How to prove $\int_{0} ^{1}{\sqrt{1-x^4}dx}=\frac{1}{12}\sqrt {\frac{2}{\pi}}(\Gamma\left(\frac{1}{4}\right))^2$?

Question

To prove ,

$$\int_{0} ^{1}{\sqrt{1-x^4}dx}=\frac{1}{12}\sqrt {\frac{2}{\pi}}(\Gamma\left(\frac{1}{4}\right))^2$$

When we substitute $x^4$ with t

we get the equation $$\frac{1}{4}\int_{0}^1t^{\frac{-3}{4}}(1-t)^{\frac{1}{2}}dt$$ the expression is in the forms of the beta functions, But I am not able to solve ahead.

Answer 1

$$I=\frac { 1 }{ 4 } \int _{ 0 }^{ 1 } t^{ \frac { -3 }{ 4 } }(1-t)^{ \frac { 1 }{ 2 } }dt$$

Now, on using Beta function, we get $$I=\frac { 1 }{ 4 } B\left( \frac { 1 }{ 4 } ,\frac { 3 }{ 2 } \right) $$

Now, we use the relation of Beta and gamma function. $$I=\frac { 1 }{ 4 } \frac { \Gamma \left( \frac { 1 }{ 4 } \right) \Gamma \left( \frac { 3 }{ 2 } \right) }{ \Gamma \left( \frac { 7 }{ 4 } \right) } $$

Now, we use the formula: $\Gamma \left( n+1 \right) =n\Gamma \left( n \right) $

$$I=\frac { 1 }{ 4 } \frac { \Gamma \left( \frac { 1 }{ 4 } \right) \Gamma \left( \frac { 1 }{ 2 } \right) \times \frac { 1 }{ 2 } }{ \frac { 3 }{ 4 } \Gamma \left( \frac { 3 }{ 4 } \right) } $$

We use this value: $\Gamma \left( \frac { 1 }{ 2 } \right) =\sqrt { \pi } $

$$I=\frac { \sqrt { \pi } }{ 6 } \frac { \Gamma \left( \frac { 1 }{ 4 } \right) }{ \Gamma \left( \frac { 3 }{ 4 } \right) } $$

$$I=\frac { \sqrt { \pi } }{ 6 } \frac { { \left( \Gamma \left( \frac { 1 }{ 4 } \right) \right) }^{ 2 } }{ \Gamma \left( \frac { 1 }{ 4 } \right) \Gamma \left( \frac { 3 }{ 4 } \right) } $$

Now, by Euler's Reflection Formula, we get $$I=\frac { \sqrt { \pi } }{ 6 } \frac { { \left( \Gamma \left( \frac { 1 }{ 4 } \right) \right) }^{ 2 } }{ \frac { \pi }{ \sin { \left( \frac { \pi }{ 4 } \right) } } } $$

$$\boxed{\therefore \int _{ 0 }^{ 1 }{ \sqrt { 1-x^{ 4 } } dx } =\frac { 1 }{ 12 } \sqrt { \frac { 2 }{ \pi } } { \left( \Gamma \left( \frac { 1 }{ 4 } \right) \right) }^{ 2 }}$$