Question

Show that the four answers given in Section 1 for $\underset{\mathbf{0}}{\overset{\frac{\mathbf{\pi }}{\mathbf{2}}}{\mathbf{\int }}}\frac{\mathbf{d\theta }}{\sqrt{\mathbf{cos\theta }}}$ are all correct. Hints: For the beta function result, use(6.4). Then get the gamma function results by using (7.1) and the various Γ function formulas. For the elliptic integral, use the hint of Problem 17 with$\mathit{\alpha }\mathbf{=}\frac{\mathbf{\pi }}{\mathbf{2}}$.

Short answer

The solution is mentioned below.

$$\begin{gathered} I=\sqrt{2}K\left(\frac{1}{\sqrt{2}}\right) \\ I=\beta (\frac{1}{2},\frac{1}{4}) \\ I=\frac{\Gamma \frac{1}{2}\Gamma \frac{1}{4}}{\Gamma \frac{3}{4}} \end{gathered}$$

Step-by-step solution

Given Information

The value of integration is $\alpha =\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{1}{\sqrt{\cos \theta }}d\theta$.

Definition of elliptic form.

The elliptic form of the integral is defined as $\mathit{K}\mathbf{\left(}\mathit{k}\mathbf{\right)}\mathbf{=}\underset{\mathbf{0}}{\overset{\frac{\mathbf{\pi }}{\mathbf{2}}}{\mathbf{\int }}}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{k}}^{\mathbf{2}}{\mathbf{sin}}^{\mathbf{2}}\mathbf{\theta }}}\mathit{d}\mathit{\theta }$.

Find the value of Integral

The value of integration is$\alpha =\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{1}{\sqrt{\cos \theta }}d\theta$.

Substitute$\cos \theta =1-2{\sin }^2\frac{\theta }{2}$

The integral becomes as follows

$\alpha =\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{1}{\sqrt{1-2{\sin }^2\frac{\theta }{2}}}d\theta$

Substitute the values given below in the above equation.

$$\begin{gathered} x^2=2{\sin }^2\frac{\theta }{2} \\ dx=\frac{\sqrt{2}}{2}\cos \frac{\theta }{2}d\theta \end{gathered}$$

The equation becomes as follows.

$$\begin{gathered} I=\sqrt{2}K\left(\frac{1}{\sqrt{2}}\right) \\ I=\beta (\frac{1}{2},\frac{1}{4}) \\ I=\frac{\Gamma \frac{1}{2}\Gamma \frac{1}{4}}{\Gamma \frac{3}{4}} \end{gathered}$$

Hence, The solution is mentioned below.

$$\begin{gathered} I=\sqrt{2}K\left(\frac{1}{\sqrt{2}}\right) \\ I=\beta (\frac{1}{2},\frac{1}{4}) \\ I=\frac{\Gamma \frac{1}{2}\Gamma \frac{1}{4}}{\Gamma \frac{3}{4}} \\ I=\sqrt{2}\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{1-x^2}\sqrt{1-\frac{1}{2x^2}}}dx \end{gathered}$$