Euler-Mascheroni constant [problem]

[Another]

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

 

[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.

 

[Another]

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]

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

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 $ln x$.

 

[Delta2]

Well that limit with the summation and the inverse factorial proves to be quite hard to handle. You can find some alternative approaches for this problem at math.stackexchange.com just check here

https://math.stackexchange.com/ques...0-infty-frac-log-x-exp-x-dx?noredirect=1&lq=1

 

[Another]

Well that limit with the summation and the inverse factorial proves to be quite hard to handle. You can find some alternative approaches for this problem at math.stackexchange.com just check here

https://math.stackexchange.com/ques...0-infty-frac-log-x-exp-x-dx?noredirect=1&lq=1

thank you very much You helped me a lot.