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I am trying to prove the following $$\lim_{a\rightarrow\infty}~\int_0^\infty\frac{f(x)\sin(ax)}{x}dx=(\pi/2)f(0)$$ for all square integrable functions $f(x)$ continuous at $0$. I tried to do the following: substitute $ax=t$ then $$\lim_{a\rightarrow\infty}~\int_0^\infty\frac{f(x)\sin(ax)}{x}dx=\lim_{a\rightarrow\infty}~\int_0^\infty\frac{f(\frac{t}{a})\sin(t)}tdt$$ Now i would like to show that $$\lim_{a\rightarrow\infty}~\int_0^\infty\frac{f(\frac t a)\sin(t)}tdt=\int_0^\infty\lim_{a\rightarrow\infty}\frac{f(\frac t a)\sin(t)}tdt$$ To be able to do this I'm trying to use arzela's dominted convergence theorem for riemann integrals which would allow me to carry limit inside the integral. But I am unable to find a dominating function $k(x)\geq0,\int_0^\infty k(x)dx<\infty,|f(t/a)\sin(t)/t|<k(x) \forall a>0$. Can someone help me out

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  • $\begingroup$ @user104254 yup I should have added a factor of $pi/2$. Also since I would be using Dirac delta in the context of quantum mechanics, I would like to know what additional assumptions must I make. $\endgroup$ – somitra Apr 8 '14 at 10:11
  • $\begingroup$ $f$ is continuous at $0$ $\endgroup$ – somitra Apr 8 '14 at 10:17
  • $\begingroup$ But how to prove it. Should I be using dominated convergence theorem or some other approach. $\endgroup$ – somitra Apr 8 '14 at 10:23
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    $\begingroup$ Perhaps the Riemann-Lebesgue lemma might prove helpful. $\endgroup$ – Lucian Apr 8 '14 at 11:06
  • $\begingroup$ Do you have some additional informations about the behavior of $f(x)$ on $[0,1]$? $\endgroup$ – Thorben Apr 8 '14 at 12:38
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I'm not really sure that there is a simple solution for that. In some sense, this limit is equivalent to the theorem about the inverse Fourier transform, so there will be no simple solutions. But use Parseval/Plancherel (I never remember which one is about the Fourier series).

Or proceeding more elementary: Consider for simplicity the symmetric extension of $f$, so that the integral is over $(-∞,∞)$, resulting in twice the required value. Then, using $$ \frac{\sin(ax)}{x}=\frac{e^{iax}-e^{-iax}}{2ix}=\frac12\int_{-a}^a e^{ixv}dv $$ one gets $$ \int_{-∞}^∞ \frac{\sin(ax)}{x}f(x)\,dx = \frac12\int_{-∞}^∞\int_{-a}^a e^{ixv} f(x)\,dv\,dx $$ Now we only need to apply Fubini to exchange the order of integration, so that $$ ...=\frac12\int_{-a}^a \sqrt{2\pi} \hat f(v) \,dv $$ and by dominated convergence, this converges to $$ \frac{\sqrt{2\pi}}2\int_{-∞}^∞\hat f(v) \,dv=\pi\,\widehat{\widehat{f}}(0)=\pi\,f(0) $$ There still are some assumptions that need to be added so that the used Fourier integrals, especially the last one, actually exist. $L^1$ and at least quadratic decay at infinity should be sufficient. The test functions, tempered or with compact support, all satisfy these conditions.

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  • $\begingroup$ I think plancherel's theorem guarantees existence of fourier transform of $f$ and that $\hat f$ is in $L^2$. But it only provides that $f=\hat{\hat f}$ almost everywhere. Does almost everyswhere include $0$ $\endgroup$ – somitra Apr 11 '14 at 2:49
  • $\begingroup$ @somitra The statement about "almost everywhere" can be made more precise: equality holds at every Lebesgue point, provided the functions are properly defined at those points. A point of continuity is a Lebesgue point. $\endgroup$ – user127096 Apr 11 '14 at 3:38
  • $\begingroup$ This feels a bit circular. Isn't the integral in question the key element of the double-Fourier-transform theorem you invoke at the end? $\endgroup$ – E.P. Apr 11 '14 at 7:39
  • $\begingroup$ Yes, that is what I wrote in the first paragraph, that this limit is equivalent to the Fourier inversion or double-Fourier theorem. However, modern proofs of the Fourier inversion theorem do not use the cut-off of the integral but exponential dampening to approximate the Fourier-transform with something controllable. $\endgroup$ – LutzL Apr 11 '14 at 8:27
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It is obvious that this cannot hold unless $f$ is continuous at $0$, because otherwise the value $f(0)$ becomes decoupled to the continuum of values that determine the integral, and the result is meaningless. Therefore, whatever route you take must be based upon this continuity condition.

