calculus - Strange Limit in Proof of the Fresnel Integral

I was playing around with the Fresnel integrals and I've come up with a proof for the fact

$$\int_0^\infty \cos x^2 dx= \int_0^\infty \sin x^2 dx= \sqrt{\frac{\pi}{8}}$$

The proof goes as follows

Use the fact that

$$\int_{-\infty}^{\infty} e^{-u^2} du = \sqrt{\pi}$$
Letting $u = \sqrt{-i} \cdot x, du = \sqrt{-i}*dx$

$$\sqrt{-i} \cdot \int_{-\infty}^{\infty} e^{ix^2} du = \sqrt{\pi}$$
$$\int_{-\infty}^{\infty} e^{ix^2} dx = \int_{-\infty}^{\infty} \cos{x^2} + i \cdot \int_{-\infty}^{\infty} \sin{x^2}dx = \sqrt{\frac{-\pi}{i}}=(1+i)\sqrt{\frac{\pi}{2}}$$

Equating real and imaginary parts (we can do this because we can assume that the two integrals have real values), we get

$$\int_{-\infty}^\infty \cos x^2 dx= \int_{-\infty}^\infty \sin x^2 dx= \sqrt{\frac{\pi}{2}}$$

Then, using the fact that $\cos{x^2}$ and $\sin{x^2}$ are even functions, we get the fact above.

My question stems from the substitution made. After the substitution is made, the bounds of the integral change from $\pm\infty$ to $\pm(1+i)\infty$

Is this integral valid? Is it possible to have a complex number in the bounds of the integral?