Step 1: Understanding the Question
The question asks for the evaluation of the definite integral of the function \(e^{-x^2}\) from \(-\infty\) to \(\infty\). This is a famous integral known as the Gaussian integral.
Step 2: Key Formula or Approach
This is a standard result in calculus and probability theory. The value is:
\[
\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}
\]
While memorizing the result is fastest, it can be derived by squaring the integral and converting to polar coordinates.
Step 3: Detailed Explanation (Derivation)
Let the integral be \(I\):
\[
I = \int_{-\infty}^{\infty} e^{-x^2}\,dx
\]
We can also write \(I\) using a different variable, say \(y\):
\[
I = \int_{-\infty}^{\infty} e^{-y^2}\,dy
\]
Now, let's calculate \(I^2\):
\[
I^2 = \left( \int_{-\infty}^{\infty} e^{-x^2}\,dx \right) \left( \int_{-\infty}^{\infty} e^{-y^2}\,dy \right) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2+y^2)}\,dx\,dy
\]
Convert this double integral to polar coordinates by setting \(x=r\cos\theta\), \(y=r\sin\theta\), which gives \(x^2+y^2=r^2\) and \(dx\,dy = r\,dr\,d\theta\). The limits for \(r\) are from 0 to \(\infty\), and for \(\theta\) from 0 to \(2\pi\).
\[
I^2 = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2} r\,dr\,d\theta
\]
First, solve the inner integral \(\int_{0}^{\infty} e^{-r^2} r\,dr\). Let \(u = r^2\), so \(du = 2r\,dr\).
\[
\int e^{-r^2} r\,dr = \frac{1}{2}\int e^{-u}\,du = -\frac{1}{2}e^{-u} = -\frac{1}{2}e^{-r^2}
\]
Evaluating the definite integral:
\[
\int_{0}^{\infty} e^{-r^2} r\,dr = \left[ -\frac{1}{2}e^{-r^2} \right]_{0}^{\infty} = \left( -\frac{1}{2}e^{-\infty} \right) - \left( -\frac{1}{2}e^{0} \right) = 0 - (-\frac{1}{2}) = \frac{1}{2}
\]
Now substitute this back into the expression for \(I^2\):
\[
I^2 = \int_{0}^{2\pi} \frac{1}{2} \,d\theta = \frac{1}{2} [\theta]_{0}^{2\pi} = \frac{1}{2}(2\pi - 0) = \pi
\]
Since \(I^2 = \pi\), and the integrand \(e^{-x^2}\) is always positive, \(I\) must be positive.
\[
I = \sqrt{\pi}
\]
Step 4: Final Answer
The value of the integral is \( \sqrt{\pi} \).