For what values of $m$ and $n$ is $\int_{0}^{1} x^{m-1}(1-x)^{n-1}\log x\ dx$ convergent?

Question

I cannot really understand how to prove this I know that $\int_{0}^{1} x^{m-1} (1-x)^{n-1}\ dx$ is $\beta$ function.

Answer 1

Near zero, the integrand is $\sim x^{m-1}\log x$. As $|\log x|=O(x^{-\epsilon})$ for small positive $x$, the integral converges near zero iff $\text{Re}(m)>0$, as for the beta integral.

Near one, the integrand is $\sim -(1-x)^n$. This converges near one iff $\text{Re}(n)>-1$.

Answer 2

The integral $$ B(m,n)=\int_{0}^{1}x^{m-1}(1-x)^{n-1}\,dx = \frac{\Gamma(m)\,\Gamma(n)}{\Gamma(m+n)} $$ is convergent as soon as $\color{red}{\text{Re}(m),\text{Re}(n)>0}$, granting the integrability of $x^{m-1}(1-x)^{n-1}$ over $(0,1)$. Under these assumptions it is a differentiable function, also because $B(\cdot,n)$ and $B(m,\cdot)$ are moments, hence log-convex functions by the Cauchy-Schwarz inequality. It follows that $$ \int_{0}^{1} x^{m-1}(1-x)^{n-1}\log(x)\,dx =\frac{\partial}{\partial m}B(m,n)$$ and by logarithmic differentiation we get: $$\begin{eqnarray*} \int_{0}^{1} x^{m-1}(1-x)^{n-1}\log(x)\,dx &=& \frac{\Gamma(m)\,\Gamma(n)}{\Gamma(m+n)}\left[\psi(m)-\psi(m+n)\right]\\&=&-\frac{\Gamma(m)\,\Gamma(n+1)}{\Gamma(m+n)}\sum_{s\geq 0}\frac{1}{(s+m)(s+m+n)}\end{eqnarray*}$$ where $\psi(x)=\frac{d}{dx}\log\Gamma(x)$ is the digamma function.