Por lo tanto, ¿cómo podemos probar su uso?

Question

Donde $\gamma=0.5772156...$ es constante de Euler y $j$ es un entero, $j\ge 1.$

$a\in \Re$

Cómo puedo Mostrar

$$\int_{0}^{\infty}\sum_{i=0}^{j}(-1)^{i +j}{j\choose i}{e^{-x^{i +a}}\over x}dx=-(-1)^j\color{red}{j!\gamma\over a(a+1)(a+2)\cdots(a+j)}?\tag1$$

Deje que un simple caso, $a=1$ y $j=1$

$$I=\int_{0}^{\infty}{e^{-x}-e^{-x^2}\over x}dx=-\color{blue}{\gamma\over 2}\tag2$$

Sustitución $u=\ln{x}\rightarrow xdu=dx$

$x=\infty\rightarrow u=\infty$ y $x=0\rightarrow u=-\infty$

$$I=\int_{-\infty}^{\infty}e^{-e^u}-e^{-e^{2u}}du=-{\gamma\over 2}\tag3$$

No estoy seguro de cómo integrar $(3)$. Necesita la ayuda, gracias.

Answer 1

Escribimos $I(a,j)=\int^\infty_0\sum_{i=0}^{j}(-1)^{i}\binom{j}{i}\frac{\exp (-x^{a+i})}{x}\,dx$.

Usando el $\binom{j}{i}=\binom{j-1}{i}+\binom{j-1}{i-1}$, concluimos $I(a,j)=I(a,j-1)-I(a+1,j-1)$.

Si podemos probar $I(a,1)=\frac{-\gamma}{a(a+1)}$, la forma esperada $I(a,j)$ sigue de la inducción.

$\begin{align*} I(a,1)&=\int^\infty_0\frac{\exp(x^{-a})-\exp (-x^{a+1})}{x}\,dx\\ &=\lim_{z\to0}\int^\infty_z\frac{\exp(x^{-a})-\exp (-x^{a+1})}{x}\,dx\\ &=\lim_{z\to0}\left(\int^\infty_z\frac{\exp(x^{-a})}{x}\,dx-\int^\infty_z\frac{\exp (-x^{a+1})}{x}\,dx\right)\\ &=\lim_{z\to 0}\left(-\frac{Ei(-z^a)}{a}+\frac{Ei(-z^{a+1})}{a+1}\right)\\ &=\lim_{z\to 0}\left(-\frac{\gamma+\log(z^a)+O(z^a)}{a}+\frac{\gamma+\log(z^{a+1})+O(z^{a+1})}{a+1}\right)\\ &=\frac{-\gamma}{a(a+1)}. \end{align*} $

Answer 2

Sugerencia de $$\color{red}{\int_{-\infty }^{\infty }{\frac{{{e}^{-{{x}^{\,\alpha }}}}-{{e}^{-{{x}^{\,\beta }}}}}{x}}dx=\gamma \frac{\alpha\beta }{\alpha \beta }}$$

Prueba

Conjunto $t=x^\alpha$ , $n=\frac{\beta}{\alpha}$, así $dx=\large {\frac{1}{\alpha x^{\alpha-1}}}$$dt$ y $$I=\int_{-\infty }^{\infty }{\frac{{{e}^{-{{x}^{\,\alpha }}}}-{{e}^{-{{x}^{\,\beta }}}}}{x}}dx=\frac{1}{\alpha}\int_{0}^{\infty}\frac{e^{-t}-e^{-t^{\large n}}}{t} dt$$ por lo tanto $$\qquad I=\frac{1}{\alpha }\left[ \left( \int_{0}^{1}{\frac{{{e}^{-t}}-1}{t}dt+\int_{1}^{\infty }{\frac{{{e}^{-t}}-1}{t}dt}} \right)-\left( \int_{0}^{1}{\frac{{{e}^{-{{t}^{n}}}}-1}{t}dt+\int_{1}^{\infty }{\frac{{{e}^{-{{t}^{n}}}}-1}{t}dt}} \right) \right. $$ $$=\frac{1}{\alpha }\left[ \left( \int_{0}^{1}{\frac{{{e}^{-t}}-1}{t}dt+\int_{1}^{\infty }{\frac{{{e}^{-t}}}{t}dt}} \right)-\left( \int_{0}^{1}{\frac{{{e}^{-{{t}^{n}}}}-1}{t}dt+\int_{1}^{\infty }{\frac{{{e}^{-{{t}^{n}}}}}{t}dt}} \right) \right].$$ De hecho, hemos $$I=-\frac{1}{\alpha }\left[ \color{red}\gamma +\left( \int_{0}^{1}{\frac{{{e}^{-{{t}^{n}}}}-1}{t}dt+\int_{1}^{\infty }{\frac{{{e}^{-{{t}^{n}}}}}{t}dt}} \right) \right]\quad (1)$$ conjunto $$J=\int_{0}^{1}{\frac{{{e}^{-{{t}^{n}}}}-1}{t}dt+\int_{1}^{\infty }{\frac{{{e}^{-{{t}^{n}}}}}{t}dt}}$$ y deje $u=t^n$, luego $$J=\frac{1}{n}\int_{0}^{1}{\frac{{{e}^{-u}}-1}{u}du+\int_{1}^{\infty }{\frac{{{e}^{-u}}}{u}du}}=-\frac{1}{n}\color{red}\gamma\qquad\quad(2)$$ $(1)$ $(2)$ $$I=-\frac{1}{\alpha }( \color{red}\gamma -\frac{1}{n}\color{red}\gamma)=-\frac{1}{\alpha }\color{red}\gamma\left(1-\frac{\alpha}{\beta}\right)=\color{red}{\gamma \frac{\alpha -\beta }{\alpha \beta }}$$

