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How to evaluate the integral $$ \int_{0}^{1}x^{m-1}(1-x)^{n-1}\log{x}dx, ~\Re(m), \Re(n)>0.$$ My idea is to use Beta function, but here the limits are already from $0$ to $1$. Which substitution will work such that limits remain same and we can also remove logarithm?

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  • $\begingroup$ Why not write $\lim\limits_{c\to0}\frac{x^c-1}c=\ln(x)$ and use the linearity of the integral? This will give you a derivative of a gamma function after integration $\endgroup$ Commented Feb 15, 2023 at 14:12
  • $\begingroup$ Are you familiar with the digamma function ? $\endgroup$
    – Hamdiken
    Commented Feb 15, 2023 at 14:14
  • $\begingroup$ Thanks, I understand it. $\endgroup$ Commented Feb 15, 2023 at 14:27

3 Answers 3

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I'll use some probability theory here.

Changing the writing yields $$B(m,n)\int_{0}^{1}\log{x}\frac{x^{m-1}(1-x)^{n-1}}{B(m,n)}dx$$ The term $\frac{x^{m-1}(1-x)^{n-1}}{B(m,n)}$ represents the density $f(x;m,n)$of a random variable $X$ that is Beta distributed , hence $$B(m,n)\int_{0}^{1}\log{x}\frac{x^{m-1}(1-x)^{n-1}}{B(m,n)}dx$$ $$=B(m,n)\mathbb E\left[\log(X)\right]$$ $$=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}\left[\psi(m)-\psi(m+n)\right]$$ where $\psi$ represents the Digamma function

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If we throw the rigorousness of math out of the window just now, it's not hard to see that this integral probably equals to \begin{align*} I_{\alpha,\beta}&=\int_0^1{x^{\alpha -1}\left( x-1 \right) ^{\beta -1}}\ln \left( x \right) \mathrm{d}x \\& =\partial _{\alpha}\left[ \mathrm{B}\left( \alpha ,\beta \right) \right] \\& =\partial _{\alpha}\left[ \frac{\Gamma \left( \alpha \right) \Gamma \left( \beta \right)}{\Gamma \left( \alpha +\beta \right)} \right] \\& =\frac{\Gamma '\left( \alpha \right) \Gamma \left( \alpha +\beta \right) \Gamma \left( \beta \right) -\Gamma \left( \alpha \right) \Gamma '\left( \alpha +\beta \right) \Gamma \left( \beta \right)}{\Gamma ^2\left( \alpha +\beta \right)} \\& =\frac{\Gamma \left( \alpha \right) \Gamma \left( \beta \right)}{\Gamma \left( \alpha +\beta \right)}\left[ \mathrm{\psi}_0\left( \alpha \right) -\mathrm{\psi}_0\left( \alpha +\beta \right) \right] \end{align*} To prove this full-on, you might want to take a look at Lebesgue's Dominated Convergence theorem in measure theorey to see when you can interchange the partial derivatives and the integral.

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$$ \begin{aligned} I & =\left.\frac{\partial}{\partial a} \int_0^1 x^a(1-x)^{n-1} d x\right|_{a=m-1} \\ & =\left.\frac{\partial}{\partial a} B(a+1, n)\right|_{a=m-1} \\ & =B(a+1, n)[\psi(a+1)-\psi(a+1+n)]_{a=m-1} \\ & =\frac{\Gamma(m) \Gamma(n)}{\Gamma(m+n)}[\psi(m)-\psi(m+n)] \end{aligned} $$

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