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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*}

 

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