Relation Between Beta And Gamma Function

The beta and gamma functions are primarily used to simplify complex integrals. Unlike real number and integrals, complex expressions are often challenging to solve. Due to the fact that a complex expression has both real and imaginary parts, it becomes difficult to handle both the parts simultaneously.

Consequently, the need for and the emergence of beta and gamma function can be traced owing to this challenging situation. With these functions, one can quickly derive even the complex functions and integrals with ease. 

Overall, they have eased the process of finding the solution to complex integrals. Thus, students must have a clear idea about both the beta gamma function, as well as calculus, to be able to solve these numericals.

What is the Relation Between Beta and Gamma Function?

To know the relation between beta and gamma function, students have to understand that neither of these are standalone concepts. In order to simplify a complex integral, both these functions have to be put into use.

In simple words, both these functions are interdependent and are applied simultaneously to derive a solution. Look at the image below that shows the relation between them clearly.

Herein, it is vital to be aware of the sign conventions for each function. Since the names of both these functions originated from the Greek letters, you are required to keep a note of the same. 

\[\begin{gathered}

  B\left( {x + 1,y} \right) = \left[ {\left( {\frac{{ - {{\left( {1 - t} \right)}^{x + y}}}}{{x + y}}} \right){{\left( {\frac{t}{{1 - t}}} \right)}^x}} \right]_0^1 + \int_0^1 {\frac{x}{{x + y}}{{\left( {1 - t} \right)}^{x + y}}{{\left( {\frac{t}{{1 - t}}} \right)}^{x - t}}\frac{1}{{{{\left( {1 - t} \right)}^2}}}dt}  \hfill \\

  {\text{                   = }}\frac{x}{{x + y}}\int_0^1 {{t^{x - 1}}} \left[ {{{\left( {1 - t} \right)}^{x + y}}{{\left( {1 - t} \right)}^{1 - x}}{{\left( {1 - t} \right)}^{ - 2}}} \right]dt \hfill \\

  {\text{                   = }}\frac{x}{{x + y}}\int_0^1 {{t^{x + 1}}{{\left( {1 - t} \right)}^{y - 1}}dt}  \hfill \\

  {\text{                   = }}\frac{x}{{x + y}}B\left( {x,y} \right). \hfill \\

  {\text{ Now let us take y to be fixed}}{\text{. To satisy }}\Gamma \left( {x + 1} \right) = x\Gamma \left( x \right),{\text{ let us set }}f\left( x \right){\text{ to be}} \hfill \\

  {\text{                                         }}f\left( x \right) = B\left( {x,y} \right)\Gamma \left( {x + y} \right) \hfill \\ 

\end{gathered} \]

Therefore, the relation between them is established to solve complex problems in both Mathematics and Physics. The beta gamma function formula is highly useful and all you have to do is be aware of the functions and their usability. 

What are Beta and Gamma Function used for?

As already mentioned, the beta gamma formula is used in solving complex problems. Solving integrals that are based on trigonometry are usually solved with these functions. Here is an example for the same for a clear understanding. 

\[\int_0^{\frac{\pi }{2}} {\frac{{\sqrt {\tan x} }}{{{{\left( {\cos x + \sin x} \right)}^2}}}dx} \]

Using the substitution \[t = \tan x \to dx = {\cos ^2}\left( x \right)dt\]therefore,

\[\int_0^{\frac{\pi }{2}} {\frac{{\sqrt {\tan x} }}{{{{\left( {\cos x + \sin x} \right)}^2}}}dx}  = \int_0^{\frac{\pi }{2}} {\frac{{\sqrt {\tan x} }}{{{{\cos }^2}x{{\left( {1 + \tan x} \right)}^2}}}dx = \int_0^\infty  {\frac{{{t^{\frac{1}{2}}}}}{{{{\left( {1 + t} \right)}^2}}}} dt} \]

And by using the alternate definition of the beta function we can rewrite the above integral to fit the defined 

\[\int_0^\infty  {\frac{{{t^{\frac{3}{2} - 1}}}}{{{{\left( {1 + t} \right)}^{\frac{3}{2} + t}}}}dt = B\left( {\frac{3}{2},\frac{1}{2}} \right)} \]

Then by switching to gamma function 

\[\frac{{\Gamma \left( {\frac{3}{2}} \right)\Gamma \left( {\frac{1}{2}} \right)}}{{\Gamma \left( 2 \right)}} = \frac{{\frac{1}{2}\Gamma \left( {\frac{1}{2}} \right)\Gamma \left( {\frac{1}{2}} \right)}}{{1!}} = \frac{{\frac{1}{2}\sqrt \pi  \sqrt \pi  }}{1} = \frac{\pi }{2}\]

The above image shows the evaluation of a trigonometric integral using beta function and gamma function. Students must remember that they have to choose the trigonometric function in a numerical in such a way that the function can be easily substituted with other functions. 

Here, tan x is substituted as it is a function that can relate both sin x and cos x in the form of -

tan x = sin x / cos x

Besides this, the beta and gamma functions can also be used to solve factorial numerical. Consider the example given below.

Example 

Compute B(5,7)

If we go by the define of beta function to compute B(5,7). We will have to solve the following integral

Which is very tedious work. Here is the relation the relation of beta   function with gamma function comes in handy

\[B\left( {5,7} \right) = \frac{{4!6!}}{{11!}} = \frac{{4!}}{{11 \times 10 \times ........... \times 7}} = \frac{1}{{2310}}.\]

Even in this example given in the above picture, you can see the beta function is used to simplify the question. 

Therefore, students are required to go through such gamma function questions and answers for a clear understanding. Also, following each step is highly essential in solving the problems on their own during the exam.

Test Your Understanding

Write True or False

1. An alpha function is used to solve complex integrals.

2. The beta function is called Euler’s integral of the first kind.

3. Trigonometric integrals cannot be solved using a beta function.

Answers - False (beta and gamma function), True, False (can be solved)

For a more detailed explanation on the beta and gamma functions, check out our online study materials. They will provide you with a more in-depth insight into the formula as well as derivation.

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