integrate(1/(e^(ix^2)), x, -infinity, infinity)

The integral you're asking about is:

\[
\int_{-\infty}^{\infty} \frac{1}{e^{ix^2}} \, dx
\]

This integral can be transformed to a more standard form through substitution and contour integration. However, upon inspection, it's worth noting that the function \( f(x) = \frac{1}{e^{ix^2}} = e^{-ix^2} \) does not have a convergent integral over the real line. The oscillatory nature of the exponential function \( e^{-ix^2} \) complicates the integral, and standard methods like contour integration do not apply straightforwardly since it lacks convergence.

Thus, the integral:

\[
\int_{-\infty}^{\infty} \frac{1}{e^{ix^2}} \, dx 
\]

**does not converge** in the usual sense.

If you intended something different, such as a specific path or a different interpretation of the exponential term, please clarify. 

Would you like more details or have any other questions?

### Related Questions:
1. What are the conditions for the convergence of integrals with oscillatory functions?
2. How is contour integration applied to integrals with complex exponents?
3. What is the role of the Fresnel integral in evaluating integrals with oscillatory terms?
4. How does the method of steepest descents apply to oscillatory integrals?
5. What is the significance of the Gaussian integral in complex analysis?

**Tip:** When dealing with complex integrals, always check the behavior of the integrand over the contour or path of integration.

Explore the convergence issues of integrals with oscillatory functions like \( \frac{1}{e^{ix^2}} \). This detailed analysis covers complex analysis concepts, contour integration methods, and conditions for integral convergence. Dive deep into the complexities of integrating functions with oscillatory behavior.

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