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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Periodicity and integral

Let f:\mathbb{R} \rightarrow \mathbb{R} be a continuous and periodic function with period T \neq 0. If \kappa, \mu, \nu \in \mathbb{N} then prove that:

  1. \displaystyle \int_{0}^{\kappa \mathrm{T}} f(x) \, \mathrm{d}x = \kappa \int_0^{\mathrm{T}} f(x) \, \mathrm{d}x
  2. \displaystyle \int_{\alpha + \mu \mathrm{T}}^{\beta + \nu \mathrm{T}} f(x) \, \mathrm{d}x = \int_{\alpha}^{\beta} f(x) \, \mathrm{d}x + \left ( \nu - \mu \right ) \int_{0}^{\mathrm{T}} f(x) \, \mathrm{d}x where \alpha, \beta \in \mathbb{R}.

Solution

  1. We have successively:

        \begin{align*} \int_{0}^{\kappa T} f(x) \, \mathrm{d}x &= \sum_{n=0}^{\kappa-1} \int_{nT}^{(n+1) T} f(x) \, \mathrm{d}x \\ &= \sum_{n=0}^{\kappa-1} \int_{0}^{T} f \left ( u + nT \right )\, \mathrm{d}u \\ &= \sum_{n=0}^{\kappa-1} \int_{0}^{T} f(x) \, \mathrm{d}x \\ &=\kappa \int_{0}^{T} f(x) \, \mathrm{d}x \end{align*}

  2. We have successively:

        \begin{align*} \int_{\alpha + \mu \mathrm{T}}^{\beta + \nu \mathrm{T}} f(x) \, \mathrm{d}x - \int_{\alpha}^{\beta} f(x) \, \mathrm{d}x &= \int_{\alpha + \mu \mathrm{T}}^{\beta + \nu \mathrm{T}} f(x) \, \mathrm{d}x - \\ & - \left ( \int_{\alpha}^{\alpha + \mu T} f(x) \, \mathrm{d}x + \int_{\alpha +\mu T}^{\beta } f(x) \, \mathrm{d}x \right )\\ &=\int_{\beta }^{\beta + \nu \mathrm{T}} f(x) \, \mathrm{d}x - \int_{\alpha}^{\alpha + \mu T} f(x) \, \mathrm{d}x \\ &=\nu \int_{0}^{\mathrm{T}} f(x) \, \mathrm{d}x - \mu \int_{0}^{\mathrm{T}} f(x) \, \mathrm{d}t \\ &= \left ( \nu - \mu \right ) \int_{0}^{\mathrm{T}} f(x) \, \mathrm{d}x \end{align*}

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On a geometric sequence

Let \alpha, \beta, \gamma , \delta be four consecutive terms of a geometric sequence. Prove that

    \[\left ( \beta -\gamma \right )^2+\left ( \gamma -\alpha \right )^2+\left ( \delta -\beta \right )^2 = \left ( \alpha -\delta \right )^2\]

Solution

We have successively:

    \begin{align*} \left ( \beta -\gamma \right )^2+\left ( \gamma -\alpha \right )^2+\left ( \delta -\beta \right )^2 &= \beta^2 -2\beta \gamma + \gamma^2 + \gamma^2 - 2 \gamma \alpha + \alpha^2 + \delta^2 - 2\delta \beta + \beta^2 \\ &= \alpha \gamma - 2\beta \gamma + \gamma^2 +\gamma^2 - 2\gamma \alpha + \alpha^2 + \delta^2 - 2 \delta \beta+ \alpha \gamma \\ &= -2\beta \gamma + 2 \delta \beta + \alpha^2+ \delta^2 - 2\delta \beta \\ &= \alpha^2 -2 \beta\gamma + \delta^2 \\ &= \alpha^2 - 2\alpha \delta + \delta^2 \\ &= \left ( \alpha - \delta \right )^2 \end{align*}

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Constant area

Let a be a positive real number. The parabolas defined by y_1=ax^2 and y_2^2=ax intersect at the points \mathrm{O} and \mathrm{A}.

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Prove that the area enclosed by the two curves is constant. Explain why.

Solution

First of all we note that

    \begin{align*} y_1 = y_2 &\Leftrightarrow \left ( ax^2 \right )^2 = ax \\ &\Leftrightarrow a^2 x^4 = ax \\ &\!\!\!\!\!\overset{a>0}{\Leftarrow \! =\! =\! \Rightarrow } a x^4 - x =0 \\ &\Leftrightarrow x \left ( ax^3 -1 \right ) =0 \\ &\Leftrightarrow \left\{\begin{matrix} x & = & 0\\ x &= & \sqrt[3]{\frac{1}{a}} \end{matrix}\right. \end{align*}

Hence,

    \begin{align*} \mathrm{E}\left ( \Omega \right ) &= \int_{0}^{\sqrt[3]{1/a}} \left | ax^2 - \sqrt{ax} \right |\, \mathrm{d}x\\ &=\int_{0}^{\sqrt[3]{1/a}} \left ( \sqrt{ax} - ax^2 \right )\, \mathrm{d}x \\ &=\frac{2}{3} - \frac{1}{3} \\ &= \frac{1}{3} \end{align*}

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Binomial sum

Let \mathbb{N} \ni k >1. Evaluate the sum

    \[\sum_{n=1}^{\infty} \frac{\mathcal{H}_n}{\binom{n+k}{k}}\]

Solution

We have successively

\begin{aligned} \sum_{n=1}^{\infty} \frac{H_n}{\binom{n+k}{k}} &= \sum_{n=1}^{\infty} \frac{k!H_n}{(n+1) \cdots (n+k)} \\ &= \sum_{n=1}^{\infty} \frac{k!H_n}{k-1} \left(\frac{1}{(n+1)\cdots (n+k-1)} - \frac{1}{(n+2) \cdots (n+k)} \right) \\ &= \frac{k!}{k-1}\sum_{n=1}^{\infty} \sum_{i=1}^n \frac{1}{i} \left(\frac{1}{(n+1)\cdots (n+k-1)} - \frac{1}{(n+2) \cdots (n+k)} \right) \\ &= \frac{k!}{k-1} \sum_{i=1}^{\infty} \frac{1}{i} \sum_{n=i}^{\infty} \left(\frac{1}{(n+1)\cdots (n+k-1)} - \frac{1}{(n+2) \cdots (n+k)} \right) \\ &= \frac{k!}{k-1} \sum_{i=1}^{\infty} \frac{1}{i(i+1) \cdots (i+k-1)} \\ &= \frac{k!}{(k-1)^2} \sum_{i=1}^{\infty} \left(\frac{1}{i(i+1) \cdots (i+k-2)} - \frac{1}{(i+1) \cdots (i+k-1)}\right) \\ &= \frac{k!}{(k-1)^2} \frac{1}{(k-1)!} \\ &= \frac{k}{(k-1)^2} \end{aligned}

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