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Is it possible to evaluate this integral in a closed form? $$\int_0^\infty\frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x\phantom{|}}\sqrt{x^2+1}}e^{-x}dx$$

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    $\begingroup$ Probably, Maple gives $(1/2)\pi\sqrt{2}(\sin(1/2)*BesselJ(0, 1/2)-\cos(1/2)BesselY(0, 1/2))$ $\endgroup$ – Jean-Sébastien Oct 15 '13 at 18:01
  • $\begingroup$ @Jean-Sébastien Mathematica cannot evaluate this integral directly. But I got the same symbolic result semi-manually, using Mathematica to justify some steps. Trying to write a complete manual proof now. $\endgroup$ – Vladimir Reshetnikov Oct 15 '13 at 18:15
  • $\begingroup$ Its value is $ 1.3895870359586739409546879631048207809234397342299.$ $\endgroup$ – user64494 Oct 15 '13 at 18:37
  • $\begingroup$ @user64494 What I gave above evaluates to $1.8660736602994162541590468656767972604020575060868...$ $\endgroup$ – Jean-Sébastien Oct 15 '13 at 18:48
  • $\begingroup$ @Jean-Sébastien: You are right. $\endgroup$ – user64494 Oct 15 '13 at 19:01
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I introduce a parameter $a$, $$ I(a)=\int_0^\infty\frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x\phantom{|}}\sqrt{x^2+1}}e^{-ax}dx $$ then took a Mellin transform from $a$ to $s$, $$ \mathcal{M}_{a\to s}[I(a)]=\Gamma(s)\int_0^\infty x^{-s-\frac{1}{2}}\frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x^2+1}}\; dx $$ which apparently equals $$ \mathcal{M}_{a\to s}[I(a)]=\sqrt{2\pi} \frac{\Gamma(s)^2}{\Gamma(s+\frac{1}{2})}\cos\left(\frac{\pi s}{2}\right)\sec(\pi s) $$ then the inverse Mellin transform gives $$ I(a)= \frac{\pi}{\sqrt{2}}\left(\sin \left(\frac a 2 \right)J_0 \left(\frac a 2 \right)-\cos\left(\frac a 2 \right)Y_0\left(\frac a 2 \right) \right) $$

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Hint:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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  • $\begingroup$ But how to simplify further? $\endgroup$ – Harry Peter Nov 19 '13 at 16:35

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