Vtyjkyuk

I am trying to prove that

$$int_(0,infty)frac1-exp(-xt^2)t^2dt <infty$$ $forall xin (0,1)$

What I have tried is using taylor series of $$exp(x)=sum_n=0^inftyfracx^nn^!$$.

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

$$int_(0,infty)frac1-exp(-xt^2)t^2dt <int_(0,infty)frac1-(1-xt^2)t^2=int_(0,infty)xdt=infty$$.........

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

$$int_(0,infty)frac1-exp(-xt^2)t^2dt <int_(0,infty)frac1-(1-xt^2+fracx^2t^42-fracx^3t^66)t^2=infty$$.....

Is there any good way to prove this?

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

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

beginalign*int_0^inftydfrac1-exp(-xt^2)t^2,dt &= int_0^1dfrac1-exp(-xt^2)t^2,dt + int_1^inftydfrac1-exp(-xt^2)t^2,dt
\
&le int_0^1dfracxt^2t^2,dt + int_1^inftydfrac1t^2,dt
endalign*

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 $$int_0^1 frac1 - e^- xt^2t^2 , dt < x.$$

On $[1, infty)$, the integrand is less than $frac1t^2$, and
$$int_1^infty frac1t^2 = 1,$$
giving that your integral is less than $x + 1$ and so clearly finite.

Use
$$int_0^x e^-ut^2 du=dfrac1-e^-xt^2t^2$$
then
$$int_0^inftydfrac1-e^-xt^2t^2 dt=int_0^xint_0^infty e^-ut^2 dt du<infty$$

We have that

$$int_0^inftyfrac1-e^-xt^2t^2dt=int_0^1frac1-e^-xt^2t^2dt+int_1^inftyfrac1-e^-xt^2t^2dt $$

and since as $xto infty$

$$frac1-e^-xt^2t^2 sim frac1t^2$$

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

$$frac1-e^-xt^2t^2 =xfrace^-xt^2-1-xt^2to x$$

also the first integral converges.

You could have computed the antiderivative using integration by parts
$$I=intfrac1-e^-xt^2 t^2,dt$$
$$u=1-e^-xt^2 implies du=2x t , e^-t^2 x,dt$$
$$dv=fracdtt^2implies v=-frac1t$$ making
$$I=-frac1-e^-t^2 xt+2 xint e^-xt^2,dt$$ For the second integral, let $$xt^2=y^2 implies t=fracysqrtximplies dt=fracdysqrtx$$ making
$$I=-frac1-e^-t^2 xt+2sqrt xint e^-y^2,dy$$ 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.