# indefinite integral using Fourier transform

Let $f(t)=\int_{0}^{1}\sqrt\omega e^{\omega^2}cos(\omega t)d\omega$.

I want to compute $\int_{-\infty}^{\infty}|f'(t)|^2 dt = I$.

By Plancherel formula, $\int_{-\infty}^{\infty}|f'(t)|^2dt = \frac{1}{2\pi}\int_{-\infty}^{\infty}\hat{f'}.\overline{\hat{f'}} \ dt$.

On the other hand, $\hat{f'}.\overline{\hat{f'}} = it\hat{f}(-it\hat{f})=t^2 (\hat{f})^2$.

Hence, it suffices to compute $\hat{f}$. In order to do so, I tried to use the Fourier Transform Inverse Formula, however I got stucked.

Any suggestions or help, please ? (in order to compute $I$)

Where $\mathbf{1}_{[0,1]}(x)$ is the indicator function on $[0,1]$. Equivalently you could write $H(x(1-x)) = \mathbf{1}_{[0,1]}(x)$ using the Heaviside step function.
Finally you can find your integral using $\int x^3 e^{2x^2}dx = \frac{1}{8}e^{2x^2}(2x^2 -1) +C$