# Euler-Mascheroni Constant

The **Euler-Mascheroni constant**, also known as Euler's constant or simply "gamma," is a constant that appears in many problems in analytic number theory and calculus. It is denoted by $\gamma,$ and the first few digits of this constant are as follows:

$\gamma \approx 0.57721566490153286060651209008240243 \ldots$

Gamma is strongly related to the natural logarithm function and the harmonic numbers, and is often defined in these terms. There is no closed form expression for the $n^\text{th}$ harmonic number, but gamma can be used to give an estimate of the $n^\text{th}$ harmonic number. Beyond this, the applications of gamma in mathematics and practical problems are wide and varied.

For as much as gamma has been studied and applied to problems in mathematics, not much is known about the properties of the number itself. It is not known if gamma is algebraic or transcendental. It is not even known if gamma is irrational like other important mathematical constants such as $\pi$ and $e.$

## Definition

Harmonic numbers are a sequence of numbers formed by summing the reciprocals of positive integers.

Let $H_n$ be the $n^\text{th}$ harmonic number, then

$\begin{aligned} H_n&=\sum\limits_{k=1}^{n}\frac{1}{k}\\ &=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ \cdots + \frac{1}{n}. \end{aligned}$

$H_n$ can also be defined by the recurrence relation

$H_n=H_{n-1}+\frac{1}{n}, \ H_1=1.\ _\square$

Gamma is most commonly defined as the limit of the difference between the $n^\text{th}$ harmonic number and the natural logarithm of $n,$ as $n$ approaches infinity.

Let $\gamma$ be the

Euler-Mascheroni constant, otherwise known as Euler's constant. It is defined as follows:$\gamma = \lim_{n\to\infty} \left(-\ln n+\sum_{k=1}^n \dfrac{1}{k} \right)\approx 0.577216.$

In integral notation,

$\gamma = \int_1^\infty \left(\dfrac{1}{\lfloor x\rfloor}-\dfrac{1}{x}\right)\, dx. \ _\square$

## Estimating Harmonic Numbers

Due to gamma's relation to harmonic numbers, it is used in many problems that require an estimation of a harmonic number. There is no closed form expression for the $n^\text{th}$ harmonic number, so it is necessary to have an accurate estimation that can be computed efficiently.

For relatively large $n,$

$H_n \approx \ln(n)+\gamma+\frac{1}{2n}.$

Note: Although $\ln(n)+\gamma$ converges to $H_n$, the addition of $\frac{1}{2n}$ gives a faster convergence, and thus a more accurate estimation.

Harmonic numbers are applicable in some famous mathematics problems:

Harmonic numbers are also applicable in some practical problems.

The amount of rain that falls in a certain town over the course of a year is recorded every year for 100 years. If the amount of rain each year is uniformly random, how many record rainfalls would you expect to see over that time period?

Assuming that the amount of rainfall every year is uniformly random,

- the $1^\text{st}$ year is guaranteed to set a record, because it is the only year recorded,
- the $2^\text{nd}$ year has a $\frac{1}{2}$ chance to be higher than the 1st year,
- the $3^\text{rd}$ year has a $\frac{1}{3}$ chance to be higher than the other 2 years,
- the $4^\text{th}$ year has a $\frac{1}{4}$ chance to be higher than the other 3 years,
- and so on.
Calculating the expected value of the number of record rainfalls involves summing these probabilities together, which gives $H_{100},$ the $100^\text{th}$ harmonic number. This can be estimated with the formula above:

$\begin{aligned} H_{100} &\approx \ln(100)+\gamma+\frac{1}{2\cdot 100} \\ &\approx 5.187. \end{aligned}$

Therefore, you would expect to see approximately $5$ record rainfalls over that time period. $_\square$

Your company manufactures elevator cables. You are attempting to discover the maximum weight that one of your cables could safely hold. To do this, you test cables in the following way:

You gradually increase the weight on a cable until it breaks, then record the weight that broke the cable as $W_1$.

You gradually increase the weight on the second cable to $W_1.$ If the cable breaks before the weight reaches $W_1,$ then you record this new maximum safe weight as $W_2.$ Otherwise, if the second cable can hold $W_1$, then you begin testing the next cable.

Each time a cable breaks under a lesser weight, you record this as the new maximum safe weight. You continue testing cables in this way until you have tested 500 cables.

If the weight that an individual cable can hold is uniformly random, how many cables would you expect to break during this testing process?

