With suitable interpretation (!), it is perfectly ok to compute as
$$
{d\over dx}\int_{-\infty}^\infty e^{itx}\;dt \;=\;
\int_{-\infty}^\infty {d\over dx}e^{itx}\;dt
\;=\; \int_{-\infty}^\infty it\cdot e^{itx}\;dt
$$
(Faux-secretly, the integral is the Fourier transform being taken as tempered _distribution_, and not as limit of Riemann sums...) In this example, one must be prepared to recognize the outcome (which, as numerical integral diverges, of course) as simply the derivative of Dirac delta, just as the original was Dirac delta itself.
While it is true that $\delta$ does not have a pointwise value at $0$, it certainly does have pointwise values away from $0$. For that matter, before people decided to formalize "function" as something that should have pointwise values, Euler and many others often treated "function" as sometimes meaning "expression". Meanwhile, an $L^2$ function doesn't really have pointwise values, so in some sense is worse off than $\delta$.
Fourier inversion on $L^2(\mathbb R)$ would seem to involve integrals that needn't converge, but with Plancherel's theorem in hand, we continue to write those integrals, but disclaim that the notation must be understood as meaning the extension-by-continuity from a smaller space.
I think much of the point of distribution theory is to be able to treat distributions not merely as functionals on classical functions, but as generalized functions, permitting the same operations, if extended suitably.
Operations such as dilation and translation are more easily notated by using "argument" notation $x\to \delta(x-y)$, although, yes, it is risky to too hastily assume that generalized functions share all the properties of classical ones.
Further, the first round of distribution theory is not the end of the story, even to make best legitimate use of $\delta$. Namely, a finer gradation of "distributions" (and classical functions) is very often useful, as measure of how far a generalized function is from being $L^2$ or continuous, etc. Thus, one can "predict" that a solution $u$ to an equation $u''+q(x)u=\delta$ (for smooth $q$, say) will be continuous.
I only recently learned that Dirac's original intuitive use of $\delta$ did actually resemble what would now be formalized as a "Gelfand triple" $H^{+1}\subset L^2\subset H^{-1}$ of Levi-Sobolev spaces on $\mathbb R$, where $\delta$ lies in $H^{-1}$ but not quite in $L^2$. Thus, general distribution theory made $\delta$ "legal", but in itself did not yet manage to account for Dirac's marvelous intuition.
(Yes, nowadays standard coursework teaches us to have a sometimes-too-narrow viewpoint, providing an excuse to merely dismiss good ideas as "unrigorous" rather than figuring out how to legitimize them. The utility of Heaviside's and Dirac's "non-rigorous" ideas out-weighed the difficulty of justification in everyone's eyes except mathematicians, perhaps. That is, if you seem to always be able to "get the right answer" (with physical corroboration), it's hard to take seriously an objection that one wasn't playing fair, all the more when "fair" is according to rules that were made up by someone.)
Best Answer
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{0}^{\infty}{x\cos\pars{x} - \sin\pars{x} \over x^{3}}\, \cos\pars{x \over 2}\,\dd x \\[5mm] = & \int_{0}^{\infty}{x\bracks{1 - 2\sin^{2}\pars{x/2}} - \sin\pars{x} \over x^{3}}\,\cos\pars{x \over 2}\,\dd x \\[5mm] & = {1 \over 2}\int_{0}^{\infty}{2x\cos\pars{x/2} - 2x\sin\pars{x}\sin\pars{x/2} - 2\sin\pars{x}\cos\pars{x/2} \over x^{3}}\,\,\,\dd x \\[5mm] & = {1 \over 2}\int_{0}^{\infty}{x\cos\pars{x/2} + x\cos\pars{3x/2} - \sin\pars{3x/2} - \sin\pars{x/2} \over x^{3}}\,\,\,\dd x \\[1cm] & = -\,{1 \over 2}\int_{0}^{\infty}{1 - \cos\pars{x/2} \over x^{2}}\,\dd x - {1 \over 2}\int_{0}^{\infty}{1 - \cos\pars{3x/2} \over x^{2}}\,\dd x \\[5mm] & - {1 \over 4}\int_{x\ =\ 0}^{x\ \to\ \infty}\bracks{2x - \sin\pars{3x/2} - \sin\pars{x/2}}\,\dd\pars{1 \over x^{2}} \end{align}
Integrating by parts the last integral: \begin{align} &\int_{0}^{\infty}{x\cos\pars{x} - \sin\pars{x} \over x^{3}}\, \cos\pars{x \over 2}\,\dd x = \\[5mm] & = -\,{1 \over 2}\int_{0}^{\infty}{1 - \cos\pars{x/2} \over x^{2}}\,\dd x - {1 \over 2}\int_{0}^{\infty}{1 - \cos\pars{3x/2} \over x^{2}}\,\dd x \\[5mm] & + {1 \over 4}\int_{x = 0}^{\infty}{2 - 3\cos\pars{3x/2}/2 - \cos\pars{x/2}/2 \over x^{2}}\,\dd x \\[1cm] & = -\,{1 \over 2}\int_{0}^{\infty}{1 - \cos\pars{x/2} \over x^{2}}\,\dd x - {1 \over 2}\int_{0}^{\infty}{1 - \cos\pars{3x/2} \over x^{2}}\,\dd x \\[5mm] & + {3 \over 8}\int_{0}^{\infty}{1 - \cos\pars{3x/2} \over x^{2}}\,\dd x + {1 \over 8}\int_{0}^{\infty}{1 - \cos\pars{x/2} \over x^{2}}\,\dd x \\[1cm] & = -\,{3 \over 8}\int_{0}^{\infty}{1 - \cos\pars{x/2} \over x^{2}}\,\dd x -\,{1 \over 8}\int_{0}^{\infty}{1 - \cos\pars{3x/2} \over x^{2}}\,\dd x \\[5mm] & = -\,{3 \over 16}\int_{0}^{\infty}{1 - \cos\pars{x} \over x^{2}}\,\dd x -\,{3 \over 16}\int_{0}^{\infty}{1 - \cos\pars{x} \over x^{2}}\,\dd x = -\,{3 \over 8}\ \int_{0}^{\infty}{1 - \cos\pars{x} \over x^{2}}\,\dd x \\[5mm] & = -\,{3 \over 4}\int_{0}^{\infty}{\sin^{2}\pars{x/2} \over x^{2}}\,\dd x = -\,{3 \over 8}\ \underbrace{\int_{0}^{\infty}{\sin^{2}\pars{x} \over x^{2}}\,\dd x} _{\ds{=\ {\pi \over 2}}}\ = \ \bbox[#ffe,10px,border:1px dotted navy]{\ds{-\,{3 \over 16}\,\pi}} \end{align} >By integrating by parts: $\ds{\int_{0}^{\infty}{\sin^{2}\pars{x} \over x^{2}}\,\dd x = \int_{0}^{\infty}{\sin\pars{x} \over x}\,\dd x = {1 \over 2}\,\pi}$.