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.
Best 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.