Prove that $$\lim_{\epsilon\to 0^+}\bigg(\frac{1}{\pi x}\sin\frac{x}{\epsilon}\bigg)=\delta (x)$$

My attempt:

$$\lim_{\epsilon\to 0^+}\bigg(\frac{1}{\pi x}\sin\frac{x}{\epsilon}\bigg)=\delta (x)$$

Re-write LHS to: $$\bigg(\frac{1}{\pi x}\sin\frac{x}{\epsilon}\bigg)=\frac{i}{2\pi}\bigg(\frac{e^{-\frac{ix}{\epsilon}}}{x}-\frac{e^{\frac{ix}{\epsilon}}}{x}\bigg)$$

Let us integrate both sides, $$\int\bigg[\frac{i}{2\pi}\bigg(\frac{e^{-\frac{ix}{\epsilon}}}{x}-\frac{e^{\frac{ix}{\epsilon}}}{x}\bigg)\bigg]\text{d}x=\int \delta(x)\text{d}x$$

$$\int\bigg[\frac{i}{2\pi}\bigg(\frac{e^{-\frac{ix}{\epsilon}}}{x}-\frac{e^{\frac{ix}{\epsilon}}}{x}\bigg)\bigg]\text{d}x=\theta(x) +C$$

Take the limit:

$$\lim_{\epsilon\to 0^+}\int\bigg[\frac{i}{2\pi}\bigg(\frac{e^{-\frac{ix}{\epsilon}}}{x}-\frac{e^{\frac{ix}{\epsilon}}}{x}\bigg)\bigg]\text{d}x=\theta(x) +C$$

$$\frac{i}{2\pi}\bigg[ \lim_{\epsilon\to 0^+}\int\bigg(\frac{e^{-\frac{ix}{\epsilon}}}{x}\bigg)\text{d}x-\lim_{\epsilon\to 0^+}\bigg(\frac{e^{\frac{ix}{\epsilon}}}{x}\bigg)\text{d}x\bigg]=\theta(x) +C$$

where we obtain:

$$\frac{i}{2\pi}\bigg[ \lim_{\epsilon\to 0^+}\text{Erf}\bigg(-\frac{ix}{\epsilon}\bigg)+C-\lim_{\epsilon\to 0^+}\text{Erf}\bigg(\frac{ix}{\epsilon}\bigg)+C\bigg]=\theta(x) +C$$


$$\frac{Ci}{\pi}\bigg[ \lim_{\epsilon\to 0^+}\text{Erf}\bigg(-\frac{ix}{\epsilon}\bigg)-\lim_{\epsilon\to 0^+}\text{Erf}\bigg(\frac{ix}{\epsilon}\bigg)\bigg]=\theta(x) +C$$

Now we can solve the problem by letting $\epsilon=0$ and obtain

$$0=\theta(x) +C$$

which gives, with $\theta(x)=-C$

Insert now in the RHS of

$$\frac{i}{2\pi}\bigg[ \lim_{\epsilon\to 0^+}\int\bigg(\frac{e^{-\frac{ix}{\epsilon}}}{x}\bigg)\text{d}x-\lim_{\epsilon\to 0^+}\bigg(\frac{e^{\frac{ix}{\epsilon}}}{x}\bigg)\text{d}x\bigg]=-C +C$$ and obtain:


But this seems like an odd result. I wonder if the entire approach is wrong.

Any suggestions are welcome.


PS: I found this in a paper, but am not sure it can be used:

enter image description here

  • 1
    $\begingroup$ Proof has to show $\lim_{\epsilon\to 0} \ \delta_\epsilon(f) = f(0)$. I ts easier to Fourier transform. Easy to show, the $\sin x /x =\text{sinc} x$ function has a unit square well step function as fourier transform. The evaluation integral with test functions in $S$ is converted to a convolution over an interval. Proof by pointwise convergence of functions to a distribution is a special art of analysis. $\endgroup$
    – Roland F
    Feb 2 at 11:16
  • $\begingroup$ How would that look more or less like? Do a Fourier transform of both sides then take the limit? Have a look at the PS part I just added. Thanks $\endgroup$ Feb 2 at 11:25

2 Answers 2


If you consider $f(x)\in{\cal D}(\mathbb{R})$ that is the space of the test functions, one can see that $$ \lim_{\epsilon\rightarrow 0^+}\int_{-\infty}^\infty\frac{1}{\pi x}\sin\left(\frac{x}{\epsilon}\right)f(x)=f(0), $$ given the proper change of variable $y=\frac{x}{\epsilon}$.

  • $\begingroup$ Thanks a lot for this $\endgroup$ Feb 2 at 11:56

We start with the following, \begin{align} I &= \lim_{\epsilon \rightarrow 0^+}\frac{1}{\pi x}\sin(\frac{x}{\epsilon})\\ &=\lim_{\epsilon \rightarrow 0^+}\frac{i}{2\pi x} \left(e^{-i(\frac{x}{\epsilon})}-e^{i(\frac{x}{\epsilon})}\right) \end{align} Then let $u=\frac{1}{\epsilon}$, \begin{align} &=\lim_{u \rightarrow \infty} \frac{i}{2\pi x} \left(e^{-iux}-e^{iux}\right) \\ &= \frac{i}{2\pi x}\left[e^{-iux}\right]_{u=-\infty}^\infty \\ &=\frac{i}{2\pi x}\int_{-\infty}^\infty -ix e^{-iux} du\\ &= \frac{1}{2\pi}\int_{-\infty}^\infty e^{-iux}du \end{align} Now we use the integral representation of the $\delta$ function to find, \begin{align} I&=\delta(x) \end{align}


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