Integral to Infinity of Exponential of -t^2/Proof 2

Proof
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 $\lambda > 0$:

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 $\lambda > 0$:

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:

On the other hand:

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:


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