In a recent paper with two colleagues http://arxiv.org/pdf/1408.5720.pdf we derived a certain integral ruling the distribution of eigenvalues spacings for a model of random matrices. The integral reads $$ J_M(s)=\int_0^\infty dt\ \frac{\sin(s t)}{t^{M-1}}[\mathrm{erf}(t)]^M\ , $$ where $\mathrm{erf}(x)=(2/\sqrt{\pi})\int_0^x\ e^{-y^2}dy$ is the Error function. $M$ is a large positive parameter. I am somehow struggling to find the asymptotic behavior of this integral as $M\to +\infty$. A saddle-point approach seemed a natural candidate, but somehow the oscillating part is getting in the way (unless I am making a silly mistake). Any help would be much appreciated.

  • $\begingroup$ Have u tried to rescale $y/t\rightarrow \tilde{y}$ and exchange the order of integration? the $t$ integral is now doable in terms of error functions and u can use their asymptotic expansions to hopefully make the resulting integration over the unit cube doable $\endgroup$
    – tired
    Jan 4, 2016 at 12:07
  • $\begingroup$ Thanks for your suggestion. I tried your suggested rescaling, but then in order to exchange the order of integration, I need to introduce an M-fold integral in $\tilde{y}_1\cdots \tilde{y}_M$, right? This is because the $\mathrm{erf}$ is raised to power $M$... if I do that, I can indeed perform the $t$-integral, but the resulting $M$-fold integral does not seem to be any nicer... $\endgroup$ Jan 4, 2016 at 12:58

3 Answers 3


Here's a more heuristic approach. For a fixed $s$, look at the graph of the integrand for increasing $M$. You will notice that the integrand all but vanishes except for a small window near $t=\epsilon$. Thus, it is reasonable to approximate the error function with its 2nd-order Taylor expansion in this limit. We also may replace the sine with its first-order Taylor expansion. Thus, the integral is approximately

$$s \left (\frac{2}{\sqrt{\pi}} \right )^M \int_0^{\epsilon} dt \, t^2 \, \left (1-\frac{t^2}{3} \right )^M $$

which, for the purposes of applying Laplace's method, may be rewritten as

$$s \left (\frac{2}{\sqrt{\pi}} \right )^M \int_0^{\epsilon} dt \, t^2 \, e^{M \log{(1-t^2/3)}} $$

which may be approximated by, with exponentially small error, as

$$s \left (\frac{2}{\sqrt{\pi}} \right )^M \int_0^{\infty} dt \, t^2 \, e^{-M t^2/3} $$

Thus, the leading asymptotic behavior of the integral is

$$\int_0^{\infty} dt \, \frac{\sin{s t}}{t^{M-1}} (\operatorname{erf}{t})^M \sim \frac12 s \left (\frac{2}{\sqrt{\pi}} \right )^{M-1} \left (\frac{M}{3} \right )^{-3/2} \quad (M \to \infty)$$

What does the $\sim$ mean? The largest error occurs in approximating the erf and the sine by their lowest-order Taylor expansions. Further expansion will result in the same power of $2/\sqrt{\pi}$, but decreasing powers of $M/3$. The next order in the expansion in fact will be $(M/3)^{-5/2}$. Thus we may write

$$\int_0^{\infty} dt \, \frac{\sin{s t}}{t^{M-1}} (\operatorname{erf}{t})^M = \frac12 s \left (\frac{2}{\sqrt{\pi}} \right )^{M-1} \left [ \left (\frac{M}{3} \right )^{-3/2} +O \left ( M^{-5/2} \right ) \right ]$$

A log-log plot of the relative error between the integral and this leading asymptotic behavior produces a line of slope $-1$ and thus verifies the behavior. I did this in Mathematica for $s=2$ and $M \in [10,100]$: (sorry, not very efficiently written code)

Plot[Log[10, Abs[(NIntegrate[ t Sin[2 t] (Erf[t]/t)^(10^m), {t, 0, Infinity}] - (2/Sqrt[Pi])^((10^m - 1)) ((10^m)/3)^(-3/ 2))/(NIntegrate[ t Sin[2 t] (Erf[t]/t)^(10^m), {t, 0, Infinity}] + (2/Sqrt[Pi])^((10^m - 1)) ((10^m)/3)^(-3/ 2))]], {m, 1, 2}]

enter image description here


Apropos the answer added here asserting that the above result may be extended to a Gaussian-like behavior in $s$, allow me to demonstrate why this is not the case. Here I will simply extend the above result by providing the next higher-order term. Note that, in doing so, we need to consider both the expansion in the error function and the expansion of the sine term.

First, consider the error function term. It so happens that the next order expansion in the exponential is

$$\log{\left ( \frac{\operatorname{erf}{t}}{t}\right )} = -\frac{t^2}{3} + \frac{2 t^4}{45} + O \left ( t^6\right )$$

Now, if we consider how we defined $\epsilon$ above, we limited the region of integration out to $m \epsilon^2 = O(1)$, or $\epsilon \sim m^{-1/2}$. Thus, for the next-order term in the exponential, $m \epsilon^4 \sim m^{-1}$. So for the large-$m$ limit, we may actually Taylor expand the exponential of the fourth-order piece.

However, when we do that, we must also match that with the equivalent Taylor expansion of the sine. This requires a match to the correct order of $m$ in the asymptotic expansion - in this case, the quadratic term in the sine matches the quartic term from the error function because the quartic term has a factor of $m$ built in. (This is commonplace - see Bender & Orszag, sec. 6.4 for more details.)

I will not do out the algebra here, which I leave to the reader. (Truth be told - I did the high-order expansion in Mathematica.) The result is

$$\int_0^{\infty} dt \, \frac{\sin{s t}}{t^{M-1}} (\operatorname{erf}{t})^M = \frac12 s \left (\frac{2}{\sqrt{\pi}} \right )^{M-1} \left [ \left (\frac{M}{3} \right )^{-3/2} + \frac{1-s^2}{4}\left (\frac{M}{3} \right )^{-5/2} +O \left ( M^{-7/2} \right ) \right ]$$

And here's the expected log-log of the relative error with the slope of $-2$:

enter image description here

For comparison, here is the result as proposed by the other answer, using all orders of the sine function combined with the first-order expansion of the error function:

$$\int_0^{\infty} dt \, \frac{\sin{s t}}{t^{M-1}} (\operatorname{erf}{t})^M \sim \frac{s}{4} \left (\frac{2}{\sqrt{\pi}} \right )^{M} \left (\frac{M}{3} \right )^{-3/2} e^{-3 s^2/4 m} $$

Here is the log-log plot of the relative error:

enter image description here

Note that, while the relative error is very slightly improved over the simple leading-order expansion, it provides absolutely no additional higher-order information (as the slope is still $-1$).

The point of all of this is that there is a lot of ways to get potentially bad results. One of them is to propose expanding the amplitude function (i.e., the sine here) without a corresponding expansion of the "phase" function (i.e., the error function here). Higher-order terms must be obtained by balancing the expansions of the two and noting that the higher-order exponentials may be Taylor expanded by the definition of the leading asymptotic expansion. Also, assertions made here should be backed up with numerical results, which software like Mathematica makes easy to provide.

  • 1
    $\begingroup$ Excellent! Thanks very much. $\endgroup$ Jan 4, 2016 at 13:24

We can re-write the integral in the form $$ J_M (s) = \left( {\frac{2}{{\sqrt \pi }}} \right)^M \int_0^{ + \infty } {\exp \left( { - M\log \left( {\frac{{2t}}{{\sqrt \pi \operatorname{erf}(t)}}} \right)} \right)t\sin (st){\rm d}t} . $$ An application of Laplace's method then yields the asymptotic expansion \begin{align*} J_M (s) & \sim \frac{s}{2}\left( {\frac{2}{{\sqrt \pi }}} \right)^{M - 1} \left( {\frac{3}{M}} \right)^{3/2} \sum\limits_{n = 0}^\infty {\frac{{a_n(s) }}{{M^n }}} \\ & = \frac{s}{2}\left( {\frac{2}{{\sqrt \pi }}} \right)^{M - 1} \left( {\frac{3}{M}} \right)^{3/2} \left( {1 - \frac{{3s^2 - 6}}{{4M}} + \frac{{45s^4 - 420s^2 + 596}}{{160M^2 }} + \ldots } \right), \end{align*} as $M\to+\infty$, where the coefficients $a_n(s)$ are given by $$ a_n(s) = \frac{{2n + 1}}{{n!}}\left( {\frac{3}{4}} \right)^n \left[ {\frac{{{\rm d}^{2n} }}{{{\rm d}t^{2n} }}\left( {\frac{{\sin (st)}}{{st}}\left( {\frac{3}{{t^2 }}\log \left( {\frac{{2t}}{{\sqrt \pi \operatorname{erf}(t)}}} \right)} \right)^{ - n - 3/2} } \right)} \right]_{t = 0} . $$ It is not difficult to show that $a_n(s)$ is a polynomial of degree $n$ in $s^2$. Therefore the asymptotic expansion is useful provided $s=o(\sqrt M)$.


