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 $$ \begin{align*} B(p,q) &= \int_{0}^{\infty}\frac{ x^{p-1} }{ (1+x)^{p-1} }\frac{ 1 }{ (1+x)^{q-1 } }\frac{ 1 }{ (1+x)^{2} }dx \\ &= \int_{0}^{\infty}\frac{ x^{p-1} }{ (1+x)^{p+q} }dx \\ &= \int_{0}^{\infty}\frac{ t^{p-1} }{ (1+t)^{p+q} }dt \end{align*} $$
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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 $B(p,q)=\dfrac{ \Gamma (p)\Gamma (q) }{ \Gamma (p+q) }$.
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