Using the Laplace Transform, I was able to calculate the value of $$\int_0^\infty \sin\bigg(\frac{1}{x^2}\bigg)dx$$ to be $$\int_0^\infty \sin\bigg(\frac{1}{x^2}\bigg)dx=\sqrt{\frac{\pi}{2}}$$ However, I can't seem to find the value of this closely related integral: $$\int_0^\infty \bigg|\sin\bigg(\frac{1}{x^2}\bigg)\bigg|dx$$ The only thing that I thought to do was to change it into an infinite sum, but I have no idea how to evaluate integrals of this function over finite intervals.
Is there a closed form? If not, can it be expressed in terms of any other widely-used functions?
Thanks!
Another method for those kind of integral is to compute the Fourier series coefficients of $|\sin(x)|$ to obtain
$$|\sin(x)| = \frac{2}{\pi} \sum_{n=1}^\infty (\cos(2 n x)-1)\frac{1}{1-4n^2}$$ Then $$2\int_0^\infty |\sin(1/y^2)| dy = \int_0^\infty |\sin(x)| x^{-3/2}dx$$ $$ = \frac{2}{\pi} \sum_{n=1}^\infty \frac{1}{1-4n^2}\int_0^\infty (\cos(2 n x)-1) x^{-3/2}dx$$ $$ = \frac{2}{\pi} \sum_{n=1}^\infty \frac{1}{1-4n^2}n^{-1/2} \int_0^\infty (\cos(2 x)-1) x^{-3/2}dx$$ Where $\int_0^\infty (\cos(2 x)-1) x^{-3/2}dx$ can be expressed in term of $\Gamma(-1/2)$
So now the question reduces to evaluating the series $$\sum_{n=1}^\infty \frac{n^{-1/2}}{1-4n^2}$$