Integrating $\int_0^1 \ln(1-t^{a}) dt . $

Question

I would like to know how to calculate this integral

$$ A= \int_0^1 \ln(1-t^{a}) dt . $$ I tried Taylor expansion for $\ln(1-t^{a})= -t^{a}$ , that gave me this : $$ A= \lim_{ x \rightarrow 0+} \int_0^1 -t^{a} dt =\dfrac{-1}{a+1} $$

is this result correct ?

What is $a$? Also you should check what the proper Taylor expansion is and make sure it is valid. — user88595, May 19 '14 at 11:51
Note that the Taylor expansion does not simply have a single term. — Fabian, May 19 '14 at 12:22
@user2161721 : Check what happens when $a \le 0$. $a$ can't be any real number. — user88595, May 19 '14 at 12:38

score 2 · Answer 1 · answered May 19 '14 at 13:08

2

For a>0,

\begin{align} \int_0^1 \ln(1-t^{a}) dt &= \sum_{n\ge 1}\int_0^1 -\frac{t^{na}}{n}\\ &= -\sum_{n\ge 1}\frac{1}{n\,(na+1)}\\ &= -\mathrm{H}_{\frac{1}{a}} \end{align}

The summation is called a Harmonic number. Some values are given in the wiki page: https://en.wikipedia.org/wiki/Harmonic_number#Special_values_for_fractional_arguments

answered May 19 '14 at 13:08

gar

4,948

I agree. This is consistent with the relationship between Harmonic numbers and the digamma function. – JJacquelin May 19 '14 at 13:12
Yes, I was thinking to include digamma formula also, but why rewrite it here when wiki says it all! – gar May 19 '14 at 13:15
can we tell about the nature of this integral using Riemann, its convergent right ? – user2161721 May 25 '14 at 22:40
@user2161721 : We can pick any test for covergence which may suit here. – gar May 26 '14 at 03:19

JJacquelin · Answer 2 · 2014-05-19T13:14:18.550

1

I don't think that $A=\frac{-1}{a+1}$ For real $a>0$ , in general, the result cannot be expressed with a finite number of elementary functions.

enter image description here http://mathworld.wolfram.com/DigammaFunction.html

edited May 19 '14 at 13:14

answered May 19 '14 at 13:09

JJacquelin

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Integrating $\int_0^1 \ln(1-t^{a}) dt . $

2 Answers2