Question

Express the following integrals as $\mathit{\beta }$functions, and then, by (7.1), in terms of $\mathit{\Gamma }$ functions. When possible, use $\mathit{\Gamma }$ function formulas to write an exact answer in terms of $\pi ,2$, etc. Compare your answers with computer results and reconcile any discrepancies.

3. $\underset{\mathbf{0}}{\overset{\mathbf{1}}{\mathbf{\int }}}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}}}\mathit{d}\mathit{x}$.

Short answer

The value of integral is $I=\frac{1}{3}\beta \left(\frac{1}{3},\frac{1}{2}\right)$ and $I=\frac{1}{3}\times \frac{\Gamma \left(\frac{1}{3}\right)\Gamma \left(\frac{1}{2}\right)}{\Gamma \left(\frac{5}{6}\right)}$.

Step-by-step solution

Given Information

The given Integral is $\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{1-x^3}}$.

Definition of Beta function

Beta function is defined as $\mathit{\beta }\mathbf{(}\mathit{p}\mathbf{,}\mathit{q}\mathbf{)}\mathbf{=}\underset{\mathbf{0}}{\overset{\mathbf{1}}{\mathbf{\int }}}{\mathit{t}}^{\mathbf{p}\mathbf{-}\mathbf{1}}{\left(1-t\right)}^{\mathbf{q}\mathbf{-}\mathbf{1}}\mathit{d}\mathit{t}$.

Express integral as Beta integral

The Integral is$\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{1-x^3}}$.

Formula for beta function is$\beta (p,q)=\underset{0}{\overset{1}{\int }}{t}^{p-1}{(1-t)}^{q-1}dt$.

Let $x^3=ya$and find its derivate with respect to x.

$\begin{array}{c}{x}^3=y\\ 3x^{2}dx=dy\\ dx=\frac{1}{3y^{\frac{2}{3}}}dy\end{array}$

Substitute the value of $x^3$ in the integral.

The equation becomes as follows.

$\begin{array}{l}I=\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{1-y}}\frac{1}{3y^{\frac{2}{3}}}dy\\ I=\underset{0}{\overset{1}{\int }}\frac{y^{\frac{-2}{3}}}{\sqrt{1-y}}\frac{1}{3}dy\\ I=\frac{1}{3}\beta \left(\frac{1}{3},\frac{1}{2}\right)\end{array}$

Express integral as gamma integral.

The relation between $\beta$and $\Gamma$is the equation mentioned below.

$\beta (m,n)=\frac{\Gamma \left(m\right)\Gamma \left(n\right)}{\Gamma (m+n)}$

Substitute the values given below in the equation mentioned above.

$\begin{array}{c}m=\frac{1}{3}\\ n=\frac{1}{2}\end{array}$

The equation becomes as follows.

$\begin{array}{l}\beta \left(\frac{1}{3},\frac{1}{2}\right)=\frac{\Gamma \left(\frac{1}{3}\right)\Gamma \left(\frac{1}{2}\right)}{\Gamma \left(\frac{1}{3}+\frac{1}{2}\right)}\\ \beta \left(\frac{1}{3},\frac{1}{2}\right)=\frac{\Gamma \left(\frac{1}{3}\right)\Gamma \left(\frac{1}{2}\right)}{\Gamma \left(\frac{5}{6}\right)}\\ I=\frac{1}{3}\times \frac{\Gamma \left(\frac{1}{3}\right)\Gamma \left(\frac{1}{2}\right)}{\Gamma \left(\frac{5}{6}\right)}\end{array}$

Hence, the value of integral is $I=\frac{1}{3}\beta \left(\frac{1}{3},\frac{1}{2}\right)$and $I=\frac{1}{3}\beta \left(\frac{1}{3},\frac{1}{2}\right)$.