Integral equation

For any $f:\mathbb{R}\rightarrow \mathbb{R}$ is a continuous function, prove that $\int_0^4f(x(x-3)^2)dx=2\int_1^3f(x(x-3)^2)dx$. I tried many things, but nothing worked out. I observed that $g(x)+g(4-x)=4$ where $g(x)=x(x-3)^2$. Hence this action $x \rightarrow 4-x$ acts in some peculiar way. Please, any help would be great.

Solutions Collecting From Web of "Integral equation"

I think the way to solve this may be to look at it sideways.

Let $g$ be a continuous strictly increasing function on $[a,b],$
let $f$ be a continuous function on $[g(a), g(b)],$
and let $h$ be the inverse of $g,$ that is,
$h(g(x)) = x$ whenever $a \leq x \leq b.$

Consider the integral
$$ \int_a^b f(g(x))\,dx. $$

If we make the substitution $y = g(x),$ then $x = h(y)$
and $dx = h'(y)\,dy,$ so
$$ \int_a^b f(g(x))\,dx = \int_{g(a)}^{g(b)} f(y) h'(y) \,dy. $$

The same equation is true if $g$ is continuous and strictly decreasing on $[a,b]$ and $f$ is continuous on $[g(b),g(a)],$ although in this case, since $g(b) < g(a),$ it may be desirable to put the bounds of the right-hand integral in the usual order, like this:
$$ \int_a^b f(g(x))\,dx = -\int_{g(b)}^{g(a)} f(y) h'(y) \,dy. $$

For $g = x(x-3)^2$ on $[0,4],$ I would consider three different domains
separately: $g_1,$ a strictly increasing function defined on $[1,0]$;
$g_2,$ a strictly decreasing function defined on $[1,3]$; and
$g_3,$ a strictly increasing function defined on $[3,4].$
Let the inverses of these functions be respectively $h_1,$ $h_2,$ and $h_3.$
\int_1^3 f(x(x-3)^2)\,dx = \int_1^3 f(g_2(x))\,dx
= – \int_0^4 f(y)h_2′(y)\,dy
\int_0^1 f(x(x-3)^2)\,dx + \int_3^4 f(x(x-3)^2)\,dx &=
\int_0^1 f(g_1(x))\,dx + \int_3^4 f(g_3(x))\,dx \\
&= \int_0^4 f(y)h_1′(y)\,dy + \int_0^4 f(y)h_3′(y)\,dy \\
&= \int_0^4 f(y)\left( h_1′(y) + h_3′(y) \right)\,dy .

So if $h_1′(y) + h_3′(y) = – h_2′(y),$ then
$$\int_0^1 f(x(x-3)^2)\,dx + \int_3^4 f(x(x-3)^2)\,dx
= \int_1^3 f(x(x-3)^2)\,dx,$$
and therefore $\int_0^4 f(x(x-3)^2)\,dx = 2\int_1^3 f(x(x-3)^2)\,dx,$
as was to be shown.

After some playing around with the inverse of $g(x) = x(x-3)^2$
in Wolfram Alpha
(see here,
and here),
it seems to me that
h_1(y) &= 2 – u – \frac1u,\\
h_2(y) &= 2 – \omega u – \frac1{\omega u}
= 2 – \omega u – \omega^2\frac1u,\\
h_3(y) &= 2 – \omega^2 u – \frac1{\omega^2 u}
= 2 – \omega^2 u – \omega\frac1u,
$$u = \sqrt[3]{\frac{\sqrt{x^2-4x} – x + 2}{2}} = e^{i(\cos(1-x/2))/3},$$
from which it follows that $h_1(y) + h_2(y) + h_3(y) = 6,$
from which you can get the result you need.

Update: The inverse functions are really laborious to compute, even with Wolfram Alpha. For the reference of anyone else who looks at this question, a far better idea is in the answer posted by Frank Zermelo, which completely avoids constructing the inverse functions.

This is just a follow-up on @DavidK’s answer. Since it seems very hard to compute the inverses of $g(x)=x(x-3)^2$, I will provide a solution where we don’t need to compute them.

