Definite Integral to Infinity of Exponential of -i x^2

Theorem

 * $\displaystyle \int_0^\infty \map \exp {-i x^2} \rd x = \frac 1 2 \sqrt {\frac \pi 2} \paren {1 - i}$

Proof
Let $R$ be a real number.

Let $C_1$ be the straight line segment from $0$ to $R$.

Let $C_2$ be the arc of the circle of radius $R$ centred at the origin connecting $R$ and $R e^{i \pi/4}$.

Let $C_3$ be the straight line segment from $R e^{i \pi/4}$ to $0$.

Let $\Gamma = C_1 \cup C_2 \cup C_3$.

Let:


 * $\map f z = \map \exp {-z^2}$

From Complex Exponential Function is Entire, $f$ is holomorphic along $\Gamma$ and inside the region that it bounds.

So, by the Cauchy-Goursat Theorem:


 * $\displaystyle \int_\Gamma \map \exp {-z^2} \rd z = 0$

From Contour Integral of Concatenation of Contours, we therefore have:

We have:

So, taking $R \to \infty$, we have:


 * $\displaystyle e^{i \pi/4} \int_0^\infty \map \exp {-i t^2} \rd t = \int_0^\infty \map \exp {-x^2} \rd x$

giving: