Evaluation the Gaussian Integral through the use of Power Series and Master Theorem

Abstract

The Gaussian Integral is an important tool, imperative to the study of probability and statistics.It and similar integrals are computed usually through the method of double integrals and spherical coordinates. This paper sets forward an avenue of exploration highlighting the versatility and applications of the Ramanujan Master Theorem, a method to find the value of infinite series using the Gamma function which is an analytic extension of the factorial. In using this theorem, we solve the Gaussian Integral and put forward a new approach to solving improper integrals of similar classes by converting them into infinite series.

I. Introduction

Commonly known as the Standard Distribution or Bell Curve, the Gaussian function

[1] is defined by the function e−x2also noted as exp x2 , where e denotes Euler’s number. The function represents a probability distribution uniformly clustered about the mean, distributed equally in both directions.

In this investigation we will attempt to calculate the definite integral of this function, known as the Gaussian Integral [1], which is expressed as (where ∞ denotes infinity)

Z

e−x2dx. 1

Figure 1: The Gaussian Function (e−x2) [3]

Although the value of this integral has already been calculated in the past, a new approach through the use of Power Series [2] will give us more insight on how we can work to approximate and evaluate this otherwise seemingly impossible integral.

II. Visualisation and Context

The Gaussian function has one turning point at (0,1) and a horizontal asymptote, denoted as the value of x goes to infinity as y goes to 0, this can be visualised in Figure 1. Calculat ing the Gaussian integral is useful for calculating expectations of continuous probability distributions related to the normal distribution, such as the log-normal distribution or binomial distribution.

III. Method | Expression through Taylor Series

An expression for e−x2 through the Taylor Series can be found using substitution into the ex Taylor series [2]. The ex Taylor series is shown below, reading as the sum from n = 0 to infinity of xn over n factorial.

ex =

n!= 1 + x +x2

2! +x3

3! +x4

xn

4! · · ·

n=0

2

If we substitute −x2into the series we can get the expression for e−x2

e−x2=

n=0

which we can simplify to

(−x2)n

n!= 1 +(−x2)1

1! +(−x2)2

2! +(−x2)3

3! · · ·

e−x2=

n!= 1 − x2 +x4 2! x6

n=0

(1)n· x2n

3! +x8

4! · · ·

We can now integrate this series, term by term to find a Taylor series expression for the indefinite Gaussian Integral

Z

e−x2dx. =

n=0

Z(1)n· x2n

n!dx. =

n=0

z}|{

x

1!

n=1

z}|{

x3

3+

n=2

z }| {

x5

5 · 2!

n=3

z }| {

x7

7 · 3! +

n=4

z }| {

x9

9 · 4! · · · + C

We can see that this follows a pattern and can express the integrated series to be

Z

e−x2dx. =

n=0

(1)n· x2n+1

(2n + 1) · n!(1)

IV. Testing for Convergence

To see if this series is convergent we can utilise the ratio test [4], which states that if we let limn→|an+1|

|an|= L (2)

where an signifies the nth term in an infinite series. Then the series is divergent if L is greater than one, absolutely convergent if L is less than one and the test is inconclusive if L is equal to one.

We can equate the ratio test in Equation (2) to

limn→|an+1| · 1|an|= L (3)

We will now substitute our integrated series from Equation (1) into Equation (3)

limn→

(1)n+1(x2n+2)

(2n + 2)(n + 1)!·(2n + 1)n! (1)n(x2n+1)

= L

3

We can cancel out the alternating (1)n values due to the absolute value function

limn→

Now combining fractions

x2n+2

(2n + 2)(n + 1)!·(2n + 1)(n!) x2n+1

Simplifying to

limn→

(x2n+2)(2n + 1)(n!) (2n + 2)((n + 1)!)(x2n+1)

limn→

2n + 1 2n + 2

x

(n + 1)

Which is

limn→

2nx + x (2n2 + 4n + 2)

As the denominator is a quadratic while the numerator is only a linear function, to further evaluate, we must apply L’Hôpital’s rule (which states that if a limit is of indeterminate form, we can differentiate the top and bottom until it is not) twice.

limn→

2

∂n2 (2nx + x)

2

∂n2 (2n2 + 4n + 2)

!

=04

Therefore the value of “L” is 0, and referring back to the ratio test, this value is under 1 and thus the series Equation (1) converges for all real values.

V. The Ramanujan Master Theorem

The Ramanujan Master Theorem [5], primarily used and introduced by mathematician Srinivasa Ramanujan is a technique that analytically expresses the Mellin transform of an analytic function.

It states that if

φ(n)

n!(x)n(4)

4

f(x) =

n=0

then the Mellin transform is given by

Z 0

xs−1f(x)dx = Γ(s)φ(−s) (5)

*where Γ denotes the Euler-Gamma function. [6]

The Mellin transform [7] in the case of this series, is the same as the evaluated series from 0 to infinity, the s is the same as the exponent of the −x, but is written as s to denote that it can also be a complex number.

The reason this is useful is that we can use this to evaluate our infinite series from Equation (1) and hence the Gaussian integral, since this is evaluating the series with respect to x from 0 to ∞ and the Mellin transform is the same as a direct evaluation of our series. As the bounds of the integral are from 0 to infinity it is practically the same as half of the Gaussian Integral (which has bounds from ∞ to ∞) as it is an even function.

We can now attempt to evaluate our integral using the Master Theorem. Using our integrated series of

f(x) =

n=0

(x2n+1)

(2n + 1) · n!

And through substitution and rearrangement, letting it equal to

φ(n)

·n!· (x)2n+1

where

f(x) =

n=0

φ(n) = 1

(2n + 1)

This is almost an identical form to the form given in the definition of the Master Theorem but a problem arises in the exponent of the x term, where the power is not simply a number rather, rather a linear equation.

