Find the Fourier Sine transform of f(x) = 1 in 0,l
F(f(x))=2π∫0∞f(x)sin(ωx) dxF(1)=2π∫0l1×sin(ωx) dx=2π∫0lsin(ωx) dx=2π∫0l−cos(ωx)ω dx=2π⋅−cos(ωx)ω∣0l=2π⋅cos(ωx)ω∣l0=2π(1−cos(lω)ω)\displaystyle \mathcal{F}(f(x)) = \sqrt{\frac{2}{\pi}}\int_0^\infty f(x)\sin(\omega x) \, \mathrm{d}x\\ \begin{aligned} \mathcal{F}(1) &= \sqrt{\frac{2}{\pi}}\int_0^l 1 \times \sin(\omega x) \, \mathrm{d}x = \sqrt{\frac{2}{\pi}}\int_0^l \sin(\omega x) \, \mathrm{d}x \\&= \sqrt{\frac{2}{\pi}}\int_0^l \frac{-\cos(\omega x)}{\omega} \, \mathrm{d}x \\&= \sqrt{\frac{2}{\pi}}\cdot \frac{-\cos(\omega x)}{\omega}\biggr\vert_0^l = \sqrt{\frac{2}{\pi}} \cdot \frac{\cos(\omega x)}{\omega}\biggr\vert_l^0 \\&= \sqrt{\frac{2}{\pi}}\left(\frac{1 - \cos(l\omega)}{\omega}\right) \end{aligned}F(f(x))=π2∫0∞f(x)sin(ωx)dxF(1)=π2∫0l1×sin(ωx)dx=π2∫0lsin(ωx)dx=π2∫0lω−cos(ωx)dx=π2⋅ω−cos(ωx)∣∣0l=π2⋅ωcos(ωx)∣∣l0=π2(ω1−cos(lω))
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