Why does the derivative of Dirichlet integral $\int_0^\infty \frac{\sin{ax}}{x} dx$ with respect to $a$ gives nonsense?

Question

If I define $I(a)=\int_0^\infty \frac{\sin{ax}}{x} dx$. We know that $\int_0^\infty \frac{\sin{ax}}{x} dx=\frac{\pi}{2}$. If I differentiate both sides by $a$ then we get $\int_0^\infty \cos{ax} dx=0$. Clearly LHS is undefined, what's going on here?

Answer 1

Let $$F(a)=\int_0^b\frac{\sin(ax)}{x}dx.$$ Let $t=ax$. Then $$F(a)=\int_0^{ab}\frac{\sin(t)}{t}dt.$$Thus $$F(a)=\operatorname{Si}(ab)\implies F'(a)=\frac{\sin(ab)}{ab}b=\frac{\sin(ab)}{a}.$$ The $b\to\infty$ limit of $\sin(ab)$ does not exist.

Answer 2

Let $f(a,x)=\frac{\sin(ax)}{x}$. What you did was interchange integral and derivative: $$\frac d{da}\mathcal I_x f = \mathcal I_x\partial_a f,$$ where $\mathcal I_x$ is shorthand for $\int_0^\infty\cdot\,dx.$ Leibniz integral rule is the relevant theorem here. The bounds are not finite, though, so one should look at this section. For the interchange to work, one needs integrable function $g\colon [0,\infty)\to \mathbb R$ such that for all $a$ and almost every $x$ we have $|\cos(ax)|\leq g(x).$ There is no such $g$ since $t\mapsto \int_0^t|\cos(ax)|\,dx$ grows without bound. So, the interchange is not justified.

That doesn't mean that one still couldn't do it. However, you explicitly showed that you can't and that is in no way problematic, since no theorem says that it can always be done.

EDIT: By the comment of Jack D'Aurizio, $\mathcal I_xf$ is not differentiable at $0$, so even if the bounds were finite, one can't interchange integration and differentiation since the assumptions of the Leibniz integral rule are not satisfied.