Evaluation of integral [MAA]

Let f be a continuous real valued function on (0, +\infty) satisfying the identity

\displaystyle f\left ( \frac{1}{x} \right ) = - f(x) \quad \text{forall} \; x \in (0, +\infty)

Evaluate the integral

\displaystyle \mathcal{J} = \int_{\sqrt{2}-1}^{\sqrt{2}+1} \frac{{\rm d}x}{\left ( 1+x^2 \right ) \left ( 1+a^{f(x)} \right )}

(D.M.Batinetu-Giurgiu , George Emil Palade)

Solution

We apply the classical sub u=\frac{1}{x} thus:

    \begin{align*} \mathcal{J} &= \int_{\sqrt{2}-1}^{\sqrt{2}+1} \frac{{\rm d}x}{\left ( 1+x^2 \right ) \left ( 1+a^{f(x)} \right )} \\ &\!\!\!\!\!\!\!\overset{u=1/x}{=\! =\! =\! =\!} \int_{\sqrt{2} -1}^{\sqrt{2}+1} \frac{1}{\left ( 1 + \frac{1}{x^2} \right )\left ( 1+a^{f\left ( \frac{1}{x} \right )} \right )} \frac{{\rm d}x}{x^2} \\ &=\int_{\sqrt{2}-1}^{\sqrt{2}+1} \frac{{\rm d}x}{\left ( 1+x^2 \right )\left ( 1+a^{-f(x)} \right )} \\ &=\int_{\sqrt{2}-1}^{\sqrt{2}+1} \frac{a^{f(x)}}{\left ( 1+x^2 \right )\left ( 1+a^{f(x)} \right )} \, {\rm d}x \\ &= \frac{1}{2} \bigg[ \int_{\sqrt{2}-1}^{\sqrt{2}+1} \frac{{\rm d}x}{\left ( 1+x^2 \right ) \left ( 1+a^{f(x)} \right )} + \\ & \quad \quad \quad \quad +\int_{\sqrt{2}-1}^{\sqrt{2}+1} \frac{a^{f(x)}}{\left ( 1+x^2 \right )\left ( 1+a^{f(x)} \right )} \, {\rm d}x \bigg]\\ &= \frac{1}{2} \int_{\sqrt{2}-1}^{\sqrt{2}+1} \frac{\cancel{1+a^{f(x)}}}{\left ( 1+x^2 \right )\cancel{\left ( 1+a^{f(x)} \right )}} \, {\rm d}x \\ &= \frac{1}{2} \int_{\sqrt{2}-1}^{\sqrt{2}+1} \frac{{\rm d}x}{x^2+1} \\ &= \frac{1}{2} \left [ \arctan x \right ]_{\sqrt{2}-1}^{\sqrt{2}+1} \\ &= \frac{\pi}{8} \end{align*}

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On an exercise with classic questions

Let f:\mathbb{R} \rightarrow \mathbb{R} be a twice differentiable function such that f''(x)>0 forall x \in \mathbb{R} . If f has a local extrema at x_0=0 and the value there is 0 and it holds that

f(-2)+f(2)=2017

then

(i) Evaluate the area bounded by the graph of f' , the lines x=-2, \; x=2 and the axis x'x.

(ii) Let \mathbb{R} \ni b >a \geq 0.  If

 \displaystyle \int_a^b f(t) \, {\rm d}t > (b-a) f(a)

holds then prove that

\displaystyle f(0)+f(1)+\cdots+f(2016)< \int_0^{2017} f(t) \, {\rm d}t

(iii) Prove that f is invertible in (0, +\infty). If you also know that the graph passes through the points (1, 8) and (10, 13) then evaluate the value of

\displaystyle \mathcal{R}=\int_1^{10} f(x) \, {\rm d}x + \int_8^{13} f^{-1}(x) \, {\rm d}x

(iv) Prove that forall x\in [1, 10] it holds that

f^2(x)-21f(x)+104\leq 0

Solution

Since f''(x)>0 forall x \in \mathbb{R} we conclude that f is convex. We also conclude that f' is strictly increasing in \mathbb{R}. Since f has a local extrema at x_0=0 this has to be a local minimum, because f'(0)=0 (it follows from Fermat’s theorem) and combining the monotony along with the root of the derivative we get that

f'(x) \geq 0 \Leftrightarrow x \geq 0

(i) The area bounbed by the graph of f' , the lines x=-2, \; x=2 and the axis x'x is equal to

    \begin{align*} {\rm E}\left ( \Omega \right ) &=\int_{-2}^{2} \left | f'(x) \right | \, {\rm d}x \\ &= \int_{-2}^{0} \left | f'(x) \right | \, {\rm d}x + \int_{0}^{2} \left | f'(x) \right | \, {\rm d}x \\ &=- \int_{-2}^{0} f'(x) \, {\rm d}x + \int_{0}^{2} f'(x) \, {\rm d}x \\ &= -\left [ f(0) - f(-2) \right ] + \left [ f(2) - f(0) \right ]\\ &= f(-2) - f(0) + f(2) - f(0) \\ &= f(-2) +f(2) \\ &= 2017 \end{align*}

since f(0)=0.

