Question

$\text{Aratati ca valoarea integralei }\displaystyle\int_{0}^1\dfrac{(1+x)^{2019}+(1-x)^{2019}}{1+x^2}dx\\ \text{este un numar irational.}$

Answer 1

$\displaystyle Avem~(1+x)^{2019}+(1-x)^{2019}=2+2C^2_{2019}x^2+2C^4_{2019}x^4+...+ \\ \\ +2C^{2018}_{2019}x^{2018}. \\ \\ Integrala~devine~ 2\int\limits^1_0 \frac{1}{1+x^2} dx+2C^2_{2019} \int\limits^1_0 \frac{x^2}{1+x^2} dx+...+ \\ \\ +2C^4_{2019}\int\limits^1_0 \frac{x^4}{1+x^2} dx+...+2C^{2018}_{2019}\int\limits^1_0 \frac{x^{2018}}{1+x^2} dx.$

$\displaystyle Acum~vom~determina~o~relatia~de~recurenta~intre~inegralele de~\\ \\ forma~ \int\limits^1_0 \frac{x^{2k}}{1+x^2} dx: \\ \\ \int\limits^1_0 \frac{x^{2k}}{1+x^2} dx= \frac{1}{2k+1}\int\limits^1_0 (x^{2k+1}})' \cdot \frac{1}{1+x^2} dx= \\ \\ = \frac{1}{2k+1}\left(\left. \frac{x^{2k+1}}{x^2+1} \right\rvert^1_0- \int\limits^1_0 {x^{2k+1} \left(\frac{1}{x^2+1}\right)^{\big '}} dx \right)= \\ \\ = \frac{1}{2k+1} \left(\frac{1}{2}+ 2\int\limits^1_0 \frac{x^{2k+2}}{x^2+1} dx \right).$

$\displaystyle Deci~relatia~de~recurenta~arata~asa: \\ \\ \int\limits^1_0 \frac{x^{2k+2}}{x^2+1} dx = \frac{2k+1}{2}\int\limits^1_0 \frac{x^{2k}}{x^2+1} dx-\frac{1}{4}. \\ \\ Datorita~acestei~relatii~de~recurenta,~conchidem~ca~fiecare \\ \\ integrala~de~forma~I_{p}= \int\limits^1_0 \frac{x^{2p}}{x^2+1} dx~se~exprima~liniar~in \\ \\ functie~de~I_0~si~o~constanta.$

$\displaystyle Prin~urmare~integrala~initiala~se~poate~pune~sub~forma~aI_0+b, \\ \\ cu~a,b \in \mathbb{Q}. \\ \\ In~fine,~I_0=arctg1-arctg0=\frac{\pi}{4},~iar~a \cdot \frac{\pi}{4}+b \notin \mathbb{Q}.$