Let $h_1(x),h_2(x),h_3(x)$ be inverses of $g(x)$ on the intervals $[0,1],[1,3],[3,4].$ We know by the monotonicity of $g(x)$ that $g:[0,1]\rightarrow [0,4]$, $g:[1,3]\rightarrow [0,4]$, $g:[3,4]\rightarrow [0,4]$ are surjective functions, also proving that the domains of definition of $h_1(x),h_2(x),h_3(x)$ is $[0,4]$ for all three of them.

By David K’s answer, it is sufficient to prove that $h_1(y)+h_2(y)+h_3(y)=6$ for all $y \in [0,4]$.

Take $y \in [0,4]$. By the above mentioned surjectivity of $g(x)$, there exists $x_1 \in [0,1], x_2 \in [1,3],x_3\in[3,4]$ with $g(x_1)=g(x_2)=g(x_3)=y$. By the property of inverse functions, we know that $h_i(g(x_i))=x_i$, for all $i=1,2,3.$ Hence $$h_1(y)+h_2(y)+h_3(y)=h_1(g(x_1))+h_2(g(x_2))+h_3(g(x_3))=x_1+x_2+x_3$$
the statement to prove is equivalent of proving that: If $x_1 \in [0,1], x_2 \in [1,3],x_3\in[3,4]$ with $g(x_1)=g(x_2)=g(x_3)$, then $x_1+x_2+x_3=6$.

Suppose $g(x_1)=g(x_2)=g(x_3)=y$ for some $y\in [0,4]$. Let $P(x)=g(x)-y=x^3-6x^2+9x-y$, a cubic polynomial in $x$. We know by hypothesis that $P(x_1)=P(x_2)=P(x_3)=0$, hence $x_1,x_2,x_3$ are the roots of $P$. By the Vieta’s formula, we get $x_1+x_2+x_3=6$, and we are done.

Using the OP’s notation, $g(x)=x(x-3)^2,$ it seems that the symmetry of the cubic on the interval in question is a key. Other functions with this symmetry can be used to construct similar functional equations.


Let $h = 4x$ on $[0,1]$, $h=-2x+6$ on $[1,3]$ and $h =4 x-12$ on $[3,4].$ Then for $h$ also, exploiting symmetry, and $f$ as in the OP,

$$ \int_0^1 f(h(x))dx +\int_3^4f(h(x))dx = \int_1^3f(h(x))dx$$ or

$$\int_0^4f(h) = 2\int_1^3 f(h). $$

For $g$ we have that

$\int_0^1 g = \int_3^4 4-g,$ and $ \int_3^4 g = \int_0^1 4-g.$

We already have by symmetry for $g$ that $\int_1^3g = \int_1^34-g$ so

$\int_1^3 g + \int_1^3 4-g = 2\int_1^3 g =\int_1^3 4 = 8.$ So for $g,$

$2\int_1^3 g = \int_0^1 g+\int_3^4 g+\int_1^3 g = \int_0^4 g.$

Then I think using the definition of $f$, the Riemann integral, continuity, we can show that on $[0,1],[3,4],$ if

$\int_0^1 g = \int_3^4 4-g$ then $\int_0^1f(g) = \int_3^4 f(4-g).$

Likewise, if $\int_1^3 g = \int_1^3 4-g,$ then $\int_1^3 f(g) = \int_1^3 f(4-g).$

These last follow from the naive observation (in this narrow context)** that if $g$ takes on a set of values in $[a,b]$ rearranging those values, loosely speaking, will not alter the value of the integral on the interval. The same is true of $f(g).$

** Edit: By “this narrow context” I mean symmetry. This conclusion is proved in the answer here.

Sample calculation for $h$. Using $f(x) = \cos x,$

$\int_0^1 \cos(4x)dx+\int_3^4 \cos(4x-12)dx = \int_1^3 \cos(-2x+6)dx\approx -.3784.$

$\int_0^1 4x dx+\int_3^4 (4x-12) dx = \int_1^3 (-2x+6) dx = 4.$

Sample calculation for $g.$ Using $f(x) = \cos x,$

$\int_0^1 \cos(x(x-3)^2)dx = \int_3^4 \cos(4-x(x-3)^2)dx\approx -.486.$