"I have to prove the following:∫01x131−xlog⁡1xdx=∑n=0∞9(3n+4)2" was answered by our experts

Carole Juarez

Answered question

2022-02-26

I have to prove the following:

\int_{0}^{1} \frac{x^{\frac{1}{3}}}{1 - x} \log \frac{1}{x} d x = \sum_{n = 0}^{\infty} \frac{9}{{(3 n + 4)}^{2}}

Capodarcod0f

Beginner2022-02-27Added 6 answers

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

Note that

\frac{d}{d t} ∣_{t = 0^{+}} x^{t} = \log (x)

. Hence, by Fubuni-Tonelli we can interchange derivative/limit and integral sign:

I = - \int_{0}^{1} \frac{x^{\frac{1}{3}}}{1 - x} \log (x) d x = - \int_{0}^{1} \frac{x^{\frac{1}{3}}}{1 - x} (\frac{d}{d t} ∣_{t = 0^{+}} x^{t}) d x

= \frac{d}{d t} ∣_{t = 0^{+}} - \int_{0}^{1} \frac{x^{\frac{1}{3} + t}}{1 - x} d x

Now expand

\frac{1}{1 - x}

I = \frac{d}{d t} ∣_{t = 0^{+}} - \int_{0}^{1} x^{\frac{1}{3} + t} \sum_{n = 0}^{\infty} x^{n} d x

Again, by Fubuni-Tonelli:

I = \frac{d}{d t} {|_{{t = 0^{+}}} - \sum_{n = 0}^{\infty} \int_{0}^{1} x^{\frac{1}{3} + t + n} d x = \frac{d}{d t} |}_{t = 0^{+}} - \sum_{n = 0}^{\infty} \frac{1}{n + t + \frac{4}{3}}

= \sum_{n = 0}^{\infty} [\frac{d}{d t} ∣_{t = 0^{+}} \frac{- 1}{n + t + \frac{4}{3}}]

This is helpful 0 Flag

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?