# Euler–Mascheroni constant How to Solution

• MHB
• Another1
In summary, I proposed a method for solving the integral $\int_0^\infty e^{-x}\ln(x)\, dx$ by using integration by parts and relating it to the limit definition of the Euler-Mascheroni constant. This method involves using a series of integrals to simplify the original integral and ultimately finding the value of the integral in terms of the constant $\gamma$. There is no specific name for this method.
Another1
Why $$\displaystyle \int_{0}^{\infty} e^{-x}ln(x)\,dx = -\gamma$$ when $$\displaystyle \gamma$$ is Euler–Mascheroni constant

My solution is ...

$$\displaystyle u = ln(x)$$ and $$\displaystyle du = \frac{dx}{x}$$
$$\displaystyle dv = e^{-x} dx$$ and $$\displaystyle v = -e^{-x}$$

so... $$\displaystyle \int_{0}^{\infty} e^{-x}ln(x)\,dx = -ln(x)e^{-x}+\int_{0}^{\infty}\frac{e^{-x}}{x} \,dx$$

when $$\displaystyle e^{-x}=\sum_{n=0}^{\infty}\frac{(-x)^n}{n!}=\sum_{n=0}^{\infty}\frac{(-1)^n(x)^n}{n!}$$
$$\displaystyle \frac{e^{-x}}{x}=\frac{1}{x}-1+\frac{x}{2!}-\frac{x^2}{3!}+...$$so...$$\displaystyle -ln(x)e^{-x}+\int_{0}^{\infty}\frac{e^{-x}}{x} \,dx = -ln(x)e^{-x}+\int_{0}^{\infty}[\frac{1}{x}-1+\frac{x}{2!}-\frac{x^2}{3!}+... ]\,dx$$
$$\displaystyle = -ln(x)e^{-x}+ln(x)+\sum_{n=1}^{\infty}\frac{(-1)^nx^n}{n\cdot n!}$$

$$\displaystyle \lim_{{x}\to{0}}\left\{-ln(x)e^{-x}+ln(x)+\sum_{n=1}^{\infty}\frac{(-1)^nx^n}{n\cdot n!}\right\}=\lim_{{x}\to{0}}\left\{-ln(x)e^{-x}+ln(x)\right\}+\lim_{{x}\to{0}}\sum_{n=1}^{\infty}\frac{(-1)^nx^n}{n\cdot n!}$$

$$\displaystyle \lim_{{x}\to{0}}\left\{-ln(x)e^{-x}+ln(x)+\sum_{n=1}^{\infty}\frac{(-1)^nx^n}{n\cdot n!}\right\}=0+0$$

and
$$\displaystyle \lim_{{x}\to{\infty}}\left\{-ln(x)e^{-x}+ln(x)+\sum_{n=1}^{\infty}\frac{(-1)^nx^n}{n\cdot n!}\right\}=\lim_{{x}\to{\infty}}-ln(x)e^{-x}+\lim_{{x}\to{\infty}}\left\{ln(x)+\sum_{n=1}^{\infty}\frac{(-1)^nx^n}{n\cdot n!} \right\}$$

$$\displaystyle \lim_{{x}\to{\infty}}\left\{-ln(x)e^{-x}+ln(x)+\sum_{n=1}^{\infty}\frac{(-1)^nx^n}{n\cdot n!}\right\}=0+\lim_{{x}\to{\infty}}\left\{ln(x)+\sum_{n=1}^{\infty}\frac{(-1)^nx^n}{n\cdot n!} \right\}$$

finally...
$$\displaystyle \int_{0}^{\infty} e^{-x}ln(x)\,dx =\lim_{{x}\to{\infty}}\left\{ln(x)+\sum_{n=1}^{\infty}\frac{(-1)^nx^n}{n\cdot n!} \right\}$$

How to solution $$\displaystyle \int_{0}^{\infty} e^{-x}ln(x)\,dx = -\gamma$$ ??

some time why someone integral $$\displaystyle \int_{0}^{\infty}\frac{e^{-x}}{x} \,dx = E{(-x)}_{1}$$ as $$\displaystyle E{(0)}_{1}=\infty$$

Last edited:
Hi Another,

Let $A_n := \int_0^\infty x^ne^{-x}\ln(x)\, dx$ for $n = 0,1,2,3,\ldots$. We desire to find $A_0$.

