Cómo evaluar $\int_{0}^{\infty}{\ln(x)\over x}\left({1+x+e^x\over e^x+e^{2x}}-{1\over \sqrt{1+ x^2}}\right)\mathrm dx?$

Question

Considere esta integral

$$\int_{0}^{\infty}{\ln(x)\over x}\left({1+x+e^x\over e^x+e^{2x}}-{1\over \sqrt{1+x^2}}\right)\mathrm dx=I\tag1$$

Mi intento:

$x=\ln(y)$ entonces $(1)$ se convierte en

$$\int_{1}^{\infty}{\ln(\ln y)\over y\ln y}\left({1+y+\ln y\over y+y^2}-{1\over \sqrt{1-\ln^2 y}}\right)\mathrm dy\tag2$$

Pruebe un split $(1)$ pero sigue siendo complicado

$$\int_{0}^{\infty}{\ln(x)\over xe^x}dx+\int_{0}^{\infty}{\ln(x)\over xe^x(1+e^x)}dx-\int_{0}^{\infty}{\ln(x)\over x\sqrt{1-x^2}}dx=I_1+I_2-I_3\tag3$$

$I_3$ parece divergir (tal vez todos parecen divergir)

¿Cómo podemos evaluar $(1)?$

Answer 1

$$\int\limits_0^\infty \frac{\ln x}{x}\left(\frac{1+x+e^x}{e^x+e^{2x}}-\frac{1}{\sqrt{1+x^2}}\right)dx=\int\limits_0^\infty \frac{\ln x}{e^x+e^{2x}}dx - \int\limits_0^\infty \frac{\ln x}{x}\left(\frac{1}{\sqrt{1+x^2}}-\frac{1}{e^x}\right)dx$$

con

$$\int\limits_0^\infty \frac{\ln x}{e^x+e^{2x}}dx=\int\limits_0^\infty \frac{\ln x}{e^x}dx-\int\limits_0^\infty \frac{\ln x}{1+e^x}dx=\frac{(\ln 2)^2-2\gamma}{2}$$

y

$$\int\limits_0^\infty \frac{\ln x}{x}\left(\frac{1}{\sqrt{1+x^2}}-\frac{1}{e^x}\right)dx=\frac{(\ln 2)^2-\gamma^2}{2}$$ .

Por lo tanto, el resultado es $\enspace\displaystyle -\frac{\gamma}{2}(2-\gamma)$ . $\enspace\gamma$ es aquí la constante de Euler-Mascheroni.

Para la segunda integral puedes utilizar la integración por partes con:

$$\frac{d}{dx}(\ln x)^2=2\frac{\ln x}{x}$$

$$\int\limits_0^\infty \frac{(\ln x)^2}{e^x}=\gamma^2+\frac{\pi^2}{6}$$

$$\int\limits_0^\infty \frac{x(\ln x)^2}{\sqrt{1+x^2}^3}=\frac{\pi^2}{6}+(\ln 2)^2$$

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Basándose en el comentario de Lucian calculamos ahora $\enspace\displaystyle \int\limits_0^\infty \frac{\ln t}{t}(\frac{1}{\sqrt{1+t^2}}-\frac{1}{e^t}) dt\,$ de una forma diferente a la anterior. Visite $\,F(x):=x\Gamma(x)\,$ . Tenemos

$\displaystyle \int\limits_0^\infty \frac{dt}{t^x e^t}=\Gamma(1-x)=F(-x)\enspace$ , $\enspace\displaystyle \int\limits_0^\infty \frac{dt}{t^x \sqrt{1+t^2}}= \frac{\Gamma(\frac{1-x}{2})\Gamma(\frac{x}{2})}{2\sqrt{\pi}} =\frac{2 F(\frac{1-x}{2})F(\frac{x}{2})}{\sqrt{\pi}x(1-x)} \,$ ,

$\enspace\displaystyle F(-\frac{x}{2})F(\frac{1-x}{2})= \frac{\sqrt{\pi}}{2^{1-x}} F(1-x)\,$ , $\,\displaystyle (\ln F(x))’ = \lim\limits_{n\to\infty}(\ln n - \sum\limits_{k=1}^n\frac{1}{k+x})\,$ , $\,\displaystyle (\ln F(x))’’ = \sum\limits_{k=1}^\infty\frac{1}{(k+x)^2}\,$ .