The advice given above can be tweaked to get more detailed information. After you have made the approximation $h(t)= erf(t)/t = \frac{2}{\sqrt{\pi}} ( 1- \frac{t^2}{3} +\ldots) \approx \frac{2}{\sqrt{\pi}} e^{-t^2/3}$ as suggested above, you can note further that the integral you are interested in is

$\int_0^{\infty} t\sin(st) h^m(t) dt \approx (\frac{2}{\sqrt{\pi}})^m \int _0^{\infty} t \sin (st) e^{- mt^2/3} dt = -(\frac{2}{\sqrt{\pi}})^m\frac{\partial}{\partial s} \int _0^{\infty} \cos (st) e^{- mt^2/3} dt$ and recognize the rightmost integral as the Fourier transform of a Gaussian function, which is therefore a Gaussian in $s$.

More precisely, in order to justify modeling $H(t)=\frac{\sqrt{\pi}}{2}h(t) = ( 1- t^2/3 +\ldots )$ and its high powers as Gaussians, observe that the Morse Lemma tells us that the function $ - \ln H(t)$ that vanishes quadratically at $t=0$ can be expressed exactly as the square of some differentiable function $u(t)$ that vanishes at the origin: $ - \ln H(t) = u^2(t) =t^2/3+\ldots$. Thus $H(t) = e^{-u^2}$ exactly. (The series that defines $u(t)=\frac{t}{\sqrt{3}}+ \ldots $ converges on some interval that contains the origin). By reversion of series we can solve for $t$ as a power series in $u$.

$ H^m(t) = e^{- m u(t)^2}$ exactly. The integral we seek is $ \int_{t=0}^{t=\infty} t \sin (s t) h^m(t) dt$ which is exactly $$((\frac{2}{\sqrt{\pi}})^m \int _{t=0}^{t=\infty} t \sin (st) e^{- m u^2(t) } dt $$

$$= ((\frac{2}{\sqrt{\pi}})^m\frac{d}{ds}\int _{t=0}^{t=\infty} \cos(st) e^{- m u^(t)^2 } dt $$ This is now ideally suited to the method of analysis by stationary phase when $m$ is large, because the last integrand has a sharp peak localized at the origin. The last integral can be expressed as an integral with respect to the variable $u$, by reversion of series. However, it is already qualitatively clear that since $t\approx u$, the asymptotic behavior will be similar to the behavior of the Fourier transform of a Gaussian.

  • 1
    $\begingroup$ I apologize up front because I do not like to comment on another answer, but I fail to see the point or even the validity of what you are showing. For example, how do account for the higher order terms that come from the next terms in the expansion of the error function? How would you validate your expansion numerically? $\endgroup$
    – Ron Gordon
    Jan 4, 2016 at 16:27
  • $\begingroup$ Looking through your explanation, I am even more baffled. At one point, you invoke Morse's Lemma (thank you for the introduction - I was not familiar with the formalism) and state that $u^2(t) = t^2/3 + \cdots$. Then, toward the end, you state that "it is qualitatively clear that since $t \approx u$..." Which is it? The bottom line is that your $u$ has higher-order terms in $t$ that must be accounted for before you can bring higher-order terms in from the sine. This will then stymie your ability to consider the integral as a FT of a Gaussian or its derivative. $\endgroup$
    – Ron Gordon
    Jan 5, 2016 at 12:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .