Convergence test for improper integral

Question

How to show the below improper integral is convergent?

$$\int_{0}^{1}x^{m-1}(1-x)^{n-1}\log(x)dx$$

Answer 1

For $ n\in\mathbb{N} $, define the function $ f_{n} $, as follows : $$ f_{n}:\mathbb{R}_{+}^{*}\rightarrow\mathbb{R},\ t\mapsto\int_{0}^{1}{x^{t-1}\left(1-x\right)^{n-1}\,\mathrm{d}x} $$

For $ t\in\mathbb{R}_{+}^{*} $ and $ k\in\mathbb{N} $, we have : \begin{aligned}f_{n-k}\left(t+k\right)&=\int_{0}^{1}{x^{t+k-1}\left(1-x\right)^{n-k-1}\,\mathrm{d}x}\\&=\left[\frac{x^{t+k}}{t+k}\left(1-x\right)^{n-k-1}\right]_{0}^{1}+\frac{n-k-1}{t+k}\int_{0}^{1}{x^{t+k}\left(1-x\right)^{n-k-2}\,\mathrm{d}x}\\ f_{n-k}\left(t+k\right)&=\frac{n-k-1}{t+k}f_{n-k-1}\left(t+k+1\right)\\ \iff \prod_{k=0}^{n-2}{\frac{f_{n-k}\left(t+k\right)}{f_{n-k-1}\left(t+k+1\right)}}&=\prod_{k=0}^{n-2}{\frac{n-k-1}{t+k}}\\ \iff\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{f_{n}\left(t\right)}{f_{1}\left(t+n-1\right)}&=\frac{\left(n-1\right)!}{\prod\limits_{k=0}^{n-2}{\left(t+k\right)}}\\ \iff\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f_{n}\left(t\right)&=\frac{\left(n-1\right)!}{\prod\limits_{k=0}^{n-1}{\left(t+k\right)}}\end{aligned}

Defining another function : $$ g_{n}:\mathbb{R}_{+}^{*}\rightarrow\mathbb{R},\ t\mapsto\prod_{k=0}^{n-1}{\left(t+k\right)} $$

Notice $ \ln{\left(g_{n}\left(t\right)\right)}=\sum\limits_{k=0}^{n-1}{\ln{\left(t+k\right)}} $, differentiating both sides leads to : $$ \frac{g_{n}'\left(t\right)}{g_{n}\left(t\right)}=\sum_{k=0}^{n-1}{\frac{1}{t+k}} $$

Now getting back to $ f_{n} $, differentiating it gives : $$ f_{n}'\left(t\right)=-\left(n-1\right)!\frac{g_{n}'\left(t\right)}{g_{n}^{2}\left(t\right)}=-\frac{\left(n-1\right)!}{\prod\limits_{k=0}^{n-1}{\left(t+k\right)}}\sum_{k=0}^{n-1}{\frac{1}{t+k}} $$

Thus for all $ t\in\mathbb{R}_{+}^{*} $, and $ n\in\mathbb{N} $, we have : \begin{aligned}\int_{0}^{1}{x^{t-1}\left(1-x\right)^{n-1}\ln{x}\,\mathrm{d}x}=\int_{0}^{1}{\frac{\partial}{\partial t}\left(x^{t-1}\left(1-x\right)^{n-1}\right)\mathrm{d}x}&=\frac{\mathrm{d}f_{n}}{\mathrm{d}t}\left(t\right)\\&=-\frac{\left(n-1\right)!}{\prod\limits_{k=0}^{n-1}{\left(t+k\right)}}\sum_{k=0}^{n-1}{\frac{1}{t+k}}\end{aligned}

Setting $ t=m\in\mathbb{N}^{*} $, expressing everything in terms of factorials and the harmonic sequence gives : $$ \int_{0}^{1}{x^{m-1}\left(1-x\right)^{n-1}\ln{x}\,\mathrm{d}x}=-\frac{\left(n-1\right)!\left(m-1\right)!}{\left(n+m-1\right)!}\left(H_{n+m-1}-H_{m-1}\right) $$