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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? –  glebovg Nov 20 '12 at 4:20
    
You may need to read this. –  Mhenni Benghorbal Nov 20 '12 at 5:00
    
Concerning the edit : just some days earlier we had the same kind of question. Set $z=x^3$ to get $\ \frac 13\int \frac{\cos(z)}z dz$ and use the demonstration there for $\int \frac{\sin(z)}z dz$. –  Raymond Manzoni Nov 20 '12 at 9:09
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2 Answers 2

up vote 3 down vote accepted

$$I = \int \dfrac{\cos(x^3)}{x} dx = \dfrac13 \int \dfrac{\cos(x^3)}{x^3} (3x^2)dx = \dfrac13 \int \dfrac{\cos(x^3)}{x^3} d(x^3) = \dfrac{\text{Ci}(x^3)}{3} + \text{constant}$$ 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? –  glebovg Nov 20 '12 at 4:21
    
Can someone explain what is "Ci"? –  learner Mar 29 '13 at 13:00
    
@learner $\operatorname{Ci}(\xi)$ stands for cosine integral. –  glebovg Apr 5 '13 at 22:01
    
@glebovg thanks a lot. –  learner Apr 6 '13 at 3:32
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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.

Disclaimer: I am in no way affiliated with Wolfram or Wolframalpha.

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