The Dirichlet integral can be evaluated with a neat trick. This post is nothing special, just a record of my own process of understanding how it works.
This is the integral:
$$\int_0^\infty \frac{\sin x}{x}dx = \frac{\pi}{2}$$
This is easy to illustrate with a Fourier transform and Parseval's theorem: $$\frac{\sin x}{x} = \operatorname{sinc} x \xleftrightarrow{\mathscr{F}} \operatorname{rect} x$$, the easiest thing to integrate. More interestingly, it can also be done with differentiation under the integral sign (see also "Liebnitz integral rule", "integration by parametric derivative", "order of integration"), by multiplying by an appropriate "auxiliary function" (a term I made up).
Define
$$I(t) = \int_0^\infty e^{-tx}\frac{\sin x}{x}dx$$
This is part one of the trick - define a problem that is more involved than the original problem, but includes it as a special case: $$I(t)$$ is equal to the Dirichlet integral for $$t=0$$. I like it because the new problem has a sort of higher dimensionality, but is also easier to solve.
Anyway, differentiate $$I(t)$$ with respect to the auxiliary variable:
$$\frac{dI}{dt} = \frac{d}{dt} \int_0^\infty \left( e^{-tx}\frac{\sin x}{x} \right) dx$$
$$\frac{dI}{dt} = \int_0^\infty \frac{d}{dt}\left( e^{-tx}\frac{\sin x}{x} \right) dx$$
$$\frac{dI}{dt} = \int_0^\infty -x \left( e^{-tx}\frac{\sin x}{x} \right) dx$$
$$\frac{dI}{dt} = -\int_0^\infty e^{-tx}\sin x dx$$
And this is the second part: by adding the exponential term, then differentiating under the integral, we get rid of the pesky denominator term.
This can be evaluated with repeated integration by parts, or complex exponentials:
$$\frac{dI}{dt} = -\frac{1}{t^2+1} $$
And this is a well-known antiderivative:
$$I(t) = -\int\frac{1}{t^2+1}dt = -\tan^{-1}(t) + c$$
To determine $$c$$, note that $$I(\infty)=0$$ (since the exponential decays infinitely fast).
$$I(\infty) = 0 = -\tan^{-1}(\infty) + c$$
$$c = \tan^{-1}(\infty)$$
$$c = \pi/2$$
And finally,
$$I(0) = -\tan^{-1}(0) + \pi/2$$
$$I(0) = \pi/2 \blacksquare$$
Sinc powers
When the sinc term is raised to a power, can we use the same method? You still get the $$1/x^k$$ cancellation, what comes after that?
$$I_n(t) = \int_0^\infty e^{-tx}\left(\frac{\sin x}{x}\right)^ndx$$
$$\frac{d^n}{dt} I_n(t) = \int_0^\infty \frac{d^n}{dt} e^{-tx}\left(\frac{\sin x}{x}\right)^ndx$$
$$\frac{d^n}{dt} I_n(t) = (-1)^n \int_0^\infty e^{-tx} (\sin x)^ndx$$
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