$\int_{-\infty}^{\infty} \frac{\sin(x)}{x} dx $ with integration by parts? The integral of the Sinc function over $\mathbb{R}$ is well-known, $$\int_{-\infty}^{\infty} \frac{\sin(x)}{x} dx = \pi$$But if I try to evaluate this using integration by parts, I get $$\int_{-\infty}^{\infty} \frac{\sin(x)}{x} dx = \left.-\frac{\cos(x)}{x} \right|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \frac{\cos(x)}{x^2} dx$$ The first part is $0$, and the second part diverges.
What's going on? Is integration by parts just not kosher here? If so, why not?
 A: The issue is the discontinuity at $0$ in the cosine terms when you perform the partial integration, which is overlooked if you integrate over $\mathbb{R}$. To get around this, consider:
$$\int_{-\infty}^{\infty}\frac{\sin x}{x}dx=2\int_0^{\infty}\frac{\sin x}{x}dx$$
By parts:
$$\int_{0}^{\infty} \frac{\sin(x)}{x} dx = \left.-\frac{\cos(x)}{x} \right|_{0}^{\infty} - \int_{0}^{\infty} \frac{\cos(x)}{x^2} dx$$ 
And here, both terms on the RHS diverge. No contradiction.
A: The definite integration by parts formula
$$\int_a^b f(x) g(x)\ dx =  F(x) g(x)\bigg|_a^b  - \int_a^b F(x) g'(x)\ dx$$
is justified by the product rule for derivative and the fundamental theorem of calculus
$$\int_a^b (F g)'(x)\ dx =  F(x) g(x) \bigg|_a^b $$
But the "fine print" there is that $F(x) g(x)$ is assumed to be continuously differentiable on the interval
$(a,b)$ and continuous on $[a,b]$.  In this case with $F(x) = -\cos(x)$ and $g(x) = 1/x$,
$F(x) g(x)$ is undefined at $x=0$, and no definition of it there will make it continuous at $x=0$, let alone differentiable.  So you can't use the integration by
parts formula for this integral on any interval containing $0$.
A: 
As others have already discussed, integrating by parts with $u=\frac1x$ and $v=-\cos(x)$ fails due to the singularity at $0$.  Here, I thought it would be instructive to present an integration by parts scheme that circumvents the difficulty of the singularity at $0$.  To that end, we proceed.


Let $I$ be the integral given by 
$$I=\int_0^\infty \frac{\sin(x)}{x}\,dx\tag1$$
Enforcing the substitution with $u=\frac1x$ and $v=1-\cos(x)$ (instead of $\displaystyle v=-\cos(x)$) in $(1)$ yields
$$\begin{align}
I&=\left.\left(\frac{1-\cos(x)}{x}\right)\right|_0^\infty+\int_0^\infty \frac{1-\cos(x)}{x^2}\,dx\\\\\
&=2\int_0^\infty \left(\frac{\sin(x/2)}{x}\right)^2\,dx\\\\
&=\int_0^\infty \left(\frac{\sin(x)}{x}\right)^2\,dx\tag2
\end{align}$$
While $(2)$ does not lead immediately to a value for $I$ it does reveal the interesting identity
$$\int_0^\infty\frac{\sin(x)}{x}\,dx=\int_0^\infty \left(\frac{\sin(x)}{x}\right)^2\,dx$$
