Problem source

Let 
$$
{\rm I}(a,b) = \int_0^1 t^{a}(1-t)^{b} \, \d t \;
\qquad (a\ge0,\ b\ge0) .$$ 
\begin{questionparts}
\item Show that  ${\rm I}(a,b)={\rm I}(b,a)$,
\item Show that ${\rm I}(a,b)={\rm I}(a+1,b)+{\rm I}(a,b+1)$.
\item Show that $(a+1){\rm I}(a,b)=b{\rm I}(a+1,b-1)$ 
when $a$ and $b$ are positive  
and hence calculate ${\rm I}(a,b)$ when $a$ and $b$ are positive integers.
\end{questionparts}

Solution source

\begin{questionparts}
\item Let $u = 1-t, \d u = -\d t$, then:
\begin{align*}
\mathrm{I}(a,b) &= \int_0^1 t^a(1-t)^b \d t \\
&= \int_{u=1}^{u=0} -(1-u)^a u^b \d u \\
&= \int_0^1(1-u)^a u^b \d u \\
&= \mathrm{I}(b, a)
\end{align*}
\item \begin{align*}
\mathrm{I}(a+1,b)+\mathrm{I}(a,b+1) &= \int_0^1 t^{a+1}(1-t)^b + t^a(1-t)^{b+1} \d t \\
&= \int_0^1 (t+(1-t))t^a(1-t)^b \d t \\
&= \int_0^1 t^a(1-t)^b \d t \\
&= \mathrm{I}(a,b)
\end{align*}
\item Integrating by parts with $\frac{du}{dt} = t^a, v = (1-t)^{b}$\begin{align*}
\mathrm{I}(a,b) &= \int_0^1 t^a (1-t)^b \d t \\
&= \left [ \frac{t^{a+1}}{a+1} (1-t)^b \right ]_0^1 + \int_0^1 \frac{t^{a+1}}{a+1} b(1-t)^{b-1} \\
&= \frac{b}{a+1} \int_0^1 t^{a+1}(1-t)^{b-1} \d t \\
&= \frac{b}{a+1} \mathrm{I}(a+1,b-1)
\end{align*}

Claim: $\mathrm{I}(a,b) = \frac{a!b!}{(a+b+1)!}$

Proof: Note that $I(a,0) = \frac{1}{a+1}$ so the formula holds for this case.

We will induct on $b$. The base case is done. Suppose that for $b = k$ our formula is true, ie: $\mathrm{I}(a,k) = \frac{a!k!}{(a+k+1)!}$ for all $a$ (and fixed $k$)

\begin{align*}
\mathrm{I}(a,k+1) &= \frac{k+1}{a+1} \mathrm{I}(a+1,k) \\
&= \frac{k+1}{a+1} \frac{(a+1)!k!}{(a+1+k+1)!} \\
&= \frac{a!(k+1)!}{(a+(k+1)+1)!}
\end{align*}

So the formula is true for $b=k+1$. Therefore, since it is true if $b=0$ and if $b=k$ is true then $b=k+1$ is true, it is true for all values of $b$.
\end{questionparts}