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Another1

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

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

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