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 $R e^{i \pi/4}$ counter-clockwise. 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:  {{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|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|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	| o = \le 	| r = \frac \pi 4 R \int_0^1 \map \exp {-R^2 \paren {1 - \theta} } \rd \theta 	| c = Shape of Cosine Function }} {{eqn	| r = \frac \pi 4 R \int_0^1 \map \exp {-R^2 \theta} \rd \theta 	| c = Integral between Limits is Independent of Direction }} {{eqn	| r = \frac \pi 4 R \intlimits {\frac {\map \exp {-R^2 \theta} } {-R^2} } 0 1 }} {{eqn	| r = \frac \pi {4 R} \paren {1 - e^{-R^2} } 	| c = Exponential of Zero }} {{eqn	| o = \to 	| r = 0 }} {{end-eqn}} 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:  {{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 45 Degrees, Cosine of 45 Degrees }} {{eqn	| r = \frac 1 2 \sqrt {\frac \pi 2} \paren {1 - i} }} {{end-eqn}} {{qed}} Category:Definite Integrals involving Exponential Function fslfn5qlwgiksteh3go002w17m830ai