Integrate :-[tex]\displaystyle \sf \large\int\limits^1_0 \dfrac{log (x + 1)}{x}[/tex] . Created by MystifiedBoy. Math

tex] ...

Questions

MystifiedBoy @MystifiedBoy

September 2021 1 70 Report

Integrate :-
[tex]\displaystyle \sf \large\int\limits^1_0 \dfrac{log (x + 1)}{x}[/tex]

Answers & Comments

JohannEuler

$\Large \mathbb{SOLUTION:}$

$\!\!\begin{array}{l} \textsf{We name our integral }I. \\ \\ I = \displaystyle \int^1_0 \dfrac{\log (x + 1)}{x}dx \\ \\ \displaystyle I = \dfrac{1}{\ln 10} \int^1_0 \dfrac{\ln (x + 1)}{x}dx \: \because \log a = \frac{\ln a}{\ln 10} \\ \\ \textsf{We know the identity } \\ \\\displaystyle \dfrac{1}{1 + x} = 1 - x + x^2 - x^3 + ... = \sum\limits_{n = 0}^{\infty} (-1)^n x^n \\ \\ \textsf{Integrating we have,} \\ \\ \small \displaystyle \ln(x + 1) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ... = \sum\limits_{n = 0}^{\infty} \dfrac{(-1)^n x^{n + 1}}{n + 1} \end{array}$

$\!\! \begin{array}{l} \textsf{The integral becomes} \\ \\ I = \displaystyle\frac{1}{\ln 10} \sum\limits_{n = 0}^{\infty} \dfrac{(-1)^n}{n + 1}\!\int^1_0 x^n dx \\ \\ \displaystyle I = \frac{1}{\ln 10}\sum\limits_{n = 0}^{\infty} \dfrac{(-1)^n}{(n + 1)^2}\: ,\:\:\: \textsf{Let }k = n + 1 \\ \\ I = \displaystyle \frac{1}{\ln 10} \sum\limits_{k = 1}^{\infty} \dfrac{(-1)^{k - 1}}{k^2} \end{array}$

$\!\! \begin{array}{l} \textsf{To evaluate the series, we need to know the }\\ \bold{Basel\ Problem} \\ \\ \displaystyle \sum\limits_{k = 1}^{\infty} \dfrac{1}{k^2} = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} ... = \frac{\pi^2}{6} \\ \\ \textsf{Using this identity, we can evaluate }I\textsf{ as follows:} \\ \\ \displaystyle I= \frac{1}{\ln 10}\left(\sum\limits_{k = 1}^{\infty} \frac{1}{k^2} - \sum\limits_{k= 1}^{\infty} \frac{2}{{(2k)}^2}\right) \\ \\ \displaystyle I = \frac{1}{\ln 10}\left(1 - \frac{2}{2^2}\right)\sum\limits_{k = 1}^{\infty} \frac{1}{k^2} \\ \\ \displaystyle I = \frac{1}{\ln 10}\left(\frac{1}{2}\right)\!\left(\frac{\pi^2}{6}\right) \\ \\ I = \boxed{\dfrac{\pi^2}{12\ln 10} \textsf{ or approx. } 0.3572} \\ \\ \end{array}$