# Euler-Mascheroni constant [problem]

• Another
In summary, the integral ##\int_{0}^{\infty} e^{-x}ln(x)\,dx = -\gamma## when ##\gamma## is Euler-Mascheroni constant can be solved using integration by parts and the definition of the Euler-Mascheroni constant. By setting ##u = ln(x)## and ##dv = e^{-x} dx##, the integral can be rewritten as ##-ln(x)e^{-x} + \int_{0}^{\infty} \frac{e^{-x}}{x} \, dx##. Using the series representation of ##e^{-x}## and simplifying, we can obtain the final result of ##ln(x) +

#### Another

Moved from a technical forum, so homework template missing
Show that ##\int_{0}^{\infty} e^{-x}ln(x)\,dx = -\gamma ## when ##\gamma## is Euler–Mascheroni constant

My solution is ...

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

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

when ##e^{-x}=\sum_{n=0}^{\infty}\frac{(-x)^n}{n!}=\sum_{n=0}^{\infty}\frac{(-1)^n(x)^n}{n!}##
##\frac{e^{-x}}{x}=\frac{1}{x}-1+\frac{x}{2!}-\frac{x^2}{3!}+...##so...##-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##
##= -ln(x)e^{-x}+ln(x)+\sum_{n=1}^{\infty}\frac{(-1)^nx^n}{n\cdot n!}##

##\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!}##

##\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\}+0##

and
##\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\}##

##\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...
##\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\}-\lim_{{x}\to{0}}\left\{-ln(x)e^{-x}+ln(x)\right\}##

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

Last edited:
Delta2
What is the definition equality in your book for the Euler-Mascheroni constant?

The limit ##\lim_{x->0}{-lnxe^{-x}+lnx}## is zero, I believe you can easily prove that.

Delta² said:
What is the definition equality in your book for the Euler-Mascheroni constant?

The limit ##\lim_{x->0}{-lnxe^{-x}+lnx}## is zero, I believe you can easily prove that.
##\lim_{x->0}{-lnxe^{-x}+lnx} = -ln(0)e^{-0}+ln(0) = -1 + 1 = 0 ## thankyou

In my book the Euler-Mascheroni constant ## \gamma = \lim_{n->∞}\left\{-ln(n)+\sum_{i=1}^{n}\frac{1}{i}\right\}##

Another said:
##\lim_{x->0}{-lnxe^{-x}+lnx} = -ln(0)e^{-0}+ln(0) = -1 + 1 = 0 ## thankyou

Er no that limit is not so straight forward afterall sorry. Your mistake here is that ##ln(0)=-\infty## and not 1. Instead work this limit as

##lnx(1-e^{-x})=-xlnx\frac{1-e^{-x}}{-x}## when we know that ##\lim_{x->0}{\frac{1-e^{-x}}{-x}}=-1## so all you left is to work
##\lim_{x->0}{xlnx}## with De L Hopital rule.

So after that all that essentially is left with is the limit ##\lim_{x->\infty}\sum_{n=1}^{\infty}\frac{(-1)^nx^n}{nn!}##.

Another
Delta² said:
Er no that limit is not so straight forward afterall sorry. Your mistake here is that ##ln(0)=-\infty## and not 1. Instead work this limit as

##lnx(1-e^{-x})=-xlnx\frac{1-e^{-x}}{-x}## when we know that ##\lim_{x->0}{\frac{1-e^{-x}}{-x}}=-1## so all you left is to work
##\lim_{x->0}{xlnx}## with De L Hopital rule.
Oh thank you very much. I am stupid. I confused between ln(0) and ln(1) hahahahahahaha thank you very much

Another said:
Show that ##\int_{0}^{\infty} e^{-x}ln(x)\,dx = -\gamma ## when ##\gamma## is Euler–Mascheroni constant

My solution is ...

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

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

when ##e^{-x}=\sum_{n=0}^{\infty}\frac{(-x)^n}{n!}=\sum_{n=0}^{\infty}\frac{(-1)^n(x)^n}{n!}##
##\frac{e^{-x}}{x}=\frac{1}{x}-1+\frac{x}{2!}-\frac{x^2}{3!}+...##so...##-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##
##= -ln(x)e^{-x}+ln(x)+\sum_{n=1}^{\infty}\frac{(-1)^nx^n}{n\cdot n!}##

##\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!}##

##\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\}+0##

and
##\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\}##

##\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...
##\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\}-\lim_{{x}\to{0}}\left\{-ln(x)e^{-x}+ln(x)\right\}##

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

I could not follow your rather cavalier reasoning. In this case the definition of the integral is
$$\lim_{a \to 0, b \to \infty} \int_a^b e^{-x} \ln(x) \, dx$$
It would be better to express the finite integral in series terms, then examine the limits as ##a \to 0## and ##b \to \infty##.

BTW: in LaTeX you should type "\ln" instead of "ln", so you get ##\ln x## instead of ##ln x##.

Another
Delta2

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

The Euler-Mascheroni constant, denoted by the symbol γ, is a mathematical constant that arises in various areas of mathematics, such as number theory and calculus. It is approximately equal to 0.57721566 and is represented by the infinite series: γ = Σ(1/n) - ln(n), where n ranges from 1 to infinity.

## 2. Who discovered the Euler-Mascheroni constant?

The Euler-Mascheroni constant is named after Leonhard Euler and Lorenzo Mascheroni, two mathematicians who independently discovered and studied the constant in the 18th century. Euler is credited with introducing the symbol γ to represent the constant.

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

The Euler-Mascheroni constant has many important applications in mathematics, such as in the study of prime numbers, the Riemann zeta function, and the Stirling's approximation formula. It also plays a role in various mathematical proofs and has connections to the famous Euler's constant, e.

## 4. Is the Euler-Mascheroni constant irrational?

Yes, the Euler-Mascheroni constant is an irrational number, meaning it cannot be expressed as a ratio of two integers. It has been proven to be a transcendental number, which means it is not a root of any polynomial equation with rational coefficients.

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

The Euler-Mascheroni constant can be approximated using various mathematical methods, such as the Euler-Maclaurin formula, the continued fraction expansion, and the Riemann zeta function. It can also be calculated using computer algorithms, which have been able to compute billions of digits of the constant.