Consider the function $f(t),$ with: $$ f(t)=\int_{0}^{1} \sqrt{y} e^{y^{2}} \cos (y t)\, dy $$ Compute $$ \int_{-\infty}^{\infty}\left|f^{\prime}(t)\right|^{2}\, dt$$ where the prime denotes differentiation with respect to $t .$
To start off how would you differentiate $f(t)$ with respect to $t$ ? Is $y$ implicitly a function of $t$?
You can differentiate under the integral sign, meaning $$ f'(t)=\int_{0}^{1} \sqrt{y} e^{y^{2}} {\partial \over \partial t}\cos (y t)\, dy $$ $$= -\int_{0}^{1} y\sqrt{y} e^{y^{2}} \sin (y t)\, dy $$ You want to be able to use Plancherel's Theorem here, so what you can do is write the integral as a symmetric sine transform as $$-{1 \over 2}\int_{-1}^{1} sgn(y)|y|^{3 \over 2} e^{y^{2}} \sin (y t)\, dy $$ $$=-{1 \over 2}\int_{-\infty}^{\infty} \chi_{[-1,1]}(y)sgn(y)|y|^{3 \over 2} e^{y^{2}} \sin (y t)\, dy $$ Can you take it from here?