La forma cerrada para $\int_0^\infty\frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x\phantom{|}}\sqrt{x^2+1}}e^{-x}dx$

Question

Es posible evaluar esta integral en forma cerrada? $$\int_0^\infty\frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x\phantom{|}}\sqrt{x^2+1}}e^{-x}dx$$

Answer 1

Sugerencia:

$\int_0^\infty\dfrac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x}\sqrt{x^2+1}}e^{-x}~dx$

$=\int_0^\infty\dfrac{\sqrt{\sinh x+\sqrt{\sinh^2x+1}}}{\sqrt{\sinh x}\sqrt{\sinh^2x+1}}e^{-\sinh x}~d(\sinh x)$

$=\int_0^\infty\dfrac{\sqrt{\sinh x+\cosh x}}{\sqrt{\sinh x}\cosh x}e^{-\sinh x}\cosh x~dx$

$=\int_0^\infty\dfrac{e^{-\sinh x}\sqrt{e^x}}{\sqrt{\dfrac{e^x-e^{-x}}{2}}}dx$

$=\sqrt2\int_0^\infty e^{-\sinh x}\sqrt{\dfrac{e^x}{e^x-e^{-x}}}~dx$

$=\sqrt2\int_0^\infty\dfrac{e^{x-\frac{e^x-e^{-x}}{2}}}{\sqrt{e^{2x}-1}}dx$

$=\sqrt2\int_0^\infty\dfrac{e^{-\frac{e^x}{2}+\frac{1}{2e^x}}}{\sqrt{e^{2x}-1}}d(e^x)$

$=\sqrt2\int_1^\infty\dfrac{e^{-\frac{x}{2}+\frac{1}{2x}}}{\sqrt{x^2-1}}dx$

$=\sqrt2\int_1^\infty e^{-\frac{x}{2}+\frac{1}{2x}}~d(\cosh^{-1}x)$

$=\sqrt2\int_0^\infty e^{-\frac{\cosh x}{2}+\frac{1}{2\cosh x}}~dx$

$=\sqrt2\int_0^\infty e^\frac{1-\cosh^2x}{2\cosh x}~dx$

$=\sqrt2\int_0^\infty e^{-\frac{\sinh^2x}{2\cosh x}}~dx$

$=\sqrt2\int_0^\infty e^{-\frac{x^2}{2\sqrt{x^2+1}}}~d(\sinh^{-1}x)$

$=\sqrt2\int_0^\infty\dfrac{e^{-\frac{x^2}{2\sqrt{x^2+1}}}}{\sqrt{x^2+1}}dx$

$=\sqrt2\int_\infty^0\dfrac{e^{-\frac{1}{2x^2\sqrt{\frac{1}{x^2}+1}}}}{\sqrt{\dfrac{1}{x^2}+1}}d\left(\dfrac{1}{x}\right)$

$=\sqrt2\int_0^\infty\dfrac{e^{-\frac{1}{2x\sqrt{x^2+1}}}}{x\sqrt{x^2+1}}dx$