Proving the Integral Representation of the Beta Function Using the Gamma Function and Complex Analysis

The integral representation of the Beta function, ( B(x, 1-x) ), can be proven through the properties of the Gamma function and complex analysis. This article explores the rigorous derivation and provides a detailed explanation of these methods.

Introduction to the Gamma Function and Beta Function

The Gamma function, denoted as ( Gamma(x) ), is a generalization of the factorial function to complex and real numbers. It is defined as:

( Gamma(x) int_0^{infty} t^{x-1} e^{-t} dt )

The Beta function, ( B(x, y) ), is closely related to the Gamma function and is defined as:

( B(x, y) int_0^1 t^{x-1} (1-t)^{y-1} dt )

Using the Hypergeometric Function to Integrate

The integral of interest is given by:

( int_0^{infty} frac{t^{x-1}}{1 t} dt )

Using the Hypergeometric2F1 function, we can express this integral in a different form. The Hypergeometric2F1 function, ( {}_2F_1(a, b; c; z) ), is a special function that appears in many integrals:

( int_0^{infty} frac{t^{x-1}}{1 t} dt t^x/x - t^{1-x} cdot {}_2F_1(1, 1; x 1; -t) / x )

This equation holds if ( 0 Re(x) 1 ).

Special Case Analysis

For ( x 1/2 ), the integral evaluates to:

( int_0^{infty} frac{t^{1/2-1}}{1 t} dt pi )

This is derived using limits as follows:

( lim_{t to 0} t^{1/2}/(1/2) - t^{1-1/2} cdot {}_2F_1(1, 1; 3/2; -t) / (1/2) 0 )

and

( lim_{t to infty} t^{1/2}/(1/2) - t^{1-1/2} cdot {}_2F_1(1, 1; 3/2; -t) / (1/2) pi )

Beta Function and Gamma Function Connection

The relationship between the Beta function and the Gamma function is given by:

( B(x, 1-x) frac{Gamma(x) cdot Gamma(1-x)}{Gamma(1)} Gamma(x) cdot Gamma(1-x) )

From Euler's Reflection Formula (the RHS is a well-known result in mathematics), we have:

( Gamma(x) cdot Gamma(1-x) frac{pi}{sin(pi x)} ) for ( 0 x 1 )

Complex Analysis Approach

We can use complex analysis to evaluate the integral. Consider the integral:

( mathcal{I} int_0^{infty} frac{t^{x-1}}{1 t} dt )

Using a keyhole contour, we can derive the same result. The key idea is to use the contour integral:

( oint_{mathcal{C}} frac{z^{x-1}}{1-z} dz )

Where the contour is a keyhole contour around the branch cut on the negative real axis. Evaluating this integral using Cauchy's Residue Theorem, we get:

( -2pi i int_{mathcal{C}_1} frac{z^{x-1}}{1-z} dz int_{mathcal{C}_2} frac{z^{x-1}}{1-z} dz int_{mathcal{C}_3} frac{z^{x-1}}{1-z} dz int_{mathcal{C}_4} frac{z^{x-1}}{1-z} dz )

Solving the contour integrals, we find that:

( int_0^{infty} frac{t^{x-1}}{1 t} dt frac{2pi i}{sin(pi x)} )

Using Euler's formula, ( e^{ixpi} cos(pi x) i sin(pi x) ), we can simplify it to:

( int_0^{infty} frac{t^{x-1}}{1 t} dt frac{pi}{sin(pi x)} )

Conclusion

The integral representation of the Beta function in terms of the Gamma function can be established using the Hypergeometric2F1 function and verified through complex analysis. This approach provides a powerful tool for understanding and solving integral equations in advanced mathematics.