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Let $F(t) = \int_{\mathbb{R}_{\ge 0}} e^{-tx^2}dx = \int_{\mathbb{R}_{\ge 0}} f$

I want to show that $F(t)$ exists and is continuous for $t > 0$ whereby $\underset{t \downarrow 0+}{lim}$ $F(t) = +\infty$

$\fbox{Idea to Prove the Existence of F(t)}$

One could use the Monotone Convergence Theorem to prove the existence of $F(t)$. Specifically, if we could find a sequence of measurable functions $\{f_n\}$ s.t. $0 \le f_n \le e^{-tx^2}$ for all $n$ and $f_n \rightarrow e^{-tx^2}$ pointwise on $\mathbb{R}_{\ge 0}$, then it would follow that $e^{-tx^2}$ is measurable with

$$ lim \int_{\mathbb{R}_{\ge 0}} f_n = \int_{\mathbb{R}_{\ge 0}} e^{-tx^2} = F(t) $$

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1 Answer 1

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The existence of $F(t)$ is given by the fact that $0\leq e^{-tx^2}\leq e^{-tx}$ for $x\geq 1$ and $x\mapsto e^{-tx}$ is integrable for each $t>0$.

Hint to the other question: do the substitution $s=\sqrt tx$ to see what $F$ is.

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Why is $e \mapsto e^{-tx}$ integrable for each $t >0$? I take it this is true if and only if $e^x$ is integrable. But why is $e^x$ integrable over $\mathbb{R}_{\ge 0}$? (I'm using Royden as a text and there is no mention of the exponential function). –  user1770201 Oct 28 '12 at 11:24
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Note that $\mathrm e^{-tx}\leqslant\min\{1,1/(t^2x^2)\}$ and $x\mapsto1/x^2$ is integrable on $x\geqslant1/t$, for every $t\gt0$. (This argument shows that the integral of $x\mapsto\mathrm e^{-tx}$ is at most $2/t$.) –  Did Oct 28 '12 at 11:27

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