How can I evaluate ;
$$\int \frac{\cos(x)}{x} \, dx $$
I tried to do partial integration but it became to an infinite loop of partial integrations
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$\begingroup$Recall the Taylor series expansion for $\cos x$\begin{align*}\cos x &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \\ \implies \frac{\cos x}{x} &= \frac{1}{x} - \frac{x}{2!} - \frac{x^3}{4!} - \frac{x^5}{6!} + \cdots \\ \therefore \int \frac{\cos x}{x} &= ln\ x - \frac{x^2}{2 \times 2!} + \frac{x^4}{4 \times 4!} - \frac{x^6}{6 \times 6!} + \cdots + \frac{x^{2k}}{(-1)^{k}\times2k \times (2k)!} + \cdots \end{align*}This sum approaches zero so that the indefinite integral is $ln(x)$ up to an integration constant. Moreover, if the terminals of integration are say $a$ and $b$ (not zero or infinity), the definite integral would be $ln(a)-ln(b)$. Where the terminals include zero or infinity, the Trigonometric Integral $Ci(x)$ or $Cin(x)$ need to be used.
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