Integral to Infinity of Exponential of -t^2/Proof 1
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Theorem
- $\ds \int_0^\infty \map \exp {-t^2} \rd t = \dfrac {\sqrt \pi} 2$
Proof
Let $\ds I = \int_0^\infty \map \exp {-t^2} \rd t$.
Let $\ds I_P = \int_0^P \map \exp {-x^2} \rd x = \int_0^P \map \exp {-y^2} \rd y$.
Then we have:
- $I = \ds \lim_{P \mathop \to \infty} I_P$
Hence:
\(\ds {I_P}^2\) | \(=\) | \(\ds \paren {\int_0^P \map \exp {-x^2} \rd x} \paren {\int_0^P \map \exp {-y^2} \rd y}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^P \int_0^P \map \exp {-\paren {x^2 + y^2} } \rd x \rd y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \iint_{\mathscr R_P} \map \exp {-\paren {x^2 + y^2} } \rd x \rd y\) |
where $\mathscr R_P$ is the square $\Box OACE$ in the figure below:
Because the integrand is positive:
\(\text {(1)}: \quad\) | \(\ds \iint_{\mathscr R_1} \map \exp {-\paren {x^2 + y^2} } \rd x \rd y\) | \(\le\) | \(\ds {I_P}^2\) | |||||||||||
\(\ds \) | \(\le\) | \(\ds \iint_{\mathscr R_2} \map \exp {-\paren {x^2 + y^2} } \rd x \rd y\) |
where $\mathscr R_1$ and $\mathscr R_2$ are the regions in the first quadrant bounded by the circles with centers $O$ and radii $P$ and $P \sqrt 2$ respectively.
Using polar coordinates, we can express $(1)$ as:
\(\text {(1)}: \quad\) | \(\ds \int_{\theta \mathop = 0}^{\frac \pi 2} \int_{r \mathop = 0}^P \map \exp {-r^2} r \rd r \rd \theta\) | \(\le\) | \(\ds {I_P}^2\) | |||||||||||
\(\ds \) | \(\le\) | \(\ds \int_{\theta \mathop = 0}^{\frac \pi 2} \int_{r \mathop = 0}^{P \sqrt 2} \map \exp {-r^2} r \rd r \rd \theta\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac \pi 4 \paren {1 - \map \exp {-P^2} }\) | \(\le\) | \(\ds {I_P}^2\) | |||||||||||
\(\ds \) | \(\le\) | \(\ds \dfrac \pi 4 \paren {1 - \map \exp {-2 P^2} }\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \lim_{P \mathop \to \infty} {I_P}^2\) | \(=\) | \(\ds I^2\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac \pi 4\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds I\) | \(=\) | \(\ds \dfrac {\sqrt \pi} 2\) |
$\blacksquare$