Euler–Mascheroni constant How to Solution

[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 \(\displaystyle E{(0)}_{1}=\infty\)

 

[Euge]

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

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

 

[Another1]

Thank you very much Euge

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?

 

[Euge]

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.'