Get the solution to "Finding integral: ∫01ln⁡(1+x)xdx"

piarepm

Answered question

2022-01-03

Finding integral:

\int_{0}^{1} \frac{\ln (1 + x)}{x} d x

Answer & Explanation

Paul Mitchell

Beginner2022-01-04Added 40 answers

Here is an elementary integration of

I = \int_{0}^{1} \frac{\ln (1 + x)}{x} d x

I = - \frac{1}{2} \int_{0}^{1} \frac{\ln (1 + x)}{x} d x + \frac{3}{2} \int_{0}^{1} \frac{\ln (1 + x)}{x} d x = - \frac{3}{2} \int_{0}^{1} \frac{\ln (1 - x + x^{2})}{x} d x

Let

J (a) = \int_{0}^{1} \frac{\ln (1 - 2 x \sin a + x^{2})}{x} d x

J^{'} (a) = - \int_{0}^{1} \frac{2 \cos a}{{(x - \sin a)}^{2} + \cos^{2} a} d x = - (\frac{π}{2} + a)

Then, width

J (0) = \int_{0}^{1} \frac{\ln (1 + x^{2})}{x} d x = \frac{12}{I}

I = - \frac{32}{J} (\frac{π}{6}) = - \frac{32}{J (0) + \int_{0}^{\frac{π}{6}} J^{'} (a) d a} = - \frac{34}{I} + {\frac{32}{\int}}_{0}^{\frac{π}{6}} (\frac{π}{2} + a) d a

which leads to

I = \frac{π^{2}}{12}

sonorous9n

Beginner2022-01-05Added 34 answers

Hint. One may recall that

\ln (1 + x) = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{x^{n}}{n}, | x | < 1

one may then divide by x and one is allowed to integrate termwise obtaining

\int_{0}^{1} \frac{\ln (1 + x)}{x} d x = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \int_{0}^{1} \frac{x^{n - 1}}{n} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n^{2}}

Can you take it from here?

user_27qwe

Skilled2022-01-11Added 375 answers

By integration by parts:
$\int_{0}^{1} \frac{\log (1 + x) d x}{x} = - \int_{0}^{1} \frac{\log x}{1 + x} d x = 2 \int_{0}^{1} \frac{- \log x}{1 - x^{2}} d x - \int_{0}^{1}$
but since $\int_{0}^{1} (- \log x) x^{k} d x = \frac{1}{(k + 1)^{2}}$ , by expanding $\frac{1}{1 - x^{2}}$ and $\frac{1}{1 - x}$ as geometric series we get:
$\int_{0}^{1} \frac{\log (1 + x) d x}{x} = 2 \sum_{k o d d} \frac{1}{k^{2}} - \sum_{k \geq 1} \frac{1}{k^{2}} = \sum_{k \geq 1} \frac{1}{k^{2}}$
$= \frac{π^{2}}{12}$

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