Assuming that the maximum weight a cable can hold is uniformly random,

- the $1^\text{st}$ cable will always be loaded to its maximum weight,
- the $2^\text{nd}$ cable will have a $\frac{1}{2}$ chance to have a lesser maximum weight than the $1^\text{st}$ cable,
- the $3^\text{rd}$ cable will have a $\frac{1}{3}$ chance to have a lesser maximum weight than the other 2 cables,
- the $4^\text{th}$ cable will have a $\frac{1}{4}$ chance to have a lesser maximum weight than the other 3 cables,
- and so on.
Calculating the expected value of the number of broken cables involves summing these probabilities together, which gives $H_{500},$ the $500^\text{th}$ harmonic number. This can be estimated with the formula above:

$\begin{aligned} H_{500} &\approx \ln(500)+\gamma+\frac{1}{2\cdot 500} \\ &\approx 6.793, \end{aligned}$

Therefore, you would expect to break approximately $7$ of the cables with this process. $_\square$

## Other Applications in Mathematics

Gamma is not just used in problems involving harmonic numbers. Like $\pi$ and $e$, it finds a place in many areas of mathematics. The following are a few of the places in which gamma makes an appearance:

Gamma is used in the Laplace transform of the natural logarithm function.

Laplace transform of the natural logarithm function:$\begin{array}{rc} \text{Time domain:} & f(t)=\ln(t) \\ s \text{ domain:} & F(s)=\mathcal{L}\{f(t)\}=-\frac{1}{s}\big[\ln(s)+\gamma\big]. \end{array}$

The digamma function is defined as the logarithmic derivative of the gamma function. The digamma function is related to the harmonic numbers through gamma.

Digamma function's relation to harmonic numbers:$\psi(n)=H_{n-1}-\gamma.$

There are a number of other identities which contain the digamma function and gamma, and these can be seen on the linked page above.

**Trigonometric integrals** are functions based on integrals that involve trigonometric functions. The cosine integrals are related by gamma.

The

cosine integralfunctions are defined as follows:$\begin{aligned} \text{Ci}(x)&=-\int_{x}^{\infty}{\frac{\cos{t}}{t}\, dt}\\\\ \text{Cin}(x)&=\int_{0}^{x}{\frac{1-\cos{t}}{t}\, dt}. \end{aligned}$

These functions are related by gamma:

$\text{Cin}(x)=\gamma+\ln(x)-\text{Ci}(x).$

Gamma is used to give bounds for the growth of the divisor function.

Growth rate of the divisor function:For all $x \ge 1,$

$\sum\limits_{n = 1}^{x}\sigma_0(n)=x\ln{x}+(2\gamma-1)x+O\left(\sqrt{x}\right),$

where $\sigma_0(n)$ is the divisor function and $O\left(\sqrt{x}\right)$ is big O notation.

In other words, if one were to count the number of divisors for each positive integer up to $x,$ and then sum those counts together, then the resulting sum, $S,$ would satisfy the following inequality:

$x\ln{x}+(2\gamma-1)x<S<x\ln{x}+(2\gamma-1)x+\sqrt{x}.$

Show that the inequality above holds true for $\sum\limits_{n=1}^{10}\sigma_0(n).$

A table with the values of the divisor function for all $n \le 10$ is given below:

$\begin{array}{c|c} n & \sigma_0(n) \\ \hline 1 & 1 \\ 2 & 2 \\ 3 & 2 \\ 4 & 3 \\ 5 & 2 \\ 6 & 4 \\ 7 & 2 \\ 8 & 4 \\ 9 & 3 \\ 10 & 4 \end{array}$

The sum of all these values is $\sum\limits_{n = 1}^{10}\sigma_0(n)=27.$

This is more than $10\ln{10}+(2\gamma-1)(10) \approx 24.570,$ but it is less than $24.570+\sqrt{10} \approx 27.732.$ $_\square$

Gamma is also part of an inequality that gives an upper bound for the sum of divisors function.

Upper bound for sum of divisors function:Assuming that the Riemann hypothesis is true, the sum of divisors function has an upper bound for sufficiently large $n$:

$\sigma_1(n)<e^\gamma n \ln\ln{n}.$

Without the assumption that the Riemann hypothesis is true, the upper bound for the sum of divisors function is slightly more conservative (also for sufficiently large $n$):

$\sigma_1(n)<e^\gamma n \ln\ln{n}+\frac{0.6483n}{\ln\ln{n}},$

where $\sigma_1(n)$ is the sum of divisors function.

**Cite as:**Euler-Mascheroni Constant.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/euler-mascheroni-constant/