(ii) Successively we have

    \begin{align*} \int_{0}^{2017} f(x) \, {\rm d}x &= \sum_{k=0}^{2016}\int_{k}^{k+1} f(t)\, {\rm d}t \\ &>\sum_{k=0}^{2016} f(k)\left ( k+1-k \right ) \\ &=\sum_{k=0}^{2016} f(k) \\ &=f(0) + f(1) + \cdots + f(2016) \end{align*}

(iii) We begin by the classical change of variables u=f^{-1}(x) thus:

\begin{aligned} \int_{1}^{10} f(x) \, {\rm d}x + \int_{8}^{13} f^{-1} (x) \, {\rm d}x &\overset{u=f^{-1}(x)}{=\! =\! =\! =\! =\!} \int_{1}^{10} f(x) \, {\rm d}x + \int_{1}^{10} x f'(x) \, {\rm d}x \\ &= \int_{1}^{10} f(x) \, {\rm d}x + \left [ x f(x) \right ]_1^{10} - \int_{1}^{10} f(x) \, {\rm d}x\\ &= 10 f(10) - f(1) +\cancel{\int_{1}^{10} f(x) \, {\rm d}x - \int_{1}^{10} f(x) \, {\rm d}x} \\ &= 10 \cdot 13 - 8 \\ &=130 -8 \\ &=122 \end{aligned}

(iv) We simply note that

f^2 (x) - 21 f(x) + 104 = [f(x) - 8] [f(x) - 13]

and the result follows.

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An inequality using AM – GM

Let x, y , z be positive numbers. Prove that

\displaystyle \frac{3xy}{xy+x+y}+ \frac{3yz}{yz+y+z}+ \frac{3zx}{zx+z+x}\leq 2+ \frac{x^2+y^2+z^2}{3}

Solution

We are invoking the AM – GM inequality

    \begin{align*} 2+ \frac{x^2+y^2+z^2}{3} &=\sum \frac{x^2+y^2+4}{6} \\ &\geq \sum \sqrt[6]{x^2y^2} \\ &=\sum \frac{xy}{\sqrt[3]{(xy)xy}} \\ &\geq \sum \frac{xy}{\frac{xy+x+y}{3}}\\ &=\sum \frac{3xy}{xy+x+y} \end{align*}

and we conclude the result.

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An integral with Gamma and digamma

Let \Gamma denote the Euler’s Gamma function and \psi denote the digamma function. Evaluate

\displaystyle \bigintsss_{0}^{2}\frac{\Gamma^2\left ( \frac{1}{t} \right )}{2t^3 \Gamma \left ( \frac{2}{t} \right )} \bigg[ t + 2\psi\left ( \frac{1}{t} \right) - 2 \psi \left ( \frac{2}{t} \right ) \bigg] \, {\rm d}t

Solution

We have successively

\begin{aligned} \int_{0}^{2}\frac{\Gamma^2\left ( \frac{1}{t} \right )}{2t^3 \Gamma \left ( \frac{2}{t} \right )} \bigg[ t + 2\psi\left ( \frac{1}{t} \right) - 2 \psi \left ( \frac{2}{t} \right ) \bigg]\, {\rm d}t & =\int_{0}^{2} \frac{\Gamma^2\left ( \frac{1}{t} \right )}{t^3 \Gamma\left ( \frac{2}{t} \right )} \bigg[ \frac{t}{2}+ \psi\left ( \frac{1}{t} \right ) - \psi\left ( \frac{2}{t} \right ) \bigg] \, {\rm d}t \\ &\!\!\overset{u=2/t}{=\! =\! =\!} \frac{1}{4}\int_{1}^{\infty} \frac{\Gamma \left ( \frac{u}{2} \right ) \Gamma \left ( \frac{u}{2} \right )}{\Gamma \left ( \frac{u}{2}+ \frac{u}{2} \right )} \bigg[ 1+ u \psi \left ( \frac{u}{2} \right ) - u \psi (u) \bigg] \, {\rm d}u\\ &= \frac{1}{4}\int_{1}^{\infty} {\rm B}\left ( \frac{u}{2}, \frac{u}{2} \right ) \left ( 1+ u \psi \left ( \frac{u}{2} \right )- u \psi (u) \right ) \, {\rm d}u\\ &=\frac{1}{4}\int_{1}^{\infty} {\rm B}\left ( \frac{u}{2}, \frac{u}{2} \right ) \bigg( 1+ u \left ( \psi \left ( \frac{u}{2} \right ) - \psi\left ( \frac{u}{2}+ \frac{u}{2} \right ) \right ) \bigg )\, {\rm d}u \\ &=\frac{1}{4}\int_{1}^{\infty } \bigg[{\rm B} \left ( \frac{u}{2}, \frac{u}{2} \right ) + u {\rm B}\left ( \frac{u}{2}, \frac{u}{2} \right ) \left ( \psi\left ( \frac{u}{2} \right ) - \psi \left ( \frac{u}{2}+ \frac{u}{2} \right ) \right )\bigg]\, {\rm d}u \\ &=-\frac{1}{4}\int_{1}^{\infty} \left ( u {\rm B}\left ( \frac{u}{2}, \frac{u}{2} \right ) \right )'\, {\rm d}u \\ &= \frac{\pi}{4} \end{aligned}

The exercise can also be found at Aops.com .

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An inequality involving harmonic number

Let \mathcal{H}_n denote the n – th harmonic number and let  n \in \mathbb{N}. Prove that

\displaystyle \int_{0}^{1} \frac{{\rm d}x}{x^{n-1} + x^{n-2} + \cdots+x+1} \geq \frac{1}{\mathcal{H}_n}

Solution

We might begin with the integral representation of the harmonic number, namely the equation:

\displaystyle \mathcal{H}_n = \int_{0}^{1} \left ( 1+x+\cdots+x^{n-1} \right ) \, {\rm d}x = \sum_{k=1}^{n} \frac{1}{k}

So we have to prove the equivelant inequality

\displaystyle \int_{0}^{1}\left ( x^{n-1} + \cdots + x +1 \right ) \int_{0}^{1} \frac{{\rm d}x}{x^{n-1} + \cdots + x +1} \geq 1

and this is obvious using the Cauchy – Schwarz inequality.

The exercise can also be found here .

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