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?
To evaluate integral using Beta function
beta functiongamma functionintegration
Related Solutions
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{1}\bracks{1 - \pars{1 - x^{3}}^{\root{2}}}^{\root{3}}x^{2}\,\dd x} \\[5mm] \stackrel{x^{3}\ \mapsto\ y}{=}\,\,\,& {1 \over 3}\int_{0}^{1} \bracks{1 - \pars{1 - y}^{\root{2}}}^{\root{3}}\dd y \\[5mm] \stackrel{1 - y\ \mapsto\ z}{=}\,\,\,& {1 \over 3}\int_{0}^{1} \bracks{1 - z^{\root{2}}}^{\root{3}}\dd z \\[5mm] \,\,\,\stackrel{\pars{1 - z}^{\root{2}}\ \mapsto\ t}{=}\,\,\,& {\root{2} \over 6}\int_{0}^{1}t^{\root{3}}\pars{1 - t}^{\root{2}/2 - 1} \,\dd t \\[5mm] = &\ \bbx{{\root{2} \over 6}\,\mrm{B}\pars{\root{3} + 1,{\root{2} \over 2}}} \approx 0.1546 \end{align}
$\ds{\mrm{B}}$ is the Beta Function.
Deriving the Surface Area of a Sphere
Start with the $1$ dimensional integral $$ \int_{-\infty}^\infty e^{-\pi x^2}\,\mathrm{d}x=1\tag1 $$ and take the product of $n$ copies: $$ \int_{-\infty}^\infty\int_{-\infty}^\infty\cdots\int_{-\infty}^\infty e^{-\pi \left(x_1^2+x_2^2+\cdots+x_n^2\right)}\,\mathrm{d}x_1\,\mathrm{d}x_2\cdots\,\mathrm{d}x_n=1\tag2 $$ Since the integrand is constant over spherical shells, we can easily write it as a spherical integral $$ \int_0^\infty e^{-\pi r^2}\omega_{n-1}r^{n-1}\,\mathrm{d}r=1\tag3 $$ where $\omega_{n-1}$ is the surface area of the $n-1$ dimensional unit sphere.
Thus, we get
$$
\begin{align}
1
&=\int_0^\infty e^{-\pi r^2}\omega_{n-1}r^{n-1}\,\mathrm{d}r\tag{4a}\\
&=\frac{\omega_{n-1}}2\int_0^\infty e^{-\pi r^2}r^{n-2}\,\mathrm{d}r^2\tag{4b}\\
&=\frac{\omega_{n-1}}2\int_0^\infty e^{-\pi r}r^{n/2-1}\,\mathrm{d}r\tag{4c}\\
&=\frac{\omega_{n-1}}{2\pi^{n/2}}\int_0^\infty e^{-r}r^{n/2-1}\,\mathrm{d}r\tag{4d}\\
&=\frac{\omega_{n-1}}{2\pi^{n/2}}\Gamma(n/2)\tag{4e}
\end{align}
$$
Explanation:
$\text{(4a)}$: copy $(3)$
$\text{(4b)}$: $r\,\mathrm{d}r=\frac12\mathrm{d}r^2$ and pull the constant out front
$\text{(4c)}$: substitute $r\mapsto r^{1/2}$
$\text{(4d)}$: substitute $r\mapsto r/\pi$
$\text{(4e)}$: definition of the Gamma function
Therefore, we get $$ \omega_{n-1}=\frac{2\pi^{n/2}}{\Gamma(n/2)}\tag5 $$
Application to the Question
We can write the integral from the question as an integral on spherical shells:
$$
\begin{align}
&\int_0^\infty\overbrace{(a+r)^me^{-\lambda r^{2k}}}^{\substack{\text{integrand on a}\\\text{shell of radius $r$}}}\overbrace{\omega_{n-1}r^{n-1\vphantom{r^2}}}^{\substack{\text{surface area}\vphantom{g}\\\text{of the shell}}}\,\mathrm{d}r\tag{6a}\\
&=\frac{\omega_{n-1}}{2k}\sum_{j=0}^m\binom{m}{j}\int_0^\infty a^{m-j}r^{\frac{j+n}{2k}-1}e^{-\lambda r}\,\mathrm{d}r\tag{6b}\\
&=\frac{\omega_{n-1}}{2k}\sum_{j=0}^m\binom{m}{j}a^{m-j}\lambda^{-\frac{j+n}{2k}}\int_0^\infty r^{\frac{j+n}{2k}-1}e^{-r}\,\mathrm{d}r\tag{6c}\\
&=\frac{\omega_{n-1}}{2k}\sum_{j=0}^m\binom{m}{j}a^{m-j}\lambda^{-\frac{j+n}{2k}}\Gamma\!\left(\frac{j+n}{2k}\right)\tag{6d}
\end{align}
$$
Explanation:
$\text{(6a)}$: integral from the question
$\text{(6b)}$: substitute $r\mapsto r^{\frac1{2k}}$ and apply the Binomial Theorem
$\text{(6c)}$: substitute $r\mapsto r/\lambda$
$\text{(6d)}$: definition of the Gamma function
Best Answer
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