I want to calculate the following integral $\int_{x_1+\ldots+x_n \geq a} \exp\left[ -\pi \left(x_1^2+\ldots+x_n^2 \right)\right] dx_1\cdots dx_n, $ as a function of $a$, in possibly the shortest and the easiest way. I need such result in a paper in mathematical psychology and I would like to avoid writing a dedicated appendix.
I know how to find the solution. However, I believe there is a simpler solution (eg. a clever change of variables, or a trick with differentiation). Do you know one?
My solution:
Let
$I(n,a,b) = \int_{x_1+\ldots+x_n \geq a} \exp\left[ -\pi \left(b x_1^2+\ldots+x_n^2 \right)\right] dx_1\cdots dx_n$ After change of variables $t = x_1+\ldots+x_n$ we obtain $I(n,a,b) = \int_{-\infty}^{\infty}\cdots \int_{-\infty}^{\infty} \int_{a}^{\infty} \exp\left[ -\pi \left(b (t-x_2-\ldots-x_n)^2+\ldots+x_n^2 \right)\right] dt dx_2\cdots dx_n.$ After integrating out $x_n$ we arrive at $I(n,a,b)=\tfrac{1}{\sqrt{1+b}}I(n-1,a,\tfrac{b}{1+b}).$ Then $I(n,a,1) = \tfrac{1}{\sqrt{2}} I(n-1,a,\tfrac{1}{2}) = \ldots = \tfrac{1}{\sqrt{k}} I(n-k+1,a,\tfrac{1}{k}) = \ldots = \tfrac{1}{\sqrt{n}} I(1,a,\tfrac{1}{n}).$ Consequently, we get the solution $I(n,a,1) = \int_{a}^\infty \frac{1}{\sqrt{n}} \exp \left[ -\pi \frac{1}{n} t^2\right]dt = \int_{a/\sqrt{n}}^\infty \exp \left[ -\pi t^2\right]dt,$ which is related to the error function (Erf).