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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?!

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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
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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
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2 Answers 2

up vote 10 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.

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Gerard, 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 (occassionally). The jstor stable link is jstor.org/stable/2588989 . –  Wadim Zudilin Aug 25 '12 at 12:54
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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
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In the Portuguese book Variável Complexa by Maria A. Carreira and Maria S. M. de Nápoles, McGraw-Hill, 1997, this is solved 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 $$\begin{equation*} 2\pi i\text{ }\mathrm{res}(f,0)=2i=\lim_{r\rightarrow +\infty }\frac{4i}{ \sqrt{\pi }}\int_{0}^{r/\sqrt{\pi }}e^{-x^{2}}dx. \end{equation*}$$

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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
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@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
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