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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 $t = 0+$ ?

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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.

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What you call $g(+0)$ is more commonly denoted $g(0+)$. –  Did Aug 14 '11 at 14:37
    
@Andrew : "if the difference $g(t)-g(+0)$ is small enough." What do you mean by this, i am not able to get. –  Rajesh D Aug 16 '11 at 5:26
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@Rajesh It is an informal phrase. I meant just what was said below. –  Andrew Aug 16 '11 at 6:22
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