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

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

Proof:
Let $R$ be a positive 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 $R e^{i \pi/4}$ anticlockwise.
Let $C_3$ be the straight line segment from $R e^{i \pi/4}$ to $0$.
{{explain|What is the context of $C_1$ and $C_2$ etc?}}
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: 
:$\ds \int_\Gamma \map \exp {-z^2} \rd z = 0$
From Contour Integral of Concatenation of Contours, we therefore have: 
{{begin-eqn}}
{{eqn | l = 0	
      | r = \int_\Gamma \map \exp {-z^2} \rd z
}}
{{eqn | r = \int_{C_1} \map \exp {-z^2} \rd z + \int_{C_2} \map \exp {-z^2} \rd z + \int_{C_3} \map \exp {-z^2} \rd z
}}
{{eqn | r = \int_0^R \map \exp {-x^2} \rd x + \int_{C_2} \map \exp {-z^2} \rd z + e^{i \pi/4} \int_R^0 \map \exp {-\paren {e^{i \pi/4} t}^2} \rd t
      | c = {{Defof|Complex Contour Integral}}
}}
{{eqn | r = \int_0^R \map \exp {-x^2} \rd x + \int_{C_2} \map \exp {-z^2} \rd z - e^{i \pi/4} \int_0^R \map \exp {-i t^2} \rd t
      | c = Reversal of Limits of Definite Integral
}}
{{end-eqn}}
We have: 
{{begin-eqn}}
{{eqn | l = \size {\int_{C_2} \map \exp {-z^2} \rd z}
      | o = \le
      | r = \int_{C_2} \size {\map \exp {-z^2} } \rd \size z
      | c = Modulus of Complex Integral
}}
{{eqn | r = \int_0^1 \size {\frac \pi 4 R i \map \exp {i \frac \pi 4 \theta} } \size {\map \exp {-\paren {R \map \exp {i \frac \pi 4 \theta} }^2} } \rd \theta
      | c = {{Defof|Complex Contour Integral}}
}}
{{eqn | r = \frac \pi 4 R \int_0^1 \size {\map \exp {-R^2 \map \exp {i \frac \pi 2 \theta} } } \rd \theta
}}
{{eqn | r = \frac \pi 4 R \int_0^1 \size {\map \exp {-i R^2 \map \sin {\frac \pi 2 \theta} } } \size {\map \exp {-R^2 \map \cos {\frac \pi 2 \theta} } } \rd \theta
      | c = Exponential of Sum, Euler's Formula
}}
{{eqn | r = \frac \pi 4 R \int_0^1 \size {\map \exp {-R^2 \map \cos {\frac \pi 2 \theta} } } \rd \theta
}}
{{eqn | r = \frac \pi 4 R \int_0^1 \size {\map \exp {-R^2 \map \sin {\frac \pi 2 \paren {1 - \theta} } } } \rd \theta
      | c = Cosine of Complement equals Sine
}}
{{eqn | o = \le
      | r = \frac \pi 4 R \int_0^1 \map \exp {-\frac 2 \pi R^2 \paren {1 - \theta} } \rd \theta
      | c = Jordan's Inequality
}}
{{eqn | r = \frac \pi 4 R \int_0^1 \map \exp {-\frac 2 \pi R^2 \theta} \rd \theta
      | c = Integral between Limits is Independent of Direction
}}
{{eqn | r = \frac \pi 4 R \intlimits {\frac \pi 2 \times \frac {\map \exp {-\frac 2 \pi R^2 \theta} } {-R^2} } 0 1
}}
{{eqn | r = \frac {\pi^2} {8 R} \paren {1 - e^{-\frac 2 \pi R^2} }
      | c = Exponential of Zero
}}
{{eqn | o = \le
      | r = \frac {\pi^2} {8 R}
}}
{{eqn | o = \to
      | r = 0
      | c = as $R \to \infty$
}}
{{end-eqn}}
So, taking $R \to \infty$, we have:
:$\ds e^{i \pi/4} \int_0^\infty \map \exp {-i t^2} \rd t = \int_0^\infty \map \exp {-x^2} \rd x$
giving: 
{{begin-eqn}}
{{eqn | l = \int_0^\infty \map \exp {-i t^2} \rd t
      | r = e^{-i \pi/4} \int_0^\infty \map \exp {-x^2} \rd x
}}
{{eqn | r = \paren {\cos \frac \pi 4 - i \sin \frac \pi 4} \frac {\sqrt \pi} 2 
      | c = Euler's Formula, Integral to Infinity of $\map \exp {-t^2}$
}}
{{eqn | r = \paren {\frac 1 {\sqrt 2} - \frac i {\sqrt 2} } \frac {\sqrt \pi} 2
      | c = Sine of $\dfrac \pi 4$, Cosine of $\dfrac \pi 4$
}}
{{eqn | r = \frac 1 2 \sqrt {\frac \pi 2} \paren {1 - i}
}}
{{end-eqn}}
{{qed}}
Category:Definite Integrals involving Exponential Function
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