A definite integral

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a continuous function such that

(1) $\begin{equation*} e^x f(x) + f \left ( e^{-x} \right ) = xe^x +e^{-x} \quad \text{for all} \;\; x \in \mathbb{R} \end{equation*}$

Evaluate the integral $\int_0^1 f(x)\, \mathrm{d}x$ .

Solution

First of all we note that the equation

(2) $\begin{equation*} e^{-x} = x \end{equation*}$

has a unique root , lets call it $a$ . Hence $e^{-a} = a$ and thus $e^{-2a}=a^2$ . We note that $(1)$ is rewritten as

(3) $\begin{equation*} f(x) + e^{-x} f \left ( e^{-x} \right ) = x +e^{-2x} \quad \text{for all} \;\; x \in \mathbb{R} \end{equation*}$

Integrating $(3)$ from $0$ to $a$ we get

$\begin{align*} \int _0^a f(x)\, \mathrm{d}x + \int_0^a e^{-x} f \left ( e^{-x} \right ) \, \mathrm{d}x = \frac {a^2}{2} - \frac{1}{2} \left ( e^{-2a} -1 \right ) & \Leftrightarrow \\ \int_{0}^{a} f(x) \, \mathrm{d}x - \int_{1}^{e^{-a}} f(t) \, \mathrm{d}t = \frac{a^2}{2} - \frac{1}{2} \left ( e^{-2a} -1 \right ) &\Leftrightarrow \\ \int_{0}^{a} f(x) \, \mathrm{d}x - \int_{1}^{a} f(x) \, \mathrm{d}x = \frac{a^2}{2} - \frac{1}{2} \left ( e^{-2a} -1 \right ) &\Leftrightarrow \\ \int_{0}^{a} f(x) \, \mathrm{d}x + \int_{a}^{1} f(x) \, \mathrm{d}x = \frac{a^2}{2} - \frac{1}{2} \left ( e^{-2a} -1 \right ) & \Leftrightarrow \\ \int_{0}^{1} f(x) \, \mathrm{d}x = \frac{a^2}{2} - \frac{a^2}{2} + \frac{1}{2} &\Leftrightarrow \\ \int_{0}^{1} f(x) \, \mathrm{d}x = \frac{1}{2} \end{align*}$

A definite integral

They might interest you:

Series with cosine integral

Limit of a multiple integral

A Riemann sum I

Logarithmic - Trigonometric integral

On the sum of inverse binomial

No map exists

A harmonic limit

A divergent series .... or maybe not?

An inequality involving harmonic number

Leave a Reply Cancel reply