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How to evaluate $$\int {\frac{{\cos {x^3}}}{x}dx}?$$ Maple evaluates this as $$\frac{{{\text{Ci}}({x^3})}}{3}.$$ Edit: If this cannot be evaluated in terms of elementary functions, is there a general strategy which allows us to deduce that this is the case?

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    Why downvote? Care to comment?2012-11-20
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    You may need to read [this](http://math.hunter.cuny.edu/ksda/papers/churchill.pdf).2012-11-20
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    Concerning the edit : just some days earlier we had the same kind of [question](http://math.stackexchange.com/questions/239105/special-integrals/239854#239854). Set $z=x^3$ to get $\ \frac 13\int \frac{\cos(z)}z dz$ and use the demonstration there for $\int \frac{\sin(z)}z dz$.2012-11-20

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$$I = \int \dfrac{\cos(x^3)}{x} dx = \dfrac13 \int \dfrac{\cos(x^3)}{x^3} (3x^2)dx = \dfrac13 \int \dfrac{\cos(x^3)}{x^3} d(x^3) = \dfrac{\text{Ci}(x^3)}{3} + \text{constant}$$ There is no expression for the above integral in terms of "elementary functions". If the limits of the integral are from $-a$ to $a$, the Cauchy principal value of the integral is $0$ since the integrand is an odd function.

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    Yes, but what is that in terms of elementary functions?2012-11-20
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    Can someone explain what is "Ci"?2013-03-29
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    @learner $\operatorname{Ci}(\xi)$ stands for *cosine integral*.2013-04-05
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    @glebovg thanks a lot.2013-04-06
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http://reference.wolfram.com/mathematica/ref/CosIntegral.html

http://www.wolframalpha.com/input/?i=cosineintegral%28x%29

That means there is a real part and a imaginary part.

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