Integral to Infinity of Exponential of -t^2

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Theorem

$\displaystyle \int_0^\infty \map \exp {-t^2} \rd t = \dfrac {\sqrt \pi} 2$


Proof 1

Let $\displaystyle I = \int_0^\infty \map \exp {-t^2} \rd t$.

Let $\displaystyle I_P = \int_0^P \map \exp {-x^2} \rd x = \int_0^P \map \exp {-y^2} \rd y$.

Then we have:

$I = \displaystyle \lim_{P \mathop \to \infty} I_P$

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:


Integral to Infinity of Exponential of -t^2.png


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