<p>Fabrizio, the question at the beginning of the thread does not require multivariable calculus.</p>
<p>However, the answer to my question -- not the antiderivative of e^-x^2, but the EXACT VALUE of the integral of e^-x^2 from -infinity to +infinity, does require multivariable calculus. (Taylor series will never give an exact value.)</p>
<p>Here is how it works (S means integral).</p>
<p>We start with S e^-x^2 dx from -infinity to infinity. Now, we square it and take the square root, so it is still the same expression:</p>
<p>Square root ( S e^-x^2 dx * S e^-x^2 dx)</p>
<p>Note that it does not matter what variable we use. We could use x, y, h, or whatever. If we change the second integral in the square root to S e^-y^2 dy), we get:</p>
<p>Square root ( S e^-x^2 dx * S e^-y^2 dy) (both integrals from -infinity to +infinith).</p>
<p>We can combine these to form a double integral in multivariable calculus:</p>
<p>Square root ( S e^-(x^2 + y^2) dy dx), with x and y going from -infinity to +infinity.</p>
<p>If we change the double integral to a double integral in polar coordinates, we get:</p>
<p>Square root ( S e^-r^2 * r dr dtheta), with r going from 0 to infinity and theta going from 0 to 2pi. (Note that this region represents the entire x-y plane, exactly what x and y from -infinity to + infinity represents.)</p>
<p>Evaluating the integral inside the square root, we get</p>
<p>Square root ( -1/2 * -1 * S dtheta), where theta is again going from 0 to 2pi, we get:</p>
<p>Square root (-1/2 * -1 * 2pi) </p>
<p>= Square root(pi).</p>
<p>You can check this result by integrating e^-x^2 numerically with a ti-83 or ti-89 from a very large negative number to a very large positive number, and you will see that the result is the numerical value of Sqrt(pi), or 1.77245... etc.</p>