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We can now substitute f(x) into our Mellin transform

Z 0

(x)(2n+1)1

n=0

φ(n)

n!dx

As the summation now does not concern x, we can move it outside of the integral for clearer analysis

(x)(2n+1)1dx

Simplifying the exponent to

n=0

φ(n) n!

Z 0

n=0

φ(n) n!

Z 0

(x)2ndx

In order to simplify the 2n exponent, we can use u-substitution. where we can let u = x2.

Hence

du

dx= 2x

dx =du

2x

we can note that x is equal to u, therefore

dx =du

u12

Hence

(u)n12u12du

Factorising and simplifying to

n=0

φ(n) n!

Z 0

un− 12 du

6

n=0

−φ(n) n!

1 2

Z 0

This is in specifically the form of Equation (5). The unterm is a part of the initial series so in term we are left with u12 . To find our s value, we can simply add 1 to the exponent to get

s =12

.

Henceforth the result of our integral would be

Z 0

e−x2dx = Γ

1 2

φ

12 (6)

*where Γ denotes the Euler-Gamma function. [6] VI. The Gamma Function

12 (7)

Z

0

We can recognise that

e−x2dx = Γ

1 2

φ

φ

12 =12

And now need to calculate

Γ

1 2

As the gamma function is defined as (n − 1)! it seems unintuitive to have a fractional factorial, although the gamma function is simply an analytic extension of the factorial, thought of to match every significant property of the factorial (meeting its value at all integer values). Hence multiple identities for the Gamma function have been derived to further understand and interpret fractional factorials. We will explore two methods to

compute the value of Γ

1 2

.

i. Euler’s Reflection Formula

First highlighted by mathematician Leonhard Euler, Euler’s Reflection formula [8] states 7

Γ(z)Γ(1 − z) = π

sin()(z/Z)

Through this, if we substitute 12for z we are given

Γ

1 2

Γ

1 2

=π π

which is

Γ

1 2

2

sin

=π1

2

giving (knowing the value is positive since the function is always positive)

1

=π

Γ

2

Hence

Z 0

e−x2dx =

π 2

As the Gaussian function is even, we can then conclude the value of the Gaussian Integral.

e−x2dx =π

ii. The Beta Function

Z

Another method to calculate the Gamma of a half is through the beta function [9], a special function with similarities to the Gamma function. The Beta function is noted to be

B(x, y) =

A key property of the beta function is

Z 1 0

tx−1· (1 − t)y−1dt

B(x, y) = Γ(x)Γ(y)

Γ(x + y)

8

Hence we can substitute x and y for 12into the beta function to find our value. 2,12 =Z 10t12 · (1 − t)12 dt

B

1

=

=

=

Z 1 0

Z 1 0

Z 1 0

1t·1

p(1 − t)dt

1

tp(1 − t)dt

1

p(t − t2)dt

Let u =t dt = 2udu

= 2 = 2

Z 1 0

Z 1 0

u

u1 − u2du

1

1 − u2du

=

arcsin u

10 · 2

= π

Henceforth we then know

1

1

π =

Γ

2

Γ

Γ(1)

2

We know Γ(1) is simply 0! which is 1, so π =

Γ

1 2

2

9

knowing our integral must be positive, we have

1

=π

Γ

2

Hence

Z 0

e−x2dx =

π 2

As the Gaussian function is even, we can then conclude the value of the Gaussian Integral.

Z

e−x2dx =π

VII. Conclusion

In conclusion, this paper highlights the solution of the Gaussian integral through a heuristic theorem, The Ramanujan Master Theorem (RMT), this highlights the versatility and applicability of the RMT to infinite series and improper integrals.

This explores a method of solving improper integrals (even those without elementary antiderivatives as shown in this case), convert the integral into an infinite series problem and through the use of the RMT, compute the value of these integrals leveraging the Euler-Gamma function, in this paper two methods to compute non-integer values of the Gamma Function are also highlighted.

This is significant as integrals such as the Gaussian Integral but also other integrals of similar classification are heavily used in the study of statistics, applied mathematics and analysis and yielding the solutions to these integrals gives great value in determining the dynamics of the integrand.

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References

[1] Eric Weisstein. Gaussian Integral – from Wolfram MathWorld. 19 Nov. 2020, https://mathworld.wolfram.com/GaussianIntegral.html.

[2] Taylor Series. Maclaurin Expansion of Ex. 18 Nov. 2020, https://blogs.ubc.ca/infiniteseriesmodule/units/unit-3-power-series/taylor series/maclaurin-expansion-of-ex/.

[3] “Desmos | Beautiful, Free Math.” Beautiful, Free Math, https://www.desmos.com. Accessed 19 Nov. 2020.

[4] Calculus II – Ratio Test. 31 May 2018, https://tutorial.math.lamar.edu/classes /calcii/RatioTest.aspx.

[5] —. Ramanujan’s Master Theorem – from Wolfram MathWorld. 19 Nov. 2020, https://mathworld.wolfram.com/RamanujansMasterTheorem.html.

[6] Eric Weisstein. 2021a. “Gamma Function – from Wolfram MathWorld.” March 15. https://mathworld.wolfram.com/GammaFunction.html.

[7] 2021b. “Mellin Transform – from Wolfram MathWorld.” March 15. https://mathworld.wolfram.com/MellinTransform.html.

[8] “Euler’s Reflection Formula | Brilliant Math Science Wiki.” Brilliant, https://brilliant.org/wiki/eulers-reflection-formula/. Accessed 28 Nov. 2020.

[9] The Gamma and Beta Functions – Mathonline. http://mathonline.wikidot.com/the gamma-and-beta-functions: :text=Definition

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