Definite Integral to Infinity of Exponential of -i x^2
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Theorem
- $\ds \int_0^\infty \map \exp {-i x^2} \rd x = \frac 1 2 \sqrt {\frac \pi 2} \paren {1 - i}$
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 $0$.
This article, or a section of it, needs explaining. In particular: What is the context of $C_1$ and $C_2$ etc? You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
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:
\(\ds 0\) | \(=\) | \(\ds \int_\Gamma \map \exp {-z^2} \rd z\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_{C_1} \map \exp {-z^2} \rd z + \int_{C_2} \map \exp {-z^2} \rd z + \int_{C_3} \map \exp {-z^2} \rd z\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^R \map \exp {-x^2} \rd x + \int_{C_2} \map \exp {-z^2} \rd z + e^{i \pi/4} \int_R^0 \map \exp {-\paren {e^{i \pi/4} t}^2} \rd t\) | Definition of Complex Contour Integral | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^R \map \exp {-x^2} \rd x + \int_{C_2} \map \exp {-z^2} \rd z - e^{i \pi/4} \int_0^R \map \exp {-i t^2} \rd t\) | Reversal of Limits of Definite Integral |
We have:
\(\ds \size {\int_{C_2} \map \exp {-z^2} \rd z}\) | \(\le\) | \(\ds \int_{C_2} \size {\map \exp {-z^2} } \rd \size z\) | Modulus of Complex Integral | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^1 \size {\frac \pi 4 R i \map \exp {i \frac \pi 4 \theta} } \size {\map \exp {-\paren {R \map \exp {i \frac \pi 4 \theta} }^2} } \rd \theta\) | Definition of Complex Contour Integral | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac \pi 4 R \int_0^1 \size {\map \exp {-R^2 \map \exp {i \frac \pi 2 \theta} } } \rd \theta\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac \pi 4 R \int_0^1 \size {\map \exp {-i R^2 \map \sin {\frac \pi 2 \theta} } } \size {\map \exp {-R^2 \map \cos {\frac \pi 2 \theta} } } \rd \theta\) | Exponential of Sum, Euler's Formula | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac \pi 4 R \int_0^1 \size {\map \exp {-R^2 \map \cos {\frac \pi 2 \theta} } } \rd \theta\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac \pi 4 R \int_0^1 \size {\map \exp {-R^2 \map \sin {\frac \pi 2 \paren {1 - \theta} } } } \rd \theta\) | Cosine of Complement equals Sine | |||||||||||
\(\ds \) | \(\le\) | \(\ds \frac \pi 4 R \int_0^1 \map \exp {-\frac 2 \pi R^2 \paren {1 - \theta} } \rd \theta\) | Jordan's Inequality | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac \pi 4 R \int_0^1 \map \exp {-\frac 2 \pi R^2 \theta} \rd \theta\) | Integral between Limits is Independent of Direction | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac \pi 4 R \intlimits {\frac \pi 2 \times \frac {\map \exp {-\frac 2 \pi R^2 \theta} } {-R^2} } 0 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\pi^2} {8 R} \paren {1 - e^{-\frac 2 \pi R^2} }\) | Exponential of Zero | |||||||||||
\(\ds \) | \(\le\) | \(\ds \frac {\pi^2} {8 R}\) | ||||||||||||
\(\ds \) | \(\to\) | \(\ds 0\) | as $R \to \infty$ |
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:
\(\ds \int_0^\infty \map \exp {-i t^2} \rd t\) | \(=\) | \(\ds e^{-i \pi/4} \int_0^\infty \map \exp {-x^2} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\cos \frac \pi 4 - i \sin \frac \pi 4} \frac {\sqrt \pi} 2\) | Euler's Formula, Integral to Infinity of $\map \exp {-t^2}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\frac 1 {\sqrt 2} - \frac i {\sqrt 2} } \frac {\sqrt \pi} 2\) | Sine of $\dfrac \pi 4$, Cosine of $\dfrac \pi 4$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \sqrt {\frac \pi 2} \paren {1 - i}\) |
$\blacksquare$