📂Functions

Representation of the Beta Function in the Form of an Improper Integral

Theorem

Beta Function: $$B(p,q)=\int_{0}^{1}t^{p-1}(1-t)^{q-1}dt\quad \cdots (1)$$

The beta function can be expressed as an improper integral as follows: $$B(p,q)=\int_{0}^{\infty}\frac{ t^{p-1} }{ (1+t)^{p+q}}dt\quad \cdots (2)$$

Explanation

Using the above formula makes it easier to obtain difficult integral values. The proof is not difficult.

Proof

Let’s substitute $(1)$ with $t=\frac{x}{1+x}$. Then, $1-t=\frac{1}{1+x}$, and the range of integration changes to $\int_{0}^{1}\rightarrow \int_{0}^{\infty}$. Also, since $\displaystyle \frac{ d t }{ d x }=\frac{1}{1+x}-\frac{x}{(1+x)^{2}}=\frac{1}{(1+x)^{2}}$, we have $dt=\dfrac{1}{(1+x)^{2}}dx$, and substituting this into $(1)$ yields $$\begin{align*} B(p,q) &= \int_{0}^{\infty}\frac{ x^{p-1} }{ (1+x)^{p-1} }\frac{ 1 }{ (1+x)^{q-1 } }\frac{ 1 }{ (1+x)^{2} }dx \\ &= \int_{0}^{\infty}\frac{ x^{p-1} }{ (1+x)^{p+q} }dx \\ &= \int_{0}^{\infty}\frac{ t^{p-1} }{ (1+t)^{p+q} }dt \end{align*}$$

■

Examples

Calculate $\displaystyle \int_{0}^{\infty}\frac{ x^{5} }{ (1+x)^{8} }dx$.

Solution

This integral, in the case of $(2)$ to $p=6$, $q=2$, results in $$\begin{align*} \int_{0}^{\infty}\frac{ x^{5} }{ (1+x)^{8} }dx &= B(6,2) \\ &= \frac{ \Gamma (6)\Gamma (2) }{ \Gamma (6+2) } \\ &= \frac{ 5!1! }{ 7! } \\ &= \frac{ 1 }{ 42} \end{align*}$$ The second equality uses the relationship $B(p,q)=\dfrac{ \Gamma (p)\Gamma (q) }{ \Gamma (p+q) }$.

■