# Variations on the Mean Value Theorem for Integrals

I know a bunch of different versions of the mean value theorem for integrals, and yet none of them are able to solve my problem, but it sure as heck looks like one of them should.

I have attempted the following versions:

$1)$ let $f$ be a continuous function on $[a,b]$. Then there is $c \in [a,b]$ such that $$\int_a^b f(x) dx = f(c)(b-a)\,.$$

$2)$ Let $f$ be a continuous function and $\alpha$ be a continuous function of bounded variation. Then there is $c \in [a,b]$ such that $$\int_a^b f(x) d \alpha (x) = f(a)\int_a^c f(x) d\alpha (x) + f(b)\int_c^bf(x) d \alpha(x)\,.$$

$3)$ the two theorems stated in this question here.

Yet, unless I'm crazy, none of these theorems are able to prove the following fact, which certainly feels like some kind of mean value theorem problem.

Problem: Suppose $f$ is bounded and continuous. Show that there is some $c \in (0, \infty)$ such that $$\int_0^\infty f(x) e^{-x} dx = f(c)\,.$$

Do I need a different version? Something else all together? Did I just miss something obvious? This was an old test problem, and I need someone to put me out of my misery over this question.

• Note that $\int_0^\infty e^{-x} = 1$. So the integral you have is a weighted mean of the function. – Myridium Aug 1 '17 at 5:47

The mean value theorems you cite apply to integrals over bounded intervals, so some additional analysis is required.

Note that the integral $\int_0^\infty f(x) e^{-x} \, dx$ is convergent by the Weierstrass test.

Let $m = \inf_{x \in[0,\infty)} f(x)$ and $M = \sup_{x \in[0,\infty)} f(x)$. Since $me^{-x} \leqslant f(x)e^{-x} \leqslant Me^{-x}$, we have

$$m = \int_0^\infty me^{-x} \, dx \leqslant I = \int_0^{-\infty}f(x) e^{-x} \, dx \leqslant \int_0^\infty Me^{-x} \, dx = M,$$

and $m \leqslant I \leqslant M$.

If $m < I < M$ then there exist $a$ and $b$ such that $m \leqslant f(a) < I < f(b) \leqslant M$. (This is a basic property of the infimum and supremum.) By the familiar intermediate value theorem -- which applies to continuous functions on compact intervals -- there exists $c \in (a,b) \subset [0,\infty)$ such that

$$f(c) = I = \int_0^\infty f(x) e^{-x} \, dx.$$

See if you can handle the cases where $I = m$ and $I = M$.

Hint: If $I = m$ then $\int_0^\infty [f(x) - m]e^{-x} \, dx = 0$ where $f(x) - m \geqslant 0$.

• That extra care about first getting $a, b$ and then dealing with $c$ is important because $m, M$ are not extreme values of $f$ but just inf and sup. +1 – Paramanand Singh Aug 1 '17 at 18:11
• @ParamanandSingh: Thank you for highlighting that point. – RRL Aug 1 '17 at 19:16

The integral $$\int_0^\infty e^{-x} \mathrm d x = 1$$ meaning that your integral $\int_0^\infty f(x) e^{-x} \mathrm d x$ is a weighted mean of $f(x)$. Consequently, its value is somewhere between the minimum and maximum values of $f$ in the domain.

The Intermediate Value Theorem completes the proof by saying that there exists a $c$ such that $f(c)$ takes on this value.

If you want more rigour, denote $f_{inf}$ and $f_{sup}$ as the infimum and supremum of $f$ respectively. These exist because $f$ is bounded. We know:

$$f_{inf} = \int_0^\infty f_{inf} \ e^{-x}\; \mathrm d x \quad \leq \quad\int_0^\infty f(x) e^{-x} \mathrm d x \quad \leq \quad \int_0^\infty f_{max} \; e^{-x} \mathrm d x = f_{max}$$ from which it follows that $$f_{inf} \leq \int_0^\infty f(x) e^{-x} \mathrm d x \leq f_{sup}.$$ If the infimum is never reached, or the supremum, then replace the respective $\leq$ signs with $<$.

The intermediate value theorem completes the proof.

• The difficulty with this problem is that you can't just invoke the intermediate value theorem for a bounded function $f$ on an unbounded interval -- for a value that is not strictly between $\inf f$ and $\sup f$ since $f$ may never attain either of those values. For example, take $f(x) = 1 - e^{-x}$ on $[0,\infty)$. In this case, $\sup f = 1$. If $k = 1$ then $\inf f \leqslant k \leqslant \sup f$ but there is no point $c \in [0,\infty)$ where $f(c) = k = 1$. – RRL Aug 1 '17 at 7:31
• ... and most of your argument is a complete duplication of my answer posted 1 hour before. – RRL Aug 1 '17 at 7:36