MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

There are several ways to compute the classical integral $$ \int_{\mathbb R}e^{-x^2}dx=\sqrt{\pi}. $$ Probably, best known are

(1) squaring the integral with subsequent change of (now two) variables to the polar form, and

(2) the reducing to the Gamma-function at $1/2$.

I am interested though in a "complex" analysis method (namely, a use of the residue theorem) to do the job. The reason is that several integrals like $$ \int_0^\infty e^{-x^2}\cos ax\ dx \qquad\text{or}\qquad \int_0^\infty\sin x^2\ dx $$ can be computed via the residue theorem and the above integral, so I would like to avoid any reference to real analysis. Is there such a complex evaluation though?!

share|cite|improve this question
1  
There is. Try the book by GJO Jameson, for example – Geoff Robinson Aug 25 '12 at 12:31
    
Geoff, do you mean: Jameson, G. J. O. A first course on complex functions. Chapman and Hall, Ltd., London (Distributed in the U.S.A. by Barnes & Noble, Inc.) 1970 xii+148 pp.? I wonder whether I can easily find it... – Wadim Zudilin Aug 25 '12 at 12:41
    
When you square the integral and change the variables to the polar form, then you get the result directly, without using $\Gamma(1/2)$. In fact your integral equals $\Gamma(1/2)$ by definition and the substitution $t:=x^2$, i.e. the indicated method calculates $\Gamma(1/2)$ rather than relies on it. – GH from MO Aug 25 '12 at 12:55
1  
GH, I meant these as two different methods: the newer version should cause no confusion. – Wadim Zudilin Aug 25 '12 at 13:08
    
@Wadim: Yes, that's the book I meant – Geoff Robinson Aug 25 '12 at 13:12
up vote 13 down vote accepted

Yes! For a long time that was thought impossible, but then it was found how to do it using a parallelogram as a contour.

Desbrow, Darrell
On evaluating $\int_{-\infty}^\infty e^{ax(x-2b)}dx$ by contour integration round a parallelogram.
Amer. Math. Monthly 105 (1998), no. 8, 726–731.

According to Desbrow, the parallelogram integral evaluation for the probability integral is due to Mirsky, 1949.

share|cite|improve this answer
    
Gerald, I believe I can upvote just for this optimistic start. The details are very welcome as well! – Wadim Zudilin Aug 25 '12 at 12:45
    
Fantastic! That's why I like MO (occasionally). The jstor stable link is jstor.org/stable/2588989. – Wadim Zudilin Aug 25 '12 at 12:54
2  
But Mirsky's 1949 short and elegant proof (jstor.org/stable/3611303) does the job even better. Note that this reference is due to Desbrow's paper... – Wadim Zudilin Aug 25 '12 at 13:10
    
The contour integral in Jameson's book is round a parallelogram. That book was certainly available in the 70s. – Geoff Robinson Aug 25 '12 at 13:14
    
@Wadim: The link does not lead to a paper :( – Igor Rivin Aug 25 '12 at 14:07

In the Portuguese book Variável Complexa by Maria A. Carreira and Maria S. M. de Nápoles, McGraw-Hill, 1997, this is evaluated in chapter 6, exercise 23. The function $$\begin{equation*} f(z)=\frac{e^{i\pi z^{2}}}{\sin \left( \pi z\right) } \end{equation*}$$ is integrated around the following paralellogram $$\gamma _{1,3}(t) =te^{i\pi /4}\pm 1/2\qquad -r\leq t\leq r$$ $$\gamma _{2,4}(t) =\mp re^{i\pi /4}+t\qquad -1/2\leq t\leq 1/2. $$ The computation shows that $$\begin{equation*} 2\pi i\text{ }\mathrm{res}(f,0)=2i=\lim_{r\rightarrow +\infty }\frac{4i}{ \sqrt{\pi }}\int_{0}^{r/\sqrt{\pi }}e^{-x^{2}}dx. \end{equation*}$$

share|cite|improve this answer
2  
Américo, muito obrigado! This is essentially Mirsky's proof I mention in my remark to Gerard's answer. I guess it is the one reproduced in Jameson's book as well. – Wadim Zudilin Aug 25 '12 at 13:29
    
@Wadim, Não tem de quê! (You are welcome!). – Américo Tavares Aug 25 '12 at 13:31
1  
@Wadim Zudilin You may find other solutions in the answers to this MSE question math.stackexchange.com/q/34767/752. – Américo Tavares Aug 25 '12 at 13:56

If you take $$g(z)=\sum_{r=0}^{m-1}e^{\pi i a (z+r)^2/m}$$ and integrate $f(z)=\frac{g(z)}{e^{2\pi iz}-1}$ around the same parallelogram, you will get reciprocity law for quadratic Gauss sums: if the product $ma$ is even then $$S(a, m)= \sqrt{\frac ma}\frac{1+i}{\sqrt{2}}\overline{ S(m, a)}, $$ where $$S(a,m)=\sum_{r=0}^{m-1}e^{\pi i a r^2/m}.$$ In particular this observation allows to calculate quadratic Gauss sum (just take $a=2$). This example is taken from Apostol, T. M. Introduction to analytic number theory Springer-Verlag, 1976, section 9.10.

share|cite|improve this answer

It seems it was Polya, not Mirsky, who gave the first published evaluation of the probability integral via contour integration in 1945: https://projecteuclid.org/euclid.bsmsp/1166219199 (George Polya, Remarks on Computing the Probability Integral in One and Two Dimensions - see chapter 5). Polya's paper was published only in 1949, so Mirsky was not aware of it.

This story is told in the book D. Mitrinovic and J.D. Keckic, The Cauchy Method of Residues: Theory and Applications, Volume 1, pp. 158-168. Interestingly, the authors write:

"Some books even state that its value cannot be found by contour methods (see, for example [1] or [2]).

In fact, this is not so. We did our best to find the earliest proof of (1) by contour methods, but we did not arrive at a decisive conclusion. In a private communication dated 3 June, 1971, Professor Copson informed us that immediately after the publication of [2] (in 1935) somebody evaluated the integral (1) by the method of residues, but that he forgot who that was".

Is it possible to find this 1935 publication, if it existed?

For one more recent variant of such a calculation, see http://ijmsa.yolasite.com/resources/60--sepp.pdf (A simplified contour integration for the Probability integral, by P.W. Gwanyama)

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.