So, how do you do this? First of all, you unpack the definition. Let $\newcommand{\eps}{\varepsilon}\eps$ be some positive number which we'll fix later. Then we know that there exists $\delta=\delta_\eps$ such that $$ |x|<\delta\Longrightarrow |f(x)-f(0)|<\eps. $$ How does this relate to your integral? Your transformed form, $$ I=\int_0^\infty f(t/a) \frac{\sin(t)}{t}\mathrm dt, $$ involves values of $x=t/a$ which are both below and above the $\delta$ threshold. The ones below, i.e. those $t$ such that $|t|<\delta a$, have $f(t/a)$ close to $f(0)$, which means that they will end up contributing the most, whereas the other ones are unconnected to $f(0)$ and must therefore decouple in some way. To make this change in behaviour explicit, you can split up your integral: $$ I=J+K= \int_{0}^{\delta a} f(t/a) \frac{\sin(t)}{t}\mathrm dt + \int_{\delta a}^\infty f(t/a) \frac{\sin(t)}{t}\mathrm dt. $$


Now, it is tempting to deal with the two chunks at the same time, but the truth is that they require different handling. Let's deal with the messy, unimportant one first. $$ K=K(a)=\int_{\delta a}^\infty f(t/a) \frac{\sin(t)}{t}\mathrm dt $$ goes to zero as $a\to\infty$. This is semi-plausible: you are taking the tails of the integral, which ought to vanish, but your function $f(t/a)$ rescales itself with $a$ to match the outgoing lower limit, so that it always equals $f(\delta)$ there. It is therefore the decay of the $\mathrm{sinc}$ function that drives the limit.

One way to make this precise is to use the Cauchy-Schwarz inequality in what is effectively an $L_2$ dot product: $$ |K(a)|^2 \leq \int_{\delta a}^\infty f(t/a)^2\mathrm dt \int_{\delta a}^\infty \frac{\sin^2(t)}{t^2}\mathrm dt. $$ The first one is constant, equal to $\int_{\delta}^\infty f(t)^2\mathrm dt$ and bounded by $\int_{-\infty}^\infty f(t)^2\mathrm dt$, while the second one converges to zero. Thus $\lim_{a\to\infty}K(a)=0$.


The interesting integral, on the other hand, $$J=J(a)=\int_{0}^{\delta a} f(t/a) \frac{\sin(t)}{t}\mathrm dt,$$ needs to converge to $\propto f(0)$, so we need to use the closeness of $f(t/a)$ to $f(0)$. You can thus unpack $f(x)$ into $f(0)+(f(x)-f(0))$ to get $$ J= f(0)\int_{0}^{\delta a} \frac{\sin(t)}{t}\mathrm dt + \int_{0}^{\delta a} \left[f(t/a)-f(0)\right] \frac{\sin(t)}{t}\mathrm dt. $$ The second term is essentially bounded by $\eps$. You could try to say that $$ \left|\int_{0}^{\delta a} \left[f(t/a)-f(0)\right] \frac{\sin(t)}{t}\mathrm dt\right| < \eps\int_{0}^{\delta a} \left|\frac{\sin(t)}{t}\right|\mathrm dt $$ but the second integral diverges. Instead, one repeats the Cauchy-Schwarz trick: $$ \begin{align} \left|\int_{0}^{\delta a} \left[f(t/a)-f(0)\right] \frac{\sin(t)}{t}\mathrm dt\right|^2 &\leq \int_{0}^{\delta a} \left|f(t/a)-f(0)\right|^2\mathrm dt\int_{0}^{\delta a} \left|\frac{\sin(t)}{t}\right|^2\mathrm dt \\ & < \delta a \eps^2\int_{-\infty}^{\infty} \left|\frac{\sin(t)}{t}\right|^2\mathrm dt. \end{align} $$ You now have enough ingredients to get the limit, and you can get the nice number by doing the Dirichlet integral $$ \int_{0}^\infty \frac{\sin(t)}{t}\mathrm dt=\frac\pi2. $$ The limiting procedure takes a bit of care, because you have a fair few quantities involved. If you package all the results so far, you can say that $$ |I-\frac\pi2 f(0)| < f(0)\left|\int_{\delta a}^\infty \frac{\sin(t)}{t}\mathrm dt\right| + \eps\sqrt{\delta a} \|\mathrm{sinc}\| + \|f\|\sqrt{\int_{\delta a}^\infty \left|\frac{\sin(t)}{t} \right|^2 \mathrm dt}, $$ and you want to bound this arbitrarily close to zero. This is a bit of a balancing act, because you need $\delta a$ to be large and $\eps\sqrt{\delta a}$ to be small, but it can be done. To do it, take $A$ such that the first and third terms are small when $\delta a=A$; then, choose $\eps$ such that the second term is small enough, for $\delta a=A$ fixed; this determines the necessary $\delta$; and finally take $a=A/\delta$. This is enough to guarantee the limit. $$\tag*{$\blacksquare$}$$


Finally, let me put in a word or two about why I went all this way instead of cracking the problem with some huge theorem. Essentially, trying to apply some form of dominated convergence theorem will never work, because there is no dominating function. In physicist language, the Dirichlet kernel converges to a Dirac delta function: $$\lim_{a\to\infty}\frac{\sin(ax)}{x}=\frac\pi2 \delta(x).$$ This is loose language but its precise statement is exactly the integral you want to prove.

In other words, this integral breaks exactly the same dominated convergence schemes that you're trying to use, and the limit - in whatever sense it exists (which it does) - is outside of the function space you're looking in. This is why, unless you have fancier tools available, you need to work it from the ground up.

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  • $\begingroup$ I think the final inequality should be $|I-\frac\pi2 f(0)| < \left|f(0)\int_{\delta a}^\infty \frac{\sin(t)}{t}\mathrm dt\right| + \epsilon\sqrt{\delta a} \|\mathrm{sinc}\| + a\|f\|\sqrt{\int_{\delta a}^\infty \left|\frac{\sin(t)}{t} \right|^2 \mathrm dt},$ $\endgroup$ – somitra Apr 10 '14 at 6:21
  • $\begingroup$ Yes, that is correct. Apologies. $\endgroup$ – E.P. Apr 10 '14 at 9:24
  • $\begingroup$ In that case the third term cannot be guaranteed to be arbitrarily small as $a\rightarrow\infty$. So the analysis breaks down I think. $\endgroup$ – somitra Apr 10 '14 at 13:47
  • $\begingroup$ Sorry, I didn't see that modification. You are right that the first term was incorrect. Your third term is not. $\endgroup$ – E.P. Apr 10 '14 at 15:12
  • $\begingroup$ I say this because $\int_{\delta a}^\infty f(t/a)^2\mathrm dt=a\int_{\delta}^\infty f(t)^2\mathrm dt$ after substituting $t/a\rightarrow t$ $\endgroup$ – somitra Apr 10 '14 at 15:30
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Im really not sure if this works but i thought I try,

As you mentioned the claim, $$lim_{a\rightarrow\infty}\int_0^{\infty}f(x)\frac{sin(ax)}{x}dx=\pi/2f(0)$$ follows after your substitution as $f(x)$ is continuous at $0$ when we can move the limit inside.

So I started to construct $g(x)\in L^1(\mathbb R_+)$ s.t $|\int_0^{\infty}f(x)\frac{sin(ax)}{x}|\leq g(x)$.

For $x\in[0\infty)$ we have,

$$|f(x)\frac{sin(ax)}{x}|\leq|\frac{f(x)}{x}|$$ and Hölders inequality yields,

$$\int_1^{\infty}|\frac{f(x)}{x}|\leq(\int_1^{\infty}|f(x)|^2dx)^{1/2}(\int_1^{\infty}|1/x|^2)^{1/2}=(\int_1^{\infty}|f(x)|^2dx)^{1/2}<\infty $$ which follows from the assumption that $f(x)$ was square integrable.

Now for $x\in[0,1]$ we have two cases,

Let $A=\{x\in[0,1]:|f(x)\frac{sin(ax)}{x}|<1\}$. Clearly for all $x\in A$ we have,

$$ |f(x)\frac{sin(ax)}{x}|\leq1$$

which is integrable since $\lambda(A)\leq\lambda([0,1])=1$,

Let $B=\{x\in[0,1]:|f(x)\frac{sin(ax)}{x}|\geq1 \}$.

For $x\in B$ we have,

$$|f(x)\frac{sin(ax)}{x}|\leq|f(x)\frac{x}{x}|\leq|f(x)|\leq|f(x)|^2$$ which is integrable.

So finally we found a bound in $L^1$ for all $x\in\mathbb R_+$. And you can move the limit inside to obtain what was to prove.

I hope i didnt miss something... do not hesitate to correct me.

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    $\begingroup$ Although we have found a bound for $|f(x)\frac{sin(ax)}{x}|$ the limit still can't be moved inside as $lim_{a\rightarrow\infty}f(x)\frac{sin(ax)}{x}$ doesn't exist. $\endgroup$ – somitra Apr 10 '14 at 5:09
  • $\begingroup$ @somitra use your substitution, then it should work... or am I missing something? $\endgroup$ – Thorben Apr 10 '14 at 8:51
  • $\begingroup$ I think for substitution to work, limit should exist. If we move the limt inside before making substitution then the resultant mathematical expression does not have a meaning as the limit does not exist. $\endgroup$ – somitra Apr 10 '14 at 9:09
  • $\begingroup$ @somitra Yes you actually right, and the bound wont work after the substitution. Sorry for that! As is said, it was just a try... $\endgroup$ – Thorben Apr 10 '14 at 9:29

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