Finding a value of the following strange improper integral confronted me in the integral calculus working.
$$ \int_{0}^{\infty}\frac{\sin(x^{2})}{(x^{2}+1)^{3/2}}dx $$
I tried to bring this integral close to the Fresnel integral but it didn't work well.
How should I calculate? I have NO idea...
By substituting $x^2=z$ we have: $$ I = \frac{1}{2}\int_{0}^{+\infty}\frac{\sin(z)}{z^{1/2}(1+z)^{3/2}}\,dz\tag{1}$$ but since $\mathcal{L}\left(\sin(z)\right)=\frac{1}{s^2+1}$ and: $$ \mathcal{L}^{-1}\left(\frac{1}{z^{1/2}(1+z)^{3/2}}\right)= se^{-s/2}\left(I_0\left(\frac{s}{2}\right)-I_1\left(\frac{s}{2}\right)\right)\tag{2}$$ it follows that: $$ I = \frac{1}{2}\int_{0}^{+\infty}\frac{se^{-s/2}}{1+s^2}\left(I_0\left(\frac{s}{2}\right)-I_1\left(\frac{s}{2}\right)\right)\,ds\\=\frac{\pi}{4}\left[\left(J_1\left(\frac{1}{2}\right)-Y_0\left(\frac{1}{2}\right)\right)\cos\left(\frac{1}{2}\right)+\left(J_0\left(\frac{1}{2}\right)+Y_1\left(\frac{1}{2}\right)\right)\sin\left(\frac{1}{2}\right)\right].\tag{3} $$