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Euler-Mascheroni constant [problem]

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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 ##??????????????????????????????????
 
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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.
 
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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.

##\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!}##.
 
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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

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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##.
 

Delta2

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