By integration by parts,

\begin{align}A_n &= x^ne^{-x}(x\ln x - x)\bigg|_{x = 0}^\infty - \int_0^\infty (x\ln x - x)\frac{d}{dx}(x^ne^{-x})\, dx\\
&= -\int_0^\infty (nx^{n-1}e^{-x} - x^ne^{-x})(x\ln x - x)\, dx\\
&= -\int_0^\infty (nx^n e^{-x}\ln x - x^{n+1}e^{-x}\ln x - nx^ne^{-x} + x^{n+1}e^{-x})\, dx\\
&= -nA_n + A_{n+1} +n\cdot n! - (n+1)!\\
&= -nA_n + A_{n+1} - n! \end{align}

Thus $(1 + n)A_n = A_{n+1} - n!$, or

$$\frac{A_n}{n!} = \frac{A_{n+1}}{(n+1)!} - \frac{1}{n+1}$$

Consequently,

$$A_0 = \frac{A_n}{n!} -1 - \frac{1}{2} - \frac{1}{3} - \cdots - \frac{1}{n}$$

Alternatively,

$$A_0 = \left(\frac{A_n}{n!} - \ln n\right) - \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} - \ln n\right)$$

Since, by definition,

$$\lim_{n\to \infty} \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} - \ln n\right) = \gamma$$

It suffices to show

$$\lim_{n\to \infty} \frac{A_n}{n!} - \ln n = 0$$

For this it should be useful to write

$$\ln n = \int_0^\infty \frac{x^n}{n!}e^{-x}\ln n\, dx$$

so that we can write

$$\frac{A_n}{n!} - \ln n = \int_0^\infty \frac{x^n}{n!}e^{-x}\ln\left(\frac{x}{n}\right)\, dx$$

Thank you very much Euge
Euge said:
Hi Another,

Let $A_n := \int_0^\infty x^ne^{-x}\ln(x)\, dx$ for $n = 0,1,2,3,\ldots$. We desire to find $A_0$.

By integration by parts,
if you do not mind
Can you tell me
What is this method?

I'm not sure if there is a name for it. This is just something I thought of. My thinking was, to relate the integral to the limit definition of $\gamma$, I ought to perform several integration by parts 'the right way.'

## 1. What is the Euler-Mascheroni constant?

The Euler-Mascheroni constant, denoted by the symbol γ, is a mathematical constant named after Leonhard Euler and Lorenzo Mascheroni. It is approximately equal to 0.5772156649 and is a fundamental constant in many areas of mathematics, including number theory, analysis, and combinatorics.

## 2. What is the significance of the Euler-Mascheroni constant?

The Euler-Mascheroni constant is significant because it appears in many important mathematical formulas and theories. It is closely related to the natural logarithm and the harmonic series, and its value has been studied for centuries by mathematicians. It also has applications in physics, engineering, and computer science.

## 3. How is the Euler-Mascheroni constant calculated?

The Euler-Mascheroni constant cannot be expressed as a simple fraction or decimal, and its exact value is unknown. However, it can be approximated using various mathematical methods, such as integrals, series, and continued fractions. The most common approximation is the one given by Leonhard Euler himself, using the infinite series 1 + 1/2 + 1/3 + 1/4 + …, which converges to γ.

## 4. What are some real-life applications of the Euler-Mascheroni constant?

The Euler-Mascheroni constant has practical applications in various fields, such as finance, statistics, and cryptography. In finance, it is used to calculate the expected value of a random variable, and in statistics, it is used to measure the accuracy of estimators. In cryptography, it is used in the RSA encryption algorithm to generate public and private keys.

## 5. How can I use the Euler-Mascheroni constant in my own research or projects?

If you are a mathematician, physicist, or computer scientist, you may encounter the Euler-Mascheroni constant in your research or projects. It can be used in various mathematical formulas and calculations, and its value can be approximated using different methods. You can also explore its connections to other mathematical constants and theories to gain a deeper understanding of its significance.

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