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How does one integrate $\cos(x^{-1})$?

I understand that the function is not defined at zero, but it is well defined, continuous, and real over the rest of $\mathbb{R}$. Nonetheless, when I put $\int \cos(x^{-1})\; dx$ into a computer algebra system, the result is in terms of imaginary units and special functions with which I'm not familiar.

Surely this function has been addressed (else how would the CAS know the answer) but I don't see it on integration tables anywhere and Google turns up empty-handed. Can someone derive the answer for me? or at least point me to a good resource for it?

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Substituting $u=1/x$ and then integrating by parts yields

$ \begin{eqnarray} \int\cos\left(\frac1x\right)\mathrm dx &=& -\int\frac{\cos u}{u^2}\mathrm du \\ &=& \frac{\cos u}{u}+\int\frac{\sin u}{u}\mathrm du\;. \\ \end{eqnarray}$

That last integral is called the sine integral and denoted by $\operatorname{Si}$. Thus we have

$ \begin{eqnarray} \int\cos\left(\frac1x\right)\mathrm dx &=& \frac{\cos u}{u}+\operatorname{Si}(u) \\ &=&x\cos\left(\frac1x\right)+\operatorname{Si}\left(\frac1x\right) \end{eqnarray}$

(plus a constant).

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    @joriki: Nope, it w$a$s a problem with my CAS. See this question (http://math.stackexchange.com/questions/91812/how-does-int-1-x-cos2-pi-t-dt-have-complex-values-for-real-values-of-x/91829#91829).2011-12-15