Integral of an odd function For an odd function $f$ the integral $\int_{-R}^R f dx$ vanishes if $f$ is continuous on $[-R,+R]$. Now, let $f(x)=\frac{\cos(x)}{x}$ which is odd but discontinuous at 0. How can I show (not) that $\int_{-R}^R f dx = 0$?
 A: The issue is that $f$ is not merely discontinuous at $0$ but unbounded near $0$.  Thus what you have is an improper integral.  In order for it to converge, by definition, you need $\lim_{a \rightarrow 0^+} \int_a^R f$ to exist. 
But this limit does not exist.  Hint: for $x$ close to zero, $\frac{ \cos x}{x}$ is close to $\frac{1}{x}$.  
Added Of course you also need $\lim_{a \rightarrow 0^-} \int_{-R}^a f$ to exist.  But on the one hand, the oddness tells you that if the first limit exists then so will the second.  On the other hand, the first limit does not exist, which is enough to show the divergence.)
A: Your function is unbounded near $0$, in fact $f(x) \approx 1/x$ for small $x$. (By a Maclaurin expansion of $\cos x$.) Hence, your improper integral diverges.
A: For $-\frac{\pi}{4}\le x\le\frac{\pi}{4}$, we have $\left|\frac{\cos(x)}{x}\right|\ge\frac{1}{\sqrt{2}|x|}$. Therefore, as $\epsilon\to0$,
$$
\begin{align}
\int_\epsilon^{\pi/4}\frac{\cos(x)}{x}\mathrm{d}x
&\ge\int_\epsilon^{\pi/4}\frac{1}{\sqrt{2}\,x}\mathrm{d}x\\
&=\frac{1}{\sqrt{2}}\log\left(\frac{\pi}{4\epsilon}\right)\\
&\to\infty\tag{1}
\end{align}
$$
and
$$
\begin{align}
\int_{-\pi/4}^{-\epsilon}\frac{\cos(x)}{x}\mathrm{d}x
&\le\int_{-\pi/4}^{-\epsilon}\frac{1}{\sqrt{2}\,x}\mathrm{d}x\\
&=-\frac{1}{\sqrt{2}}\log\left(\frac{\pi}{4\epsilon}\right)\\
&\to-\infty\tag{2}
\end{align}
$$
Thus,
$$
\int_{-R}^{R}\frac{\cos(x)}{x}\mathrm{d}x\tag{3}
$$
diverges.
Cauchy Principal Value
Note that the 
$$
\int_{-R}^{-\epsilon}\frac{\cos(x)}{x}\mathrm{d}x+\int_{\epsilon}^{R}\frac{\cos(x)}{x}\mathrm{d}x=0\tag{4}
$$
as $\epsilon\to0$. Equation $(4)$ says that the Cauchy Principal Value of 
$$
\int_{-R}^{R}\frac{\cos(x)}{x}\mathrm{d}x=0\tag{5}
$$
