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?
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