$\int_{0}^{1} \left(1-x^a\right)\cdot \log\left(1-x^a \right)\,dx$ for a positive

Question

I want to calculate $$\int_{0}^{1} \left(1-x^a\right)\cdot \log\left(1-x^a \right)\,dx,$$ but no luck. Wolfram gives me a very complicated answer in terms of hypergeometric functions. Is there a better answer or a way to tackle this to get a solution (even if involves special functions)

Answer 1

$$\eqalign{ & I = \int\limits_0^1 {\left( {1 - {x^a}} \right)\ln \left( {1 - {x^a}} \right)dx} ,{\text{ let}}:t = {x^a} \Rightarrow x = {t^{\frac{1}{a}}} \Rightarrow dx = \frac{1}{a}{t^{\frac{1}{a} - 1}}dt \cr & \Rightarrow I = \frac{1}{a}\int\limits_0^1 {{t^{\frac{1}{a} - 1}}\left( {1 - t} \right)\ln \left( {1 - t} \right)dt} = {\left. {\frac{\partial }{{\partial b}}} \right|_{b = 2}}\frac{1}{a}\int\limits_0^1 {{t^{\frac{1}{a} - 1}}{{\left( {1 - t} \right)}^{b - 1}}dt} \cr & = {\left. {\frac{1}{a}\frac{\partial }{{\partial b}}} \right|_{b = 2}}{\rm B}\left( {\frac{1}{a},b} \right) = {\left. {\frac{1}{a}\frac{\partial }{{\partial b}}} \right|_{b = 2}}\frac{{\Gamma \left( {\frac{1}{a}} \right)\Gamma \left( b \right)}}{{\Gamma \left( {\frac{1}{a} + b} \right)}} = \frac{a}{{a + 1}}\left( { - {\psi ^{(0)}}\left( {2 + \frac{1}{a}} \right) - \gamma + 1} \right) \cr & = \frac{a}{{a + 1}}\left( {1 - {H_{1 + \frac{1}{a}}}} \right) \cr} $$

Answer 2

let, $$ I = \int_0^1 (1-x^a) \cdot \ln(1-x^a) \, \mathrm{d}x \tag{1}$$

Now just use the Taylor series expansion of $\ln(1-x)$

$$ I = -\int_0^1 (1-x^a) \cdot \sum_{n=1}^\infty \frac{x^{an}}{n} \, \mathrm{d}x \tag{2} $$

Now as we know that $x^an$ converges uniformly in $0$ to $1$ so we can interchange summation and integral sign, we get

$$ I = -\sum_{n=1}^\infty \frac{1}{n} \int_0^1 (x^{an} - x^{a(n+1)}) \, \mathrm{d}x \tag{3}$$

$$ I = -\sum_{n=1}^\infty \frac{1}{n} \left( \frac{1}{an+1} - \frac{1}{a(n+1)+1} \right) \tag{4} $$

$$ I = \sum_{n=1}^\infty \left( \frac{1}{n+\frac{1}{a}} - \frac{1}{n+\frac{1+a}{a}} \right) \tag{5} $$

Now from Here we know that $$ \psi(1+n) = \sum_{k=1}^\infty \left( \frac{1}{k} - \frac{1}{k+n} \right) - \gamma \tag{6} $$

So we can see that,

$$ \psi(1+n) - \psi(1+m) = \sum_{k=1}^\infty \left( \frac{1}{k+m} - \frac{1}{k+n} \right) \tag{7}$$

Now using $7$ in $5$ we get,

$$ I = \psi \left(1 + \frac{1+a}{a} \right) - \psi \left( 1 + \frac{1}{a} \right) \tag{8} $$

We know that,

$$ \psi(1+n) = \frac{1}{n} + \psi(n) $$

Now

$$I = \frac{a}{1+a} + \psi \left( 1+ \frac{1}{a} \right) - \psi \left(1+ \frac{1}{a} \right)$$

$$ I = \frac{a}{1+a} $$

Answer 3

Denoting $y=1-x^a$ you get $$I=a\int_0^1 (1-y)^{(1-a)/a}y\, \log y \, dy=a\frac{d}{ds}\left[\int_0^1 (1-y)^{(1-a)/a}y^{s+1} \, dy\right]_{s=0}=a\frac{d}{ds}\left[\frac{\Gamma(s+2)\Gamma(1/a)}{\Gamma(s+2+\frac{1}{a})}\right]_{s=0}$$

Answer 4

Noting that $$ I=\int_0^1\left(1-x^a\right) \ln \left(1-x^a\right) d x=J^{\prime}(1), $$ where $\displaystyle J(b)=\int_0^1\left(1-x^a\right)^b d x= \frac{\Gamma\left(1+\frac{1}{a}\right) \Gamma(b+1)}{\Gamma\left(b+\frac{1}{a}+1\right)} \textrm{ by the }$post .

By logarithmic differentiation on $b$ at $b=1$, we get

$$ \begin{aligned} J’(1) & =J(1) \left[\psi(b+1)-\psi\left(b+\frac{1}{a}+1\right)\right]_{b=1} \\ & =\frac{\Gamma\left(1+\frac{1}{a}\right) \Gamma(2)}{\Gamma\left(2+\frac{1}{a}\right)}\left[\psi(2)-\psi\left(2+\frac{1}{a}\right)\right] \\ \therefore I& =\frac{a}{a+1}\left[1-\gamma-\psi\left(2+\frac{1}{a}\right)\right] \end{aligned} $$