# Integrals and Fourier Transforms

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 have a $t$ in the factor $\cos(yt)$: $$f'(t) = =\int_{0}^{1} \sqrt{y} e^{y^{2}} \frac{\partial}{\partial t} \cos (y t)\, dy$$ Commented Jan 1, 2020 at 17:01

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?
• I'm assuming that $\chi_{[-1,1]}$ is just the inner integral. Why introduce an outer integral from positive infinity to negative infinity, and why are you permitted to? Commented Jan 1, 2020 at 17:51
• the $\chi_{[-1,1]}(y)$ is the function that is one from $-1$ to $1$ and zero elsewhere. If we insert such a factor into the integral, then we can view the integral as an integral from $-\infty$ to $\infty$, and then use Plancherel's theorem to compute the integral you seek since we now have written the expression for $f'(t)$ as the sine transform of a function. Commented Jan 1, 2020 at 18:22
• So we use the fact that $$\int_{-\infty}^{\infty}\left|f^{\prime}(t)\right|^{2}\, dt=\int_{-\infty}^{\infty}\left|F_{sin}(g(y))\right|^{2}\, dt=\int_{-\infty}^{\infty}\left|g(y)\right|^{2}\, dy$$ Wuth $F_{sin}(g(y))$ being the last line of your answer. After computing everything I seem to get an answer of $0$, is that correct? Commented Jan 2, 2020 at 12:59