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I know that:

$$\int\limits_0^\infty \left(\sqrt{x}-\sqrt{\sqrt{1+x^2}-1}\,\right)\sin{x}\,\text{d}x = \frac{1}{2}\,\sqrt{\frac{\pi}{2}}\,\frac{e-1}{e}$$

but 

Integrate[(Sqrt[x] - Sqrt[Sqrt[1 + x^2] - 1]) Sin[x], {x, 0, Infinity}]

yields the integral itself in output after a few seconds.

Is there a way to get the correct result or can MMA provide only an approximation?

I found a way to fool MMA, that is to expand the double radical:

Integrate[(Sqrt[x] - Sqrt[(-1 + I x)/2] - Sqrt[(-1 - I x)/2]) Sin[x], {x, 0, Infinity}]

((-1 + E) Sqrt[\[Pi]/2])/(2 E)

Is there a command that does this automatically?

Deepening further, I noticed that by writing:

Plot[Sqrt[Sqrt[1 + x^2] - 1] - (Sqrt[(-1 + I x)/2] + Sqrt[(-1 - I x)/2]), {x, -10, 10}]

we get:

enter image description here

and this, if I am not mistaken, indicates that MMA does not have an algorithm that can establish its equivalence.

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  • $\begingroup$ For the approximation, you could use NIntegrate. $\endgroup$
    – JungHwan Min
    Commented Jul 12, 2018 at 17:42
  • $\begingroup$ Ok I correct the comment. You can check with NIntegrate code: Sqrt[Pi/2]/2*((E - 1)/E) - NIntegrate[(Sqrt[x] - Sqrt[Sqrt[1 + x^2] - 1])*Sin[x], {x, 0, Infinity}, WorkingPrecision -> 200] almost zero. $\endgroup$
    – Mariusz Iwaniuk
    Commented Jul 12, 2018 at 18:56
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    $\begingroup$ @MariuszIwaniuk You can verify the solution like that, but you cannot compute it (plus, being close does not necessarily mean they're equal). The assumption here is that you don't know the solution $\frac{1}{2}\,\sqrt{\frac{\pi}{2}}\,\frac{e-1}{e}$ a priori. $\endgroup$
    – JungHwan Min
    Commented Jul 12, 2018 at 19:28
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    $\begingroup$ The questions arise: isn't it art for art's sake? where is this integral applied? $\endgroup$
    – user64494
    Commented Jul 12, 2018 at 19:32
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    $\begingroup$ How you expand the double radical? Could you explain and update the question? $\endgroup$
    – Mariusz Iwaniuk
    Commented Jul 13, 2018 at 21:45

3 Answers 3

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FWIW, Wolfram|Alpha can denest this radical:

radical = WolframAlpha[
  "Simplify[Sqrt[Sqrt[1 + x^2] - 1]]", 
  {{"Result", 1}, "ComputableData"}
][[1]]

enter image description here

Integrate[(Sqrt[x] - radical) Sin[x], {x, 0, ∞}]

((-1 + E) Sqrt[\[Pi]/2])/(2 E)
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FWIW, you can de-nest roots using the identity

$$\sqrt{\frac{1}{2} (-a+i b x)}+\sqrt{\frac{1}{2} (-a-i b x)}=\sqrt{\sqrt{a^2+b^2 x^2}-a}$$ from which it follows that $$ \int_{\mathbb R^+}\sqrt{\sqrt{a^2+b^2 x^2}-a}\ \sin x\mathrm dx=\frac{1}{2} \sqrt{\frac{\pi }{2}} \sqrt{b} \left(1-e^{-\frac{a}{b}}\right) $$ as given by

Integrate[
          (Sqrt[b x] - Sqrt[(-a + I x b)/2] - Sqrt[(-a - I x b)/2]) Sin[x]
         , {x, 0, Infinity}, Assumptions -> a > 0 && b > 0
]

Your integral corresponds to $a=b=1$,

% /. {a -> 1, b -> 1}
(* 1/2 (1 - 1/E) Sqrt[Pi/2] *)

as expected.

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    $\begingroup$ In general, radical de-nesting is a non-trivial problem. You can find some more details on the wikipedia page, en.wikipedia.org/wiki/Nested_radical. $\endgroup$
    – AccidentalFourierTransform
    Commented Jul 14, 2018 at 22:08
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    $\begingroup$ Doesn't the OP already have the de-nesting in the question, and asks, "Is there a command that does this automatically?" $\endgroup$
    – Michael E2
    Commented Jul 14, 2018 at 22:10
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1 2 ( − a + i b x ) + 1 2 ( − a − i b x ) = a 2 + b 2 x 2 − a

It seems that Mathematica does not know this way of expanding the nested square root. For example, changing the integrate to something else will still result in the unevaluated form. I think this is a bug or something.

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    $\begingroup$ Yes is something, but not a bug. $\endgroup$
    – Mariusz Iwaniuk
    Commented Jul 15, 2018 at 14:57

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