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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 $t$?

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  • $\begingroup$ You have a $t$ in the factor $\cos(yt)$: $$f'(t) = =\int_{0}^{1} \sqrt{y} e^{y^{2}} \frac{\partial}{\partial t} \cos (y t)\, dy$$ $\endgroup$
    – md2perpe
    Commented Jan 1, 2020 at 17:01

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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?

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  • $\begingroup$ 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? $\endgroup$
    – David
    Commented Jan 1, 2020 at 17:51
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    $\begingroup$ 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. $\endgroup$
    – Zarrax
    Commented Jan 1, 2020 at 18:22
  • $\begingroup$ 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? $\endgroup$
    – David
    Commented Jan 2, 2020 at 12:59

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