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How can I evaluate the integral of $$\int_0^1\sin\left(\frac{1}{x}\right)dx.$$ Maybe it needs the cosine integral to evaluate it, but I cannot understand it very well.

Thanks a lot.

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    Make a substitution so that you have $\sin t$ in there and then see what you get.2012-04-16

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This integral cannot be done in terms of elementary functions, in general. That being said, we can try to get somewhere:
Let $\frac{1}{x} = t$. Then $-\frac{1}{x^2} dx = dt$, which means that $dx = -\frac{1}{t^2} dt$. So $$\int_0^1 \sin\!\left(\frac{1}{x}\!\right)\, dx = \int_\infty^1 -\frac{\sin(t)}{t^2}\, dt = \int_1^\infty \frac{\sin(t)}{t^2}\,dt = \sin(1) + \int_1^\infty \frac{\cos(t)}{t}\, dt$$ by integration by parts. Use the substitution:
$dv = \frac{1}{t^2} dt$, $u = \sin(t)$,
$v = -\frac{1}{t}$, $du = \cos(t) dt$

Using the definition of the cosine integral $$\mathrm{ci}(x) = -\int_x^\infty \frac{\cos(t)}{t}\, dt$$ we get that $$\int_0^1 \sin\!\left(\frac{1}{x}\!\right)\, dx = \sin(1) - \mathrm{ci}(1)$$

And I don't believe this is number going to be elementarily expressible, but it can be approximated. (The approximate value of the integral is around 0.504)

(Thank you to J.M. for his correction of a piece of misused terminology.)

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    thanks a lot ,but do you know how to prove that it cannot be done in closed form,maybe it is very hard.add you to facebook.2012-04-16
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    I'm trying to recall how to prove that. The only thing that comes to mind is Liouville's Theorem, which is the fundamental theorem from Differential Galois Theory. Unfortunately, I can't recall all of the details, but I suspect that the integral in question cannot be expressed in closed form because $si(x) = -\int_x^\infty \frac{\sin(t)}{t}\, dt$ cannot be expressed in closed form in terms of elementary functions.2012-04-16
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    And it is (almost a direct consequence) true that $ci(x)$ also cannot be expressed in closed form in terms of elementary functions. But again, I can't quite recall the details of the proof.2012-04-16
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    Whoa, time out! Let's fix your wording a bit, mmkay? OP's integral *does* have a closed form; what it *doesn't* have is an expression in terms of elementary functions, which is why OP mentioned the need for the sine integral and cosine integral. That these special functions cannot be expressed elementarily can be shown via, say, Risch's algorithm.2012-04-17
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    @J.M., my apologies for that slip-up. What I meant to say was that they couldn't be expressed elementarily, but in typing it up I misused the terminology. Thanks for the correction, and the reference.2012-04-17
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    Isn't there another issue here? Just because $\operatorname{ci}(x)$ is not expressible with elementary functions, it's still conceivable that $\operatorname{ci}(1)$ is a "nice" number. A "nice" number could mean lots of things, like the output of some elementary function for some input in $\mathbb{Q}(\pi)$.2012-04-17
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    The answer to [this question](http://math.stackexchange.com/questions/76650) deals with the use of Liouville's theorem to show that the exponential integral is non-elementary. Since the trigonometric integrals are in fact linear combinations of exponential integrals with complex argument (in much the same way $\sin\,x$ and $\cos\,x$ are linear combinations of $\exp(\pm ix)$), the Liouville route for the trigonometric integrals proceeds similarly.2012-04-17
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    Much better. :) +1, of course.2012-04-17
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    Although the function ci is not elementary, it could be that ci(1) has a nice closed form. Not likely, but not disproved.2012-04-17