I had this on an exam yesterday, and I'm not entirely convinced that the statement is true. We were asked to show that the function $\delta (x) = \int_{-∞}^{∞} \frac{1}{t(t-x)} dt$ is a dirac delta function by demonstrating that $I=\int_{-∞}^{∞} f(x)\delta(x) dx$ holds all the necessary properties.

There are three things I believe should be shown: 1) The function should be infinite at a single point. (this function is infinite at $t=x$) 2) It should be zero everywhere else 3) It should satisfy $\int_{-∞}^{∞} f(x)\delta(x) dx=f(0)$.

I showed 2 is true by demonstrating that the Cauchy Principal Value is zero for that integral which means that it's zero everywhere save that one point we avoid.... but I don't see how 3 holds in general. I see that it holds for some functions, like $f(x)=x$, but what about $f(x)=1$, for example. Anyway, is this a delta function or not.... if so, why?

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    $\begingroup$ No! See this is a tricky point, lol. This is a function of $x$! So we are integrating out the $t$, which means it's NOT a function of $t$ so to say it's infinite at $t=0$ would be a misunderstanding (I believe) in the sense that all one needs to do to compute that integral is go into the complex plane and do a contour integral and get arbitrarily close to $t=0$ and $t=x$. $\endgroup$ – Squirtle Dec 20 '12 at 19:52
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    $\begingroup$ Someone deleted their comment, I should point out that I'm not having a conversation with myself. Previous comment was "isn't it also infinite at t=0?". $\endgroup$ – Squirtle Dec 20 '12 at 19:54
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    $\begingroup$ If $g$ is any function that is $0$ except at one point, then $\int_{-\infty}^\infty f(x)g(x) dx = 0$, no matter what $f(x)$ is and no matter what $g$ does at that one point. So you cannot possibly prove that your function has the three properties, because no function at all has those three properties. $\endgroup$ – Carl Mummert Dec 20 '12 at 19:56
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    $\begingroup$ The dirac function is formally defined by $\int_{-\infty}^\infty f(x)\delta(x)dx=f(0)$, but it's not really a function. It's a measure; the Lebesque integral with respect to $\delta$ returns $f(0)$. $\endgroup$ – Alexander Gruber Dec 20 '12 at 21:01
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    $\begingroup$ Okay, but does my function satisfy the requirements to be a Dirac delta? Function or otherwise? Ie.... Call it a function or whatever, is my thing a Dirac delta. $\endgroup$ – Squirtle Dec 20 '12 at 21:14

The question as posted is not well defined (due to multiple reasons). The fact that Dirac's $\delta$ is not a function but rather a distribution was pointed out before. What is also (maybe even more) relevant is that there is no prescription given how to handle the divergency due to the poles at $t=0$ and $t=x$. In fact, the question can be read as: is $I$ defined via $$I = \int_{-∞}^{∞} \frac{1}{(t+ i \epsilon_1) (t-x+ i \epsilon_2)} dt$$ where $\epsilon_1,\epsilon_2 \to 0$ the $\delta$-distribution. It turns out that the answer to this question depends on the fact whether $\epsilon_1$ (and $\epsilon_2$) approach 0 from above or from below. That this is important is exemplified by Sokhotski's formula.

To get more insight into the problem, we can apply the partial fraction expansion $$\frac{1}{(t+ i \epsilon_1) (t-x+ i \epsilon_2)} = \frac{1}{x+ i(\epsilon_1-\epsilon_2) } \left(\frac{1}{t-x+i\epsilon_2}- \frac{1}{t+ i \epsilon_1} \right).$$

If we perform the integral over $t$ (which is now possible as we have regularized the integral by the small imaginary parts), we obtain $$\int_{-\infty}^\infty \left(\frac{1}{t-x+i\epsilon_2}- \frac{1}{t+ i \epsilon_1} \right)dt = \begin{cases} 2\pi i & \epsilon_1 >0 , \epsilon_2 <0,\\ -2\pi i & \epsilon_1 <0 , \epsilon_2 >0,\\ 0 & \text{else}. \end{cases}$$ We see that if $\epsilon_1$ has the same sign as $\epsilon_2$, $I$ cannot be the $\delta$-distribution as it vanishes identically.

For $\epsilon_1 \epsilon_2 <0$, we have $$I = \frac{2\pi i \mathop{\rm sgn}(\epsilon_1) }{x+ i(\epsilon_1-\epsilon_2) } = 2\pi i \mathop{\rm sgn}(\epsilon_1) \left( \mathcal{P} \frac{1}{x} + i \pi \mathop{\rm sgn}(\epsilon_2-\epsilon_1) \delta(x) \right)$$ via Sokhotki's formula.

