In a book I have (Angus E. Taylor, Advanced Calculus, Blaisdell Publishing Company, 1955) there are a series of two exercises to prove the reflection formula Γ(x)Γ(1−x) = π csc(πx) within the reals.

Exercise 10 (page 683). Show (…) that B(a,1-a)=int_{0}^{1} (x^(a-1)+x^(-a))/(1+x) dx.

Exercise 11 (page 683). Use the geometric series for (1+x)^(-1) in Exercise 10 and integrate term by term. (…) Show that B(a,1-a)= π csc(πa).

The problem with turning the integral into Γ(1/2) is that you then need to evaluate Γ(1/2), which is normally done by turning it back into a Gaussian integral again!
But, as you hinted, Γ(1/2) can be obtained from the Beta function identity
B(x,1−x) = Γ(x)Γ(1−x) = π csc(πx) evaluated at x=1/2. This identity is proved via contour integration (and a version is available on my wiki, via my homepage).

I wonder what other approaches people might come up with?

Consider the binomial distribution B(n, 0.5)(x). Its standard deviation is 0.5SQRT(n), so we’re concerned with the function B(n, 0.5)(0.5SQRT(n)x + 0.5n)). At an integer x, the value of B(n, 0.5) is n!/(x!(n-x)!2^n), so for x + SQRT(n) an integer multiple of 2SQRT(1/n), the probability density we need is p(n, x) = SQRT(n)n!/(y!(n-y)!2^(n+1)) where y = (x + SQRT(n))SQRT(n)/2 = n/2 + xSQRT(n)/2.

Consider the value of (n-k)!(n+k)!/n!n!. It’s equal to (n+k)/n * (n+k-1)/(n-1) * … * (n+1)/(n-k+1) = (1 + k/n)(1 + k/(n-1))…(1 + k/(n-k+1)). Let k ~ mSQRT(n), m much smaller than SQRT(n). For large n, mSQRT(n)/(n – SQRT(n)) = m/(SQRT(n) – 1) = m/SQRT(n) + m/(n – SQRT(n)) and 1/n goes to 0 in all monomials. So the value of the expression approaches (1 + m/SQRT(n))^mSQRT(n) which approaches e^(m^2). In our case, y = n/2 + xSQRT(n/2)/SQRT(2), turning the function into e^(-x^2/2)*SQRT(n)n!/(n/2)!(n/2)!2^(n+1).

It suffices to find the constant, i.e. the limit of SQRT(n)n!/(n/2)!(n/2)!2^(n+1) as n goes to infinity. Replacing n by n+2 multiplies the expression by SQRT((n+2)/n)(n+1)(n+2)/(n+2)^2 = SQRT((n+2)/n)(n+1)/(n+2). Squaring, we get ((n+2)/n)(n+1)^2/(n+2)^2 = (n+1)^2/n(n+2). It suffices to prove that the limit of the product of (n+1)^2/n(n+2) over all even n is 4/pi (i.e. that the limit of Prod(n(n+2)/(n+1)^2) is pi/4). That will make the square of the limit we need to find equal to 4/8pi = 1/2pi as desired…

http://problemasteoremas.wordpress.com/2008/07/24/problem-of-the-week-2-from-walking-randomly-blog-submission/