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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.

J. M. ain't a mathematician
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noname1014
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1 Answers1

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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.)

J. M. ain't a mathematician
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Nicholas Stull
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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. – noname1014 Apr 16 '12 at 03:52
  • 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. – Nicholas Stull Apr 16 '12 at 04:07
  • 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. – Nicholas Stull Apr 16 '12 at 04:08
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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. – J. M. ain't a mathematician Apr 17 '12 at 01:03
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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. – Nicholas Stull Apr 17 '12 at 01:11
  • 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 $\mathbb{Q}(\pi)$. – 2'5 9'2 Apr 17 '12 at 01:13
  • The answer to this question 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. – J. M. ain't a mathematician Apr 17 '12 at 01:19
  • Much better. :) +1, of course. – J. M. ain't a mathematician Apr 17 '12 at 10:54
  • Although the function ci is not elementary, it could be that ci(1) has a nice closed form. Not likely, but not disproved. – GEdgar Apr 17 '12 at 12:17