A question on limit of a Dirichlet's integral

Question

I want to know the idea/intuition behind the Dirichlet integral. For example consider $$I(\alpha) = \int_0^{\delta} g(t) \frac{\sin(\alpha t)}{t} dt.$$ Why would $I(\alpha)$ tend to some constant times $g(0+)$ as $\alpha$ tends to $\infty$ when $g(t)$ staisfies certain conditions in $(0,\delta)$ like for example Jordan's test or Dini's test. My interest in this question is that this idea is central to the convergence of Fourier series. Can we derive any such similar result based on degree of differentiability of $g(t)$ at $t = 0+$ ?

Answer 1

Writing the integral as $$ I(\alpha)=g(+0)\int_0^{\delta} \frac{\sin(\alpha t)}{t} dt+\int_0^{\delta}\frac{g(t)-g(+0)}t \sin\alpha t\; dt=I_1(\alpha)+I_2(\alpha) $$ we have $$ \lim_{\alpha\to+\infty}I_1(\alpha)=g(+0)\int_0^{\infty} \frac{\sin(t)}{t}dt=g(+0)\frac\pi2 $$ and the second one tends to zero if the difference $g(t)-g(+0)$ is small enough. The sine function oscillates quickly as $\alpha\to\infty$. If say function $f(t)={\bf 1}_{[0,\delta]}\frac{g(t)-g(+0)}t\ $ is absolutely integrable then $\lim_{\alpha\to+\infty}I_2(\alpha)=0\;$ because Fourier transform of a function $f\in L_1(\mathbb R)$ vanishes at infinity. Differentiability of $g$ on $[0,\delta]$ is a sufficient condition, but in this case the result follows from the Dini's test.

Answer 2

The criteria in Jordan's and Dini's tests are not compatible in the sense that there are cases for which Jordan's test applies while Dini's does not and others for which Dini's test applies and Jordan's does not. Their difference can be summarized as follows: Dini's requires a global property (integrability), whereas Jordan's requires local regularity (bounded variation in a neighborhood of a point).

Here I discuss Jordan's test and Dini's test separately and some examples to illustrate their incompatibility.

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Jordan's integral result: If $f$ is bounded variation on $[0,\delta]$ for some $0<\delta$, then \begin{align} \lim_{a\rightarrow\infty}\int^\delta_0 f(t)\frac{\sin\alpha t}{t}\,dt = \frac{\pi}{2} f(0+)\tag{0}\label{zero}\end{align}

The bounded variation condition guarantees that $f$ has left and right limits at every point in $[0,\delta]$ since $f$ can be expresses as the difference of monotone nondecreasing functions. The fast oscillations of $t^{-1} \sin(\alpha t)$ near $0$ as $\alpha\rightarrow$ average out to a constant; away from $0$, we have integrability and negligible effect due to Riemann-Lebesgue's lemma.

More explicitly, notice that the function $F(t)=f(t)-f(0+)$ for $0<t\leq \delta$ and $F(0)=0$ is of bounded variation on $[0,\delta]$, and continuous at $x=0$. Then there are monotone non decreasing functions $G$ and $F$ such that $F=G-H$, $G,F\geq0$ continuous at $t=0$, and $G(0)=H(0)=0$. To obtain \eqref{zero} it is then enough to assume that $F$ monotone nondecreasing with $F(0)=0=F(0+)$.

Since the improper integral $\int^\infty_0\frac{\sin t}{t}\,dt$ converges, it follows that $$0<M:=\sup_{0<a<b<\infty}\Big|\int^b_a\frac{\sin t}{t}\,dt\Big|<\infty$$ Given $\varepsilon>0$, choose $\delta_1$ such that $0\leq F(t)<\frac{\varepsilon}{M}$ whenever $0<t\leq\delta_1$. Then \begin{align} \int^\delta_0F(t)\frac{\sin \alpha t}{t}\,dt&=\int^{\delta_1}_0 F(t)\frac{\sin \alpha t}{t}\,dt+\int^\delta_{\delta_1}F(t)\frac{\sin\alpha t}{t}\,dt\\ &= I_\alpha(0,\delta_1) + I_\alpha(\delta_1,\delta) \end{align}

The function $t\mapsto\frac{F(t)}{t}\mathbb{1}_{(\delta_1,\delta]}(t)$ is integrable; hence, by the Riemann-Lebsesgue lemma $I_\alpha(\delta_1,\delta)=\int^\delta_{\delta_1}F(t)\frac{\sin \alpha t}{t}\,dt\xrightarrow{\alpha\rightarrow\infty}0$.

For the integral $I_\alpha(0,\delta_1)$ we are in a position to use the second mean value theorem for integrals (Bonnet's theorem). There is $\xi=\xi(\alpha)\in[0,\delta_1]$ such that $$I_\alpha(0,\delta_1)=\int^{\delta_1}_0F(t)\frac{\sin \alpha t}{t}\,dt = F(\delta_1-)\int^{\delta_1}_{\xi}\frac{\sin\alpha t}{t}\,dt=F(\delta_1-)\int^{\alpha\delta_1}_{\alpha \xi}\frac{\sin t}{t}\,dt$$ Hence $$\lim_{\alpha\rightarrow\infty}|I_\alpha(0,\delta_1)|\leq \varepsilon$$ Putting things together, $$\limsup_{\alpha\rightarrow\infty}\Big|\int^\delta_0 F(t)\frac{\sin \alpha t}{t}\,dt\Big|=0$$ and the conclusion follows.

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Dini's integral result: Suppose $f$ has right limit at $t=0$ and that the function $\phi(t):=\frac{f(t)-f(0+)}{t}$, $0<t\leq \delta$ is intégrable (in the sense of Lebesgue). Then \eqref{zero} holds.

This is a much more direct result since any problem due to a potential singularity of $\phi$ at $0$ is remove just by integrability through the Riemann-Lebesgue lemma. Indeed \begin{align} \int^\delta_0 f(t)\frac{\sin \alpha t}{t}\,dt &=\int^\delta_0\phi(t)\sin(\alpha t)\,dt+f(0+)\int^\delta_0\frac{\sin \alpha t}{t}\,dt\\ &=J_1(\alpha)+J_2(\alpha) \end{align}

The term $J_1(\alpha)=\int^\delta_0\phi(t)\sin\alpha t\,dt\xrightarrow{\alpha\rightarrow\infty}0$ by the Riemann-Lebesgue lemma; as for the second term, $$J_2(\alpha)=f(0+)\int^\delta_0\frac{\sin \alpha t}{t}\,dt=f(0+)\int^{\alpha \delta}_0\frac{\sin t}{t}\,dt\xrightarrow{\alpha\rightarrow\infty}\frac{\pi}{2}f(0+)$$

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For functions that are smooth enough, for example piecewise continuously differentiable functions, both tests (Dini's and Jordan's) are applicable. This is not the case in more general situations.

Example 1. The $2\pi$-peridic function $f(-t)=f(t)=\frac{1}{|\log\big(\frac{t}{2\pi}\big)|}$ for $0<t\leq \pi$ and $f(0)=0$ is of finite variation near $0$ and $f(0)=f(0+)$. The function $\phi(t)=f(t)/t$ is not integrable on any interval $[0,\delta]$. Jordan's test is applicable here, Dini's is not.

Example 2. The $2\pi$-peridic function $f(-t)=f(t)=\sqrt{t}\sin\big(\frac{1}{t}\big)$ for $0<t\leq \pi$ and $f(0)=0$ is not of bounded variation near $0$. The function $\phi(t) =f(t)/t$ is integrable in any interval $[0,\delta]$. Dini's is applicable here, while Jordan's is not.