1. $\enspace\displaystyle \int\limits_0^\infty \frac{1}{t^{1-x}} ( \frac{1}{\sqrt{1+t^2}} -\frac{1}{e^t})dt= \frac{1}{x}(\frac{2 F(\frac{1-x}{2}) F(\frac{x}{2})}{\sqrt{\pi}(1-x)}-F(x)) $$ \displaystyle =\frac{1}{x}(\frac{2^x F(-x) F(\frac{x}{2}) }{ F(-\frac{x}{2}) }-F(x))$

2. $\enspace\displaystyle \int\limits_0^\infty \frac{\ln t}{t^{1-x}} ( \frac{1}{\sqrt{1+x^2}} -\frac{1}{e^x})dt= \frac{d}{dx} \left( \frac{1}{x} ( \frac{2^x F(-x) F(\frac{x}{2}) }{ F(-\frac{x}{2}) } -F(x))\right)= \frac{1}{x^2}\left(- \frac{2^x F(-x) F(\frac{x}{2}) }{ F(-\frac{x}{2})} +F(x) + x (\frac{2^x F(-x) F(\frac{x}{2}) }{ F(-\frac{x}{2})} -F(x))’ \right) $

3. $\enspace$ La regla de L'Hôpital dos veces
   $\enspace\displaystyle \int\limits_0^\infty \frac{\ln t}{t} ( \frac{1}{\sqrt{1+t^2}} -\frac{1}{e^t})dt=\lim\limits_{x\to 0} \int\limits_0^\infty \frac{\ln t}{t^{1-x}} ( \frac{1}{\sqrt{1+t^2}} -\frac{1}{e^t})dt =$$ \displaystyle =\limits_{x\to \pm 0} \frac{1}{x^2} \left(- \frac{2^x F(-x) F(\frac{x}{2}) }{ F(-\frac{x}{2})} +F(x) + x (\frac{2^x F(-x) F(\frac{x}{2}) }{ F(-\frac{x}{2})} -F(x))'' \derecha) = \frac{1}{2} \left(- \frac{2^x F(-x) F(\frac{x}{2}) }{ F(-\frac{x}{2})} +F(x) + x (\frac{2^x F(-x) F(\frac{x}{2}) }{ F(-\frac{x}{2})} -F(x))'' \derecha)''|_{x=0} $ $ \displaystyle = \frac{1}{2}\left(\left(\frac{2^x F(-x) F(\frac{x}{2}) }{ F(-\frac{x}{2})}\right)'' -F''(x) + x \left(\frac{2^x F(-x) F(\frac{x}{2}) }{ F(-\frac{x}{2})} -F(x) \right)'''\right)|_{x=0}$

4. $\displaystyle \left(\frac{2^x F(-x) F(\frac{x}{2}) }{ F(-\frac{x}{2})}\right)’’=$$ \displaystyle = \frac{2^x F(-x) F(\frac{x}{2})}{ F(-\frac{x}{2})} ((\ln 2+ ( \ln F(-x) )'+( \ln F(\frac{x}{2}) )'-(\ln F(-\frac{x}{2})')^2 $$\hspace{3.5cm}\displaystyle + ( \ln F(-x) )’’+ (\ln F(\frac{x}{2}) )’’-(\ln F(-\frac{x}{2}))’’)$

$\hspace{8mm}$ => $\enspace\displaystyle \left(\frac{2^x F(-x) F(\frac{x}{2}) }{ F(-\frac{x}{2})}\right)’’|_{x=0}=$

$\hspace{2cm}\displaystyle =(\ln 2 + \gamma -\frac{\gamma}{2}-\frac{\gamma}{2} )^2+(\frac{\pi^2}{6}+\frac{\pi^2}{24}-\frac{\pi^2}{24})=(\ln 2)^2+ \frac{\pi^2}{6} $

5. $\displaystyle F’’(x)=F(x)((\ln F(x))’^2+(\ln F(x))’’) \enspace$ => $\enspace\displaystyle F’’(0)=\gamma^2 + \frac{\pi^2}{6}$

6. $\enspace\displaystyle \left(x \left(\frac{2^x F(-x) F(\frac{x}{2}) }{ F(-\frac{x}{2})} -F(x))’ \right)’’’\right)|_{x=0} = 0\enspace$ desde $\enspace\displaystyle \left(\frac{2^x F(-x) F(\frac{x}{2}) }{ F(-\frac{x}{2})} -F(x))’ \right)’’’|_{x=0} \enspace$ es limitado

7. $\enspace\displaystyle \int\limits_0^\infty \frac{\ln t}{t}(\frac{1}{\sqrt{1+t^2}}-\frac{1}{e^t}) dt = \frac{1}{2}\left (((\ln 2)^2+\frac{\pi^2}{6})-(\gamma^2+\frac{\pi^2}{6})+0\right)=\frac{(\ln 2)^2-\gamma^2}{2}$