This question asks for the value of a famous definite integral known as the Gaussian integral. It is a cornerstone of probability theory (related to the normal distribution) and advanced calculus.
Step 1: Understanding the Question:
We need to evaluate the improper integral of the function \(f(x) = e^{-x^2}\) over the entire real line, from \(-\infty\) to \(\infty\).
Step 2: Key Formula or Approach:
Since \(e^{-x^2}\) has no elementary antiderivative, a direct application of the Fundamental Theorem of Calculus is not possible. The standard method for evaluating this integral is the "Poisson trick" or "Gaussian trick", which involves squaring the integral and switching to polar coordinates.
Step 3: Detailed Explanation:
Let \(I = \int_{-\infty}^{\infty} e^{-x^2} \, dx\). The integrand is an even function, so \(I>0\).
Consider \(I^2\):
\[ I^2 = \left( \int_{-\infty}^{\infty} e^{-x^2} \, dx \right) \left( \int_{-\infty}^{\infty} e^{-y^2} \, dy \right) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2+y^2)} \, dx \, dy \]
This is a double integral over the entire xy-plane. We convert to polar coordinates, where \(x^2+y^2 = r^2\) and the area element \(dx\,dy\) becomes \(r\,dr\,d\theta\). The limits become \(0 \le r<\infty\) and \(0 \le \theta<2\pi\).
\[ I^2 = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2} (r \, dr) \, d\theta \]
We can now evaluate the integral. The inner integral with respect to \(r\) can be solved with a u-substitution (\(u = -r^2\), \(du = -2r\,dr\)):
\[ \int_{0}^{\infty} e^{-r^2} r \, dr = \left[ -\frac{1}{2} e^{-r^2} \right]_{0}^{\infty} = -\frac{1}{2}(0 - 1) = \frac{1}{2} \]
Now, we evaluate the outer integral with respect to \(\theta\):
\[ I^2 = \int_{0}^{2\pi} \left( \frac{1}{2} \right) d\theta = \frac{1}{2} [\theta]_{0}^{2\pi} = \frac{1}{2}(2\pi) = \pi \]
Since \(I^2 = \pi\) and we know \(I>0\), we take the positive square root.
Step 4: Final Answer:
The value of the integral is \( \sqrt{\pi} \).