Nota $$\color{blue}{\gamma =\int_{0}^{1}{\frac{1-{{e}^{-y}}} de{y}dy-\int_{1}^{\infty }{\frac{{{e}^{-y}}} de{y}dy}}}\qquad\qquad\qquad\qquad(3)$$

Answer 3

$\newcommand{\ángulos}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\llaves}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\mitad}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{j = 1,2,3,\ldots}$. Con $\ds{\epsilon > 0\ \mbox{and}\ \ul{a > 0},\ }$ permite considerar

\begin{align} &\color{#f00}{\int_{\epsilon}^{\infty} \sum_{k = 0}^{j}\pars{-1}^{k + j}{j \choose k}{\expo{-x^{k +a}} \over x}\,\dd x} = \sum_{k = 0}^{j}\pars{-1}^{k + j}{j \choose k} \int_{\epsilon}^{\infty}{\expo{-x^{k +a}} \over x}\,\dd x\tag{1} \end{align} y \begin{align} \fbox{%#%#%} &\ \stackrel{x^{k + a}\phantom{aa}\mapsto\ x}{=}\ \int_{\epsilon}^{\infty}{\expo{-x} \over x^{1/\pars{k + a}}} \,{1 \over k + a}\,x^{1/\pars{k + a} - 1}\,\dd x = {1 \over k + a}\int_{\epsilon^{k + a}}^{\infty}{\expo{-x} \over x}\,\dd x \\[3mm] & = {1 \over k + a}\bracks{% -\pars{k + a}\ln\pars{\epsilon}\expo{-\epsilon^{k + a}} + \int_{\epsilon^{k + a}}^{\infty}\ln\pars{x}\expo{-x}\,\dd x} \\[3mm] & = -\ln\pars{\epsilon}\expo{-\epsilon^{k + a}} + {1 \over k + a}\int_{\epsilon^{k + a}}^{\infty}\ln\pars{x}\expo{-x}\,\dd x \\[3mm] & = -\ln\pars{\epsilon}\expo{-\epsilon^{k + a}} + {1 \over k + a}\ \underbrace{\int_{0}^{\infty}\ln\pars{x}\expo{-x}\,\dd x} _{\ds{-\gamma}}\ -\ {1 \over k + a}\int_{0}^{\epsilon^{k + a}}\ln\pars{x}\expo{-x}\,\dd x \end{align}

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En el $\ds{\ \int_{\epsilon}^{\infty}{\expo{-x^{k +a}} \over x}\,\dd x\ }$ límite, tendremos $\ds{\epsilon \to 0^{+}}$ \begin{align} &\color{#f00}{\int_{0}^{\infty} \sum_{k = 0}^{j}\pars{-1}^{k + j}{j \choose k}{\expo{-x^{k +a}} \over x}\,\dd x} = \sum_{k = 0}^{j}\pars{-1}^{k + j}{j \choose k}\pars{-\,{\gamma \over k + a}} \tag{2} \\[3mm] &\ \mbox{because}\quad \sum_{k = 0}^{j}\pars{-1}^{k + j}{j \choose k} = \delta_{j0}\quad \mbox{and}\quad j \not= 0 \end{align}

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$\ds{\pars{~\mbox{see expression }\ \pars{1}~}}$ se convierte en \begin{align} &\color{#f00}{\int_{0}^{\infty} \sum_{k = 0}^{j}\pars{-1}^{k + j}{j \choose k}{\expo{-x^{k +a}} \over x}\,\dd x} = -\gamma\,\pars{-1}^{\,j}\sum_{k = 0}^{j}\pars{-1}^{k}{j \choose k} \int_{0}^{1}x^{k + a - 1}\,\dd x \\[3mm] = &\ -\gamma\,\pars{-1}^{\,j}\int_{0}^{1}x^{a - 1} \sum_{k = 0}^{j}{j \choose k}\pars{-x}^{k}\,\dd x = -\gamma\,\pars{-1}^{\,j}\int_{0}^{1}x^{a - 1}\pars{1 - x}^{j}\,\dd x \\[3mm] = &\ -\gamma\,\pars{-1}^{\,j}\, {\Gamma\pars{a}\Gamma\pars{j + 1} \over \Gamma\pars{a + j + 1}} = -\gamma\,\pars{-1}^{\,j}\,{j! \over \pars{a}_{j + 1}} \end{align} donde $\pars{2}$ es un Pochammer Símbolo: $$ \pars{a}_{j + 1} = \pars {+1}\pars {+ 2}\ldots\pars{a + j} $$

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Finalmente, $$ \color{#f00}{\int_{0}^{\infty} \sum_{k = 0}^{j}\pars{-1}^{k + j}{j \elegir k}{\expo{-x^{k +a}} \over x}\,\dd x} = \color{#f00}{% -\gamma\,\pars{-1}^{\,j}\,{j! \sobre\pars {+1}\pars {+ 2}\ldots\pars{a + j}}} $$