Prove that $\int_{(0,\infty)}\frac{1-\exp(-xt^2)}{t^2}dt <\infty$ $\forall x\in (0,1)$

Question

I am trying to prove that

$$\int_{(0,\infty)}\frac{1-\exp(-xt^2)}{t^2}dt <\infty$$ $\forall x\in (0,1)$

What I have tried is using taylor series of $$\exp(x)=\sum_{n=0}^{\infty}\frac{x^n}{n^!}$$.

Since $\exp(-xt^2)\geq 1-xt^2$,

$$\int_{(0,\infty)}\frac{1-\exp(-xt^2)}{t^2}dt <\int_{(0,\infty)}\frac{1-(1-xt^2)}{t^2}=\int_{(0,\infty)}xdt=\infty$$.........

I thought my approximation was too rough. So I also tried...

$$\int_{(0,\infty)}\frac{1-\exp(-xt^2)}{t^2}dt <\int_{(0,\infty)}\frac{1-(1-xt^2+\frac{x^2t^4}{2}-\frac{x^3t^6}{6})}{t^2}=\infty$$.....

Is there any good way to prove this?

Even Wolfram alpha cannot calculate any of this integral.. so I have really no idea.....

Answer 1

Hint: Break up the integral into two pieces, and bound each piece seperately:

\begin{align*}\int_{0}^{\infty}\dfrac{1-\exp(-xt^2)}{t^2}\,dt &= \int_{0}^{1}\dfrac{1-\exp(-xt^2)}{t^2}\,dt + \int_{1}^{\infty}\dfrac{1-\exp(-xt^2)}{t^2}\,dt \\ &\le \int_{0}^{1}\dfrac{xt^2}{t^2}\,dt + \int_{1}^{\infty}\dfrac{1}{t^2}\,dt \end{align*}

Answer 2

It's additionally true for $x \geq 0$. To show this, split the region up into $(0,1)$ and $[1, \infty)$. On $(0,1)$ the integrand is continuous save for a removable singularity at 0, at which its value is $x$. Additionally, it is decreasing in this range and non-negative, so $$\int_0^1 \frac{1 - e^{- xt^2}}{t^2} \, dt < x.$$

On $[1, \infty)$, the integrand is less than $\frac{1}{t^2}$, and $$\int_1^\infty \frac{1}{t^2} = 1,$$ giving that your integral is less than $x + 1$ and so clearly finite.

Answer 3

Use $$\int_0^x e^{-ut^2}\ du=\dfrac{1-e^{-xt^2}}{t^2}$$ then $$\int_0^\infty\dfrac{1-e^{-xt^2}}{t^2}\ dt=\int_0^x\int_0^\infty e^{-ut^2}\ dt\ du<\infty$$

Answer 4

We have that

$$\int_0^{\infty}\frac{1-e^{-xt^2}}{t^2}dt=\int_0^{1}\frac{1-e^{-xt^2}}{t^2}dt+\int_1^{\infty}\frac{1-e^{-xt^2}}{t^2}dt $$

and since as $x\to \infty$

$$\frac{1-e^{-xt^2}}{t^2} \sim \frac{1}{t^2}$$

the second integral converges, and since as $x \to 0^+$

$$\frac{1-e^{-xt^2}}{t^2} =x\frac{e^{-xt^2}-1}{-xt^2}\to x$$

also the first integral converges.

Answer 5

You could have computed the antiderivative using integration by parts $$I=\int\frac{1-e^{-xt^2 }}{t^2}\,dt$$ $$u=1-e^{-xt^2 }\implies du=2x t \, e^{-t^2 x}\,dt$$ $$dv=\frac{dt}{t^2}\implies v=-\frac{1}{t}$$ making $$I=-\frac{1-e^{-t^2 x}}{t}+2 x\int e^{-xt^2}\,dt$$ For the second integral, let $$xt^2=y^2 \implies t=\frac{y}{\sqrt{x}}\implies dt=\frac{dy}{\sqrt{x}}$$ making $$I=-\frac{1-e^{-t^2 x}}{t}+2\sqrt x\int e^{-y^2}\,dy$$ Now, for the integration between $0$ and $\infty$, the first term is $0$ if $x >0$ and the remaining integral is quite well known.

This is less elegant than @user 108128's answer but it works.