We see that the distribution defined by $I$ depends very sensitively on the signs and magnitude of $\epsilon_1$ and $\epsilon_2$. And in non of the many different limits it is in fact a simple $\delta$-distribution!

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    $\begingroup$ Nice post, shows typical problems with distributions and the order (or even kind of) limits. $\endgroup$ – Jonas Teuwen Dec 21 '12 at 0:20
  • $\begingroup$ Can I ask how to evaluate the second integral in you post? $\endgroup$ – Vladimir Mar 10 '13 at 1:26
  • $\begingroup$ @Vladimir: the second integral I evaluate using the Residue theorem (just choose a contour along the real line and close it with a big semicircle in the upper or lower complex plane). $\endgroup$ – Fabian Mar 10 '13 at 12:36
  • $\begingroup$ Just to add that Sokhotski's formula is frequently attributed to Plemelj. (HTH to ease searches). $\endgroup$ – Giuseppe Negro Apr 8 '15 at 15:24

This is a charming example! To be able to really prove something, it is best to be more circumspect about what the "function" given by the integral "is", since, as noted in comments, it's not a function in a classical (=late 19th century post-Cauchy-Weierstrass) sense, although it is of course very productive to think of it as a generalized (=post-L.Schwartz) function. There are choices. As in the first comment by the OP, we could take $\delta_\epsilon(x)=\int_{-\infty}^\infty {dt\over (t+i\epsilon)(t+i\epsilon +x)}$, and say that the integral with $\epsilon=0$ is _some_kind_of_ limit as $\epsilon\rightarrow 0^+$. In particular, it is a weak limit, meaning that we only require that limits $\lim_{\epsilon\rightarrow 0^+}\int_{-\infty}^\infty f(x)\,\delta_{\epsilon}(x)\,dx\;$ exist for nice-enough functions $f$.

Indeed, for $f(0)=0\;$, this limit certainly exists and is $0$. Assuming that the limit exists in general, this already proves that the limit must be a scalar multiple of $\delta$.

To determine the scalar, and/or to see that the limit exists for all Schwartz functions (for example), the usual trick is to take a specific, convenient $f_o$ with $f_o(0)=1$, and first note that for general $f$ we have $\delta_\epsilon(f)=\delta_\epsilon(f-f(0)\cdot f_o)+f(0)\cdot \delta_\epsilon(f_o)\;$. Thus, it suffices to evaluate $\delta_\epsilon(f_o)$ explicitly. Taking $f_o(x)=e^{-x^2}$ succeeds here...


This is a little bit rough and ready, but could hopefully be tidied up. First, as @paul points out, we need to agree on what we mean by a distribution. I'll be a little less generous and say that it is an object that acts on test functions to give a number. This action takes the form $$<\Delta,u> = \int_{-\infty}^{+\infty} \Delta(x)u(x) \,dx,$$ where $\Delta$ is the distribution and $u$ is a test function - that is, a smooth function ($C^\infty$) which vanishes outside some compact set. (Hence, there is a number $K>0$ such that $u$ and all its derivatives equal zero for $|x|>K$.) We also need to agree that if we can define functions $\Delta_\epsilon, \epsilon>0$ with the property that $$\lim_{\epsilon\to 0}<\Delta_\epsilon,u> \quad \hbox{exists for all test functions }\,u,$$ then we say that $\Delta=\lim_{\epsilon\to 0}\Delta_\epsilon$ is a distribution with $$<\Delta,u>=\lim_{\epsilon\to 0}<\Delta_\epsilon,u>.$$ (These are standard definitions.)

So taking up your principal value argument, define, for $\epsilon>0$, $$\delta_\epsilon(x) = \int_{-\infty}^{x-\epsilon}\frac{dt}{t(t-x)}+\int_{x+\epsilon}^{+\infty}\frac{dt}{t(t-x)}.$$ We can calculate that $$\delta_\epsilon(x) = \frac{1}{x}\log\left|\frac{x+\epsilon}{x-\epsilon}\right|.$$ [Oops. This doesn't deal correctly with $x=0$, so the rest of this answer is pretty much useless.]

(This - $\delta_\epsilon$ - is locally integrable on $\mathbb{R}$, which is the condition required of the 1-parameter ($\epsilon$) family of functions $\Delta_\epsilon$ above. That is, $\Delta_\epsilon$ is not required to be a classical function, which is useful because $\delta_\epsilon$ blows up at $x=\epsilon$. However the blow-up is very mild.)

