Trying to integrate $$\int \frac{\cos(x)}{x} dx = \int \frac{1}{x}\sin'(x) dx$$ by substituting $\sin(x)$, but it either becomes more complicated or I end up with a $\frac{1}{x}$ still in the integral.
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It is not possible to find an antiderivative of $\frac{\cos x}{x}$ in term of "elementary functions".
This is a consequence of Liouville's theorem. See link to article for details.
mathcounterexamples.net
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Does that mean that integrating $$\int \frac{1-\cos(x)}{x}dx$$ with substitution is also impossible? Because I split it up and thought it would be easier to integrate $\frac{1}{x}$ and $\frac{\cos(x)}{x}$ separately... – Hrach Jan 04 '19 at 14:16
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You might want to refer to them as elementary functions, because simple functions often refer to other categories (e.g. in measure theory). – edmz Jan 04 '19 at 14:46
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As noted the indefinite integral $$ \int \frac{\cos x}{x}\;dx $$ is not an elementary function. But it is useful enough that it has been given a name, the "cosine integral" function, $\mathrm{Ci}(x)$. It is conventional to fix the constant of integration so that $\lim_{x \to +\infty} \mathrm{Ci}(x) = 0$. So we may define $$ \mathrm{Ci}(x) = -\int_x^\infty\frac{\cos t}{t}\;dt $$ In fact, this definition makes sense for $x$ in the complex plane, with a cut along the negative real axis.
GEdgar
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