Expert Answer to "I would like to show that ∫01x−1ln⁡(x)dx=ln⁡2"

Joanna Benson

Answered question

2022-01-04

I would like to show that

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

Answer & Explanation

Mary Nicholson

Beginner2022-01-05Added 38 answers

This is a classic example of differentiating inside the integral sign.
In particular, let

J (α) = \int_{0}^{1} \frac{x^{α} - 1}{\log (x)} d x

Then one has that

\frac{d}{d α} J (α) = \int_{0}^{1} \frac{d}{d a} \frac{x^{α} - 1}{\log (x)} d x = \int_{0}^{1} x^{α} d x = \frac{1}{α + 1}

and so we know that

J (α) = \log (α + 1) + C

. Noting that

J (0) = 0

tells us that

C = 0

and so

J (α) = \log (α + 1)

Jacob Homer

Beginner2022-01-06Added 41 answers

\int_{0}^{1} \frac{x - 1}{\log x} d x = \int_{0}^{1} \int_{0}^{1} x^{t} d t d x

= \int_{0}^{1} x^{t} d x d t

= \int_{0}^{1} \frac{1}{1 + t} d t

= \log (2)

karton

Expert2022-01-11Added 613 answers

Making the substitution $u = \ln x$ , we get
$I = \int_{- \infty}^{0} \frac{e^{u} - 1}{u} e^{u} d u = - \int_{0}^{+ \infty} \frac{e^{- 2 s} - e^{- s}}{s} d s = \ln \frac{2}{1} = \ln 2$
since we recognize a Frullani integral type.

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