So for the programme above to work, we have to check that (i) $<\delta_\epsilon,u>$ exists for all test functions $u$; (ii) the limit $\lim_{\epsilon\to 0}<\delta_\epsilon,u>$ exists. If in addition we can show that this limit evaluates to $cu(0)$ for some constant $c$ and all test functions $u$, we can then legitimately say that $\delta=\lim_{\epsilon\to 0}\delta_\epsilon$ is indeed equal to $c\delta_{Dirac}=c\times$ the Dirac delta.

So pick a test function $u$. Since $u$ is smooth everywhere, we can apply Taylor's theorem and write $$u(x)=u(0)+u'(0)x+\frac{u''(\xi_x)}{2}x^2,$$ where $\xi_x$ depends continuously on $x$. Then $$<\delta_\epsilon,u> = I_1 + I_2 + I_3,$$ where \begin{eqnarray*}I_1&=&u(0)\int_{-\infty}^{\infty}\delta_\epsilon(x)dx,\\ I_2&=&u'(0)\int_{-\infty}^{\infty}x\delta_\epsilon(x)dx,\\ I_3&=&\int_{-\infty}^{\infty}\frac{u''(\xi_x)}{2}x^2\delta_\epsilon(x)dx.\\ \end{eqnarray*}

Notice that $\delta_\epsilon$ is even, so $I_2=0$ and $I_1$ and $I_3$ can be written as integrals over the positive half-line. Now recall that $u$ and its derivatives vanish for all $|x|>K$. Let $M$ be the maximum of the second derivative of $u$ on $[-K,K]$. Then $$|I_3|\leq M \int_0^K x^2\delta_\epsilon(x)dx =M\int_0^K x\log\left|\frac{x+\epsilon}{x-\epsilon}\right|dx.$$ We can evaluate this explicitly (bit of work...) and show that (i) it exists for $\epsilon>0$ and (ii) vanishes in the limit $\epsilon\to 0$.

It remains to deal with the integral $I_1$. We have \begin{eqnarray*} \int_0^\infty\frac{1}{x}\log\left|\frac{x+\epsilon}{x-\epsilon}\right| &=& \int_0^\epsilon\frac{1}{x}\log\left(\frac{\epsilon+x}{\epsilon-x}\right) + \int_\epsilon^\infty \frac{1}{x}\log\left(\frac{x+\epsilon}{x-\epsilon}\right)\\ &=& 2\int_0^1\frac{1}{t}\log\left(\frac{1+t}{1-t}\right)dt,\end{eqnarray*} by using $x=\epsilon t$ in the first integral and $x=\epsilon/t$ in the second. This integral is finite, with value $c\simeq 4.93$ (probably something much more elegant, but this is what Wolfram Alpha gives).

So for any test function $u$, it seems that $$\lim_{\epsilon\to0} <\delta_\epsilon,u> = c u(0),$$ and so your integral is $c\delta_{Dirac}$.

  • $\begingroup$ I am not sure how you integrate $\frac{1}{t(t-x)}$ to obtain the formula for $\delta_\epsilon$? It looks like the function behaves like $t^{-1}$ for $t\approx0$ so the integral does not converge (unless you implicitly assume that you also take the principal value there). $\endgroup$ – Fabian Dec 20 '12 at 23:51
  • $\begingroup$ Yes - I missed that important point. The $\delta_\epsilon(x)$ I used is not defined at $x=0$. A 'double regularization' is required to deal with the problem as you pointed out in your answer. $\endgroup$ – user12477 Dec 21 '12 at 0:06

Reading the question again, I noticed that you mention principal value. There is of course another interpretation of your distribution $$I = \int_{-\infty}^\infty \mathcal{P} \frac{1}{t(t-x)} dt$$ which involves a principle value integral; but the statement of my earlier post remains that the integral as stated in the question is just not well defined such that one is forced to find an interpretation. ${{{}}}$ To prove that $I$ is in fact proportional to the $\delta$-distribution, we introduce $$\begin{align} I_\epsilon&= \int_{-\infty}^\infty \mathcal{P} \frac{1}{t(t-x + i\epsilon)}dt =\int_{0}^\infty \left[\frac{1}{t(t-x + i\epsilon)} + \frac{1}{t(t+x - i\epsilon)}\right]dt\\ &= \int_0^\infty\frac{2}{t^2 - (x-i \epsilon)^2}\\ &= \frac{\pi}{i(x-i\epsilon)} \end{align}$$ with $I = \lim_{\epsilon\to 0} \mathop{\rm Re}I_\epsilon$.

We have $$\mathop{\rm Re}I_e= \frac{\pi \epsilon}{x^2+\epsilon^2} $$ which converges towards $$ I = \pi^2 \delta(x).$$

  • $\begingroup$ In this interpretation, the result is follows from the Poincaré-Bertrand formula for an iterated principal value integral. $\endgroup$ – Fabian Dec 21 '12 at 1:42

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