# Euler-Mascheroni constant [problem]

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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

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Delta2
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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.

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 $\gamma = \lim_{n->∞}\left\{-ln(n)+\sum_{i=1}^{n}\frac{1}{i}\right\}$

Delta2
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$\lim_{x->0}{-lnxe^{-x}+lnx} = -ln(0)e^{-0}+ln(0) = -1 + 1 = 0$ 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
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

Ray Vickson
Homework Helper
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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
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Another
Delta2