Is $F(t)=\int_0^{\infty} \frac{e^{-tx^3}}{1+x^4}dx$ well defined on $(0,\infty)$? I've $F(t)=\int_0^{\infty} \frac{e^{-tx^3}}{1+x^4}dx$ and I have to see that it is well defined on the interval $(0,\infty)$.
For that, I have defined $f(x,t)=\frac{e^{-tx^3}}{1+x^4}, x,t\in(0,\infty) $ so I have to see if $f$ is integrable in $(0,\infty)$.
We know that $f$ is integrable on $(0,\infty)$ $\leftrightarrow$ $\int_{(0,\infty)}|f|d\mu<\infty$
$f(x,t)=|\frac{e^{-tx^3}}{1+x^4}|=\frac{e^{-tx^3}}{1+x^4}\le e^{-tx^3}$
But hoe can I bound this? I have to bound it with an integrable function... but I don't know how to calculate the integral of $e^{-tx^3}$... Is there any other easier way to bound that? Or how can I solve my problem?
 A: Hint: $\int_0^{1}e^{-tx^{3}} dx \leq \int_0^{1} 1 dx$ and $\int_1^{\infty} e^{-tx^{3}} dx \leq \int_1^{\infty} e^{-tx} dx $
To prove continuity of $F$ it is enough to prove continuity on $(r,\infty)$ for each $r>0$. When $t>r$ we have $|f(x,t)| \leq \max \{1_{0<x<1}, e^{-rx}\}$. Now apply DCT to prove sequential continuity of $F$.
A: You can bound the integrand with $\frac{1}{1+x^4}$, or with $1$ on $[0,\,1)$, $x^{-4}$ on $[1,\,\infty)$.
A: Simply put, whenever $t,x\in(0,\infty)$
$$\frac{e^{-tx^3}}{1+x^4}\leq \frac{1}{1+x^4}$$
Hence
$$\int_0^\infty \frac{e^{-tx^3}}{1+x^4}\mathrm dx\leq\int_0^\infty\frac{1}{1+x^4}\mathrm dx \\ F(t)=\int_0^\infty \frac{e^{-tx^3}}{1+x^4}\mathrm dx\leq \frac{\pi}{2^{3/2}}\\ \forall t\in(0,\infty)$$
Showing continuity is a bit harder. If you want to prove $F$ is continuous at say, $t_0$ you need to show that $\forall \epsilon>0$, $\exists \delta>0$ such that
$$|t-t_0|<\delta\implies |F(t)-F(t_0)|<\epsilon$$
Let's see if we can come up with some bounds for $|F(t)-F(t_0)|$. First, it is easy to see that $F$ is decreasing. So, assume $t<t_0$ first. Then
$$|F(t)-F(t_0)|=F(t)-F(t_0) \\ =\int_0^\infty\frac{\exp(-tx^3)-\exp(-t_0x^3)}{1+x^4}\mathrm dx \\ \leq\int_0^\infty\left(\exp(-tx^3)-\exp(-t_0x^3)\right)\mathrm dx$$
So it is pertinent to look at the integral
$$\int_0^\infty e^{-tx^3}\mathrm dx$$
Make the change of variable $$z=tx^3 \\ \implies \mathrm dz=3tx^2\mathrm dx\implies \mathrm dx=\frac{\mathrm dz}{3tx^2} \\ \implies \mathrm dx=\frac{\mathrm dz}{3t((z/t)^{1/3})^2}=\frac{z^{-2/3}\mathrm dz}{3t^{1/3}}$$
So
$$\int_0^\infty e^{-tx^3}\mathrm dx=\frac{1}{t^{1/3}}\frac{1}{3}\int_0^\infty z^{1/3-1}e^{-z}\mathrm dz$$
So we have a proportionality rule for this integral
$$\int_0^\infty e^{-tx^3}\mathrm dx \propto t^{-1/3}$$
In fact this proportionality constant is
$$\frac{1}{3}\int_0^\infty z^{1/3-1}e^{-z}\mathrm dz=\frac{1}{3}\Gamma(1/3)=\Gamma(4/3)$$
Hence
$$F(t)-F(t_0)\leq \Gamma(4/3)\left(t^{-1/3}-{t_0}^{-1/3}\right)$$
This should already be enough to prove continuity. Next just do this work identically for the $t_0<t$ case and you are done.
