Integral to Infinity of Exponential of -t^2
Theorem
- $\ds \int_0^\infty \map \exp {-t^2} \rd t = \dfrac {\sqrt \pi} 2$
Proof 1
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:
\(\text {(1)}: \quad\) | \(\ds \int_{\theta \mathop = 0}^{\frac \pi 2} \int_{r \mathop = 0}^P \map \exp {-r^2} r \rd r \rd \theta\) | \(\le\) | \(\ds {I_P}^2\) | |||||||||||
\(\ds \) | \(\le\) | \(\ds \int_{\theta \mathop = 0}^{\frac \pi 2} \int_{r \mathop = 0}^{P \sqrt 2} \map \exp {-r^2} r \rd r \rd \theta\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac \pi 4 \paren {1 - \map \exp {-P^2} }\) | \(\le\) | \(\ds {I_P}^2\) | |||||||||||
\(\ds \) | \(\le\) | \(\ds \dfrac \pi 4 \paren {1 - \map \exp {-2 P^2} }\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \lim_{P \mathop \to \infty} {I_P}^2\) | \(=\) | \(\ds I^2\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac \pi 4\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds I\) | \(=\) | \(\ds \dfrac {\sqrt \pi} 2\) |
$\blacksquare$
Proof 2
Let $\lambda$ be a non-negative real number.
Then, we have:
- $\ds \size {\frac {e^{-\lambda^2 \paren {1 + x^2} } } {1 + x^2} } \le \frac 1 {1 + x^2}$
for each $x \in \R$.
Note that from Definite Integral to Infinity of $\dfrac 1 {x^2 + a^2}$:
- $\ds \int_0^\infty \frac 1 {x^2 + 1} \rd x = \frac \pi 2$
So by the Comparison Test for Improper Integral:
- $\ds \int_0^\infty \frac {e^{-\lambda^2 \paren {1 + x^2} } } {1 + x^2} \rd x$ converges.
We can therefore define a real function $I : \hointr 0 \infty \to \R$ by:
- $\ds \map I \lambda = \int_0^\infty \frac {e^{-\lambda^2 \paren {1 + x^2} } } {1 + x^2} \rd x$
for each $\lambda \in \hointr 0 \infty$.
We then have from Definite Integral of Partial Derivative, for $\lambda > 0$:
\(\ds \map {I'} \lambda\) | \(=\) | \(\ds \frac \d {\d \lambda} \paren {\int_0^\infty \frac {e^{-\lambda^2 \paren {1 + x^2} } } {1 + x^2} \rd x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^\infty \frac \partial {\partial \lambda} \frac {e^{-\lambda^2 \paren {1 + x^2} } } {1 + x^2} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -2 \lambda \int_0^\infty e^{-\lambda^2 \paren {1 + x^2 } } \rd x\) | Derivative of Exponential Function | |||||||||||
\(\ds \) | \(=\) | \(\ds -2 \lambda e^{-\lambda^2} \int_0^\infty e^{-\paren {\lambda x}^2} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -2 e^{-\lambda^2} \int_0^\infty e^{-t^2} \rd t\) | substituting $t \mapsto \lambda x$ |
We have, using Definite Integral to Infinity of $\dfrac 1 {x^2 + a^2}$ again:
- $\ds \map I 0 = \int_0^\infty \frac 1 {1 + x^2} \rd x = \frac \pi 2$
We also have, for $\lambda > 0$:
\(\ds 0\) | \(\le\) | \(\ds \int_0^\infty \frac {e^{-\lambda^2 \paren {1 + x^2} } } {1 + x^2} \rd x\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \int_0^\infty e^{-\lambda^2 \paren {1 + x^2} } \rd x\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \int_0^\infty e^{-\lambda x} \rd x\) | since $x \le 1 + x^2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 \lambda\) | Laplace Transform of Real Power |
so:
- $\ds \lim_{\lambda \to \infty} \int_0^\infty \frac {e^{-\lambda^2 \paren {1 + x^2} } } {1 + x^2} \rd x = 0$
Now we have:
\(\ds \int_0^\infty \map {I'} \lambda \rd \lambda\) | \(=\) | \(\ds \lim_{R \to \infty} \int_0^R \map {I'} \lambda \rd \lambda\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{R \to \infty} \paren {\map I R - \map I 0}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\frac \pi 2\) |
On the other hand:
\(\ds \int_0^\infty \map {I'} \lambda \rd \lambda\) | \(=\) | \(\ds -2 \lim_{R \to \infty} \int_0^R \map {I'} \lambda \rd \lambda\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -2 \int_0^\infty e^{-t^2} \rd t \lim_{R \to \infty} \int_0^R e^{-\lambda^2} \rd \lambda\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -2 \paren {\int_0^\infty e^{-t^2} \rd t}^2\) | Definition of Improper Integral on Closed Interval Unbounded Above |
So:
- $\ds \paren {\int_0^\infty e^{-t^2} \rd t}^2 = \frac \pi 4$
Since:
- $\ds \int_0^\infty e^{-t^2} \rd t \ge 0$
we have:
- $\ds \int_0^\infty e^{-t^2} \rd t = \frac {\sqrt \pi} 2$
$\blacksquare$
Sources
- 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Solved Problems: The Gamma Function: $29$