# 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:

$\ds \int_0^\infty e^{-t^2} \rd t = \frac {\sqrt \pi} 2$

$\blacksquare$