calculus - Generalized Fresnel Integral using Laplace

So I wanted to solve the following integral:

$$\int_0^\infty \sin{(x^2) dx}$$

I did it by using the Laplace transform of the function:

$$I(t) = \int_0^\infty \sin{(tx^2) dx}$$

$$\mathcal{L} [I(t)] = \int_0^\infty \frac {x^2}{s^2+x^4} dx$$

This last integral is easily solvable for x, and then I took the inverse Laplace Transform to get:

$$I(t) = \sqrt{\frac{\pi}{8}} t^{-1/2}$$

But afterwards I googled about it, and I found this thread. The best answer easily got the generalized Fresnel Integral by applying a LT and an ILT, and I have no idea why he did this, but it worked! (I solved the generalized integral using the Mellin transform and I got the same result).

If I try my previous method here, I would have to solve the integral:

$$\int_0^\infty \frac {x^p}{s^2+x^{2p}} dx$$

And this seems much harder than the mysterious method. So can someone please explain me what is going on there ? Thanks in advance.

Answer

There is a theorem that states that

$$\begin{align} \int_0^\infty f(x) \, g(x)\,dx=\int_0^{\infty}\mathcal{L}\{f(x)\}(s) \,\mathcal{L}^{-1}\{g(x)\}(s)\,ds \tag1 \end{align}$$
To compute the generalized integral
$$\int_0^\infty \sin (x^p) \,dx$$
You can first substitute $u=x^p$, and then use $(1)$, choosing $f(u)=\sin u$ and $g(u)=u^{\frac{1}{p}-1}$.

The resulting integral can be calculated by using one of the integral representation of the Beta function.