How to calculate ∫01xxdx using series? I read from a book that ∫01xxdx=1−122+133+…+(−1)n1(n+1)n+1+…

Just write xx=exln⁡x=∑n=0∞(xln⁡x)nn! and use that ∫01(xln⁡x)ndx=(−1)n∫0∞yne−(n+1)ydy, To show the last formula, make the change of variables x=e−y so that ∫01(xln⁡x)ndx=(−1)n∫0∞yne−(n+1)ydu, which is clearly expressible in terms of the Gamma function.

Question: "How to calculate ∫01xxdx using series? I read..."

Nylah Church

Answered question

2022-01-23

How to calculate

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

using series? I read from a book that

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

Answer & Explanation

Palandriy0

Beginner2022-01-24Added 14 answers

Just write

x^{x} = e^{x \ln x} = \sum_{n = 0}^{\infty} \frac{{(x \ln x)}^{n}}{n!}

and use that

\int_{0}^{1} {(x \ln x)}^{n} d x = {(- 1)}^{n} \int_{0}^{\infty} y^{n} e^{- (n + 1) y} d y

,
To show the last formula, make the change of variables

x = e^{- y}

so that

\int_{0}^{1} {(x \ln x)}^{n} d x = {(- 1)}^{n} \int_{0}^{\infty} y^{n} e^{- (n + 1) y} d u

,
which is clearly expressible in terms of the Gamma function.

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