Proving convergence of improper integral, g(t)

Question

I want to prove convergence of the improper integral,

$$\int_1^\infty g(t) dt$$ Where $g(t)$ is given by: $g(t)=(\frac{1-e^{-2t^2}}{t^2})$ , $t\neq0$

I have thought about using different tests, but I keep getting stuck at the fact that as t approaches zero from the right hand side, the function value increases towards infinity, so I can't figure out a limit from the right hand side.

I hope you can help.

Best regards, Christoffer

Answer 1

If you look at the integrand, it can be rewritten as $$ f(t) = 2 \int_{0}^{1} e^{-2t^2x} dx $$ So the whole expression (with $\varepsilon \to \infty$) $$ L = 2 \int_{1}^{\varepsilon}\int_{0}^{1}e^{-2xt^2}dxdt = 2\int_{0}^{1}\int_{1}^{\varepsilon} e^{-2xt^2}dt dx $$ Use the $\texttt{erf}$ function for the second integral: $$ L = \sqrt{2} \int_{0}^{1}\frac{1}{\sqrt{x}}[\texttt{erf}(\varepsilon) - \texttt{erf}(1)] $$ Now, compute the simple integral, and then take the limit as $\varepsilon \to \infty$. Can you handle from here?

Answer 2

We use the special function $$\text {erf}(x)=\sqrt{\frac{2}{\pi}}\int_0^xe^{-t^2}dt$$ and the fact that $\lim_{x \to \infty}\text{erf}(x)=1.$ First we integrate from 1 to $b$ and then we take the limit as $b \to \infty$. Integration by parts gives $$\int_1^b\frac{1-e^{-2t^2}}{t^2}dt=\int_1^b(-\frac{1}{t})'(1-e^{-2t^2})$$ $$=\frac{-(1-e^{-2t^2})}{t}|_1^b+4\int_1^be^{-2t^2}dt$$ $$=\frac{-(1-e^{-2b^2})}{b}+1-e^{-2}+\frac{4}{\sqrt 2}\int_{\sqrt 2}^{\sqrt 2 b}e^{-u^2}du$$ after making the substitution $u=\sqrt 2 t$. Thus the inegral is $$\frac{-(1-e^{-2b^2})}{b}+1-e^{-2}+\frac{4}{\sqrt 2}(\int_0^{\sqrt 2 b}e^{-u^2}du-\int_0^{\sqrt 2}e^{-u^2}du)$$ $$=\frac{-(1-e^{-2b^2})}{b}+1-e^{-2}+2\sqrt{\pi}(\text{erf}(\sqrt 2 b)-\text{erf}(\sqrt 2))$$ Taking the limit as $b \to \infty$, the given integral converges to $$1-e^{-2}+2\sqrt\pi(1-\text{erf}(\sqrt 2)).$$

Answer 3

$0<\frac{1−e^{−t^2}}{t^2} < \frac{1}{t^2}$ which is integrable on $[1,∞)$, so by comparison, your integral converges.