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In my solution, for the proof of $$\int_0^\infty \frac{\sin(ax)}{x}dx=\frac{\pi}{2} \operatorname{sgn}(a),$$

they do as under. The only important thing is where the red square is (the rest is as usual). The says that $\sin(t)\geq \frac{t}{2}$ when $[0,\pi/6]$ and $\sin(t)\geq \frac{1}{2}$ when $t\in [\pi/6,\pi/2]$. But why don't the simply use the fact that $\sin(t)\geq \frac{t}{2}$ on $[0,\pi/2]$, what gives $$\int_0^{\pi/6}e^{-R\sin(t)}dt\leq \int_0^{\pi/2}e^{-Rt/2}dt=\frac{-2}{R}\left[e^{-Rt/2}\right]_{0}^{\pi/2}=\frac{-2}{R}(e^{-R\pi/4}-1)\underset{R\to \infty }{\longrightarrow }0.$$

Is there something I didn't get ? Because taking $\sin(t)\geq t/2$ for all $t\in [0,\pi/2]$ instead of taking $\sin(t)\geq t/2$ on $[0,\pi/6]$ and $\sin(t)\geq 1/2$ on $[\pi/6,\pi/2]$ looks to work great, no ?

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user349449
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1 Answers1

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I can't tell why the author(s) of the solution chose to split the integral. Splitting it and not splitting it both work, so there's also no reason to not split, if one likes to do that. But splitting the integral introduces marginally more work since then one has to estimate two integrals instead of one. Of course estimating one of them is utterly trivial, hence "marginally more work" is basically just the additional writing. The interesting (not totally trivial, but not at all difficult once that point is reached) thing happens close to $0$, and can be summarised as

For all $A > 0$ we have $$\int_0^A e^{-Kt}\,dt \sim \int_0^{\infty} e^{-Kt}\,dt = \frac{1}{K}$$ as $K \to +\infty$.

Daniel Fischer
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