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:

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

Limit of a sequence

Derivative at 0

Integral inequality (I)

No such function exists

Double series

A Riemann sum IV

A limit with Euler's totient function

Inscribed quadrilateral

Generalization of Pythagoras' theorem

One thought on “Evaluation of integral [MAA]”

John Khan says:

February 18, 2017 at 1:09 pm

Hello T,

here is an evaluation for the last equation. Since

$\begin{align*} \arctan x - \arctan y &= \arctan \left ( \frac{x-y}{1+xy} \right ) \end{align*}$

holds, then we have that:

$\begin{align*} \arctan \left ( 1+\sqrt{2} \right ) - \arctan \left ( \sqrt{2}-1 \right ) &= \arctan \left ( \frac{1+\sqrt{2} -\sqrt{2}+1}{1- \left ( \sqrt{2}-1 \right ) \left ( \sqrt{2} +1 \right )} \right ) \\ &=\arctan \left ( \frac{2}{2} \right ) \\ &= \arctan 1 \\ &= \frac{\pi}{4} \end{align*}$

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