{-\infty}^{\infty} i(x_3)dx_3 \int_b^\infty f(x_1)dx_1, \end{equation}

as desired.

Note the we could have avoided using a new function $i$ by defining $\tilde \varphi(x) := (x_1, x_1 - x_2)$ and $F(m, n) := f(m)g(n)$ and doing the same steps as before. Indeed, if we set $z = (z_1, z_2) := \tilde \varphi(x)$, we get in the last equation

\begin{equation} \int_{\tilde \varphi(U)} F(z)dz = \int_{-\infty}^\infty \int_b^\infty f(z_1)g(z_2)dz_1dz_2 = \int_{-\infty}^{\infty} g(z_2)dz_2 \int_b^\infty f(z_1)dz_1, \end{equation}

so there's no need to introduce a new function $i$ here.

So to answer your two questions directly:

  1. Yes, introducing a new variable is correct, and the new variable is in fact a function that stands for the expression you want to substitute (in our case the function is $\varphi$, and we name it with the variable $z$). It is wrong though that the substitution doesn't affect all variables, the substitution affects every variable in the integral. In your example it just seems that $x_3$ is independent of $x_1$ because you split the integral before calculating the new integration bounds, but in fact the bounds of $x_1$ are needed to calculate the new bounds after introducing $x_3$.

  2. If you look at your new variable as a function $\varphi$, the new integration bounds are given as the image of the old integration bounds under the function $\varphi$ (as stated in the integration by substitution rule). So in the multi-dimensional case it's $\varphi(U)$, $U \subseteq \mathbb{R}^n$ an open set, and in the one dimensional case it's $\varphi(a)$ and $\varphi(b)$, $[a,b] \subseteq \mathbb{R}$ being an intervall.

Note: If we look closer to your function $h(x_2)$, we see that this looks very similar to the convolution $(f*g)(x_2)$ (the only difference is that $g$ is reflected), and in fact it is a general rule that the integral over a convolution $\int_\mathbb{R}(f * g)(t)dt$ can be split into $\int_\mathbb{R}f(t)dt \int_\mathbb{R}g(t)dt$ by Fubini's theorem (see e.g. here). In fact, we have just proven (indirectly) that this holds.


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