integrate(sqrt(e^x^2)/(e^x^2), x, -infinity, infinity)

The integral you're asking to evaluate is:

\[
\int_{-\infty}^{\infty} \frac{\sqrt{e^{x^2}}}{e^{x^2}} \, dx
\]

First, let's simplify the integrand:

\[
\frac{\sqrt{e^{x^2}}}{e^{x^2}} = \frac{e^{x^2/2}}{e^{x^2}} = e^{-x^2/2}
\]

So, the integral becomes:

\[
\int_{-\infty}^{\infty} e^{-x^2/2} \, dx
\]

This is a standard Gaussian integral with a coefficient in the exponent. The general formula for the Gaussian integral is:

\[
\int_{-\infty}^{\infty} e^{-ax^2} \, dx = \sqrt{\frac{\pi}{a}}
\]

Here, \( a = \frac{1}{2} \). Therefore, the integral evaluates to:

\[
\int_{-\infty}^{\infty} e^{-x^2/2} \, dx = \sqrt{\frac{2\pi}{1}} = \sqrt{2\pi}
\]

Thus, the final answer is:

\[
\sqrt{2\pi}
\]

Would you like further details or have any questions? Here are some related questions to expand your understanding:

1. How do Gaussian integrals relate to probability theory?
2. What is the significance of the standard normal distribution in statistics?
3. How can we evaluate integrals involving \(e^{-x^2}\) using polar coordinates?
4. How would the integral change if the exponent were \(e^{-ax^2}\) with \(a \neq \frac{1}{2}\)?
5. What are the applications of Gaussian functions in physics?

**Tip:** When dealing with integrals involving exponentials, always look for a Gaussian form, as these integrals have well-known solutions.

Learn how to evaluate the integral \( \int_{-\infty}^{\infty} \frac{\sqrt{e^{x^2}}}{e^{x^2}} \, dx \) step-by-step. Understand the simplification of the integrand and apply the Gaussian integral formula to find the solution. Suitable for advanced undergraduate students and anyone interested in integral calculus and Gaussian functions.

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