The Gaussian Integral
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Because the single-variable function \(e^{-x^2}\) lacks an elementary antiderivative, evaluating the Gaussian integral over the real line requires a classic mathematical maneuver: changing the dimension to change the perspective. By evaluating its square in a two-dimensional Cartesian system, we can shift our perspective to radial symmetry.
The Fundamental Objective
Let $$I$$ represent the continuous definite integral over the entire domain:
By introducing a secondary duplicate integral under an independent dummy variable \(y$, we can structure the squared quantity\)I^2$$ as a bivariate double integral representing the volume beneath an infinite 3D surface:
\[I^2 = \left(\int_{-\infty}^{\infty} e^{-x^2}\,dx\right)\left(\int_{-\infty}^{\infty} e^{-y^2}\,dy\right) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)}\,dx\,dy\]This expression maps a double integral over the entire Cartesian plane \(\mathbb{R}^2\).
Shifting to Radial Space
The term \((x^2 + y^2)\)indicates uniform radial symmetry across the\(xy\)-plane. We transform the system into polar variables \((r, \theta)\), scaling the differential area element by the transformation Jacobian (\(dx\,dy = r\,dr\,d\theta\)):
\[x = r\cos\theta, \quad y = r\sin\theta \implies x^2 + y^2 = r^2\]The infinite planar boundaries map cleanly to the polar space limits \(r \in [0, \infty)\)and\(\theta \in [0, 2\pi)\):
\[I^2 = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2} \, r\,dr\,d\theta\]Since the inner integration contains no explicit angular dependence, we can isolate and solve the angular variable completely, yielding a clean multiplier of \(2\pi\):
\[I^2 = \left(\int_{0}^{2\pi} d\theta\right) \int_{0}^{\infty} r e^{-r^2}\,dr = 2\pi \int_{0}^{\infty} r e^{-r^2}\,dr\]Evaluation via Substitution
Applying a standard \(u\)-substitution where \(u = r^2\)and\(du = 2r\,dr\) transforms the calculation into an elementary calculus antiderivative:
\[I^2 = 2\pi \cdot \frac{1}{2} \int_{0}^{\infty} e^{-u}\,du = \pi \Big[ -e^{-u} \Big]_{0}^{\infty}\]Evaluating the limits reveals the exact volume under the surface:
\[I^2 = \pi (0 - (-1)) = \pi\]Analytical Result
Because the real-valued function $$e^{-x^2}$$ is strictly positive everywhere, we extract the positive root:
