Answer to Question #213360 in Civil and Environmental Engineering for Elisha

Question

Find the Fourier sine and Fourier cosine integral for the following function 

f(x) = (x2, 0 < x ≤ 10

0, x > 10.

Answer 1

$\displaystyle (i)\\ \textsf{Fourier sine transform of}\,f(x) \\ \begin{aligned} f_s(\omega) &= \sqrt{\frac{2}{\pi}}\int_{-\infty}^{\infty} f(x) \sin(\omega x) \mathrm{d}x\\ &=\sqrt{\frac{2}{\pi}}\int_0^{10} x^2 \sin(\omega x) \mathrm{d}x + \sqrt{\frac{2}{\pi}}\int_{10}^{\infty} 0 \times \sin(\omega x) \mathrm{d}x\\ &= \frac{1}{\omega}\sqrt{\frac{2}{\pi}}\int_0^{10} x^2 \mathrm{d}(-\cos(\omega x))\\ &= \sqrt{\frac{2}{\pi \omega^2}}\left(-x^2 \cos(\omega x)\vert_0^{10} + 2\int_0^{10} x\cos(\omega x) \mathrm{d}x\right)\\ &= \sqrt{\frac{2}{\pi \omega^2}}\left(-100 \cos(100\omega) + \frac{2}{\omega}\int_0^{10} x\mathrm{d}(\sin(\omega x))\right)\\ &= \sqrt{\frac{2}{\pi \omega^2}}\left(-100 \cos(100 \omega) + \frac{2}{\omega} x\sin(\omega x)\vert_0^{10} - \frac{2}{\omega}\int_0^{10} \sin(\omega x) \mathrm{d}x\right)\\ &= \sqrt{\frac{2}{\pi \omega^2}}\left(-100 \cos(100 \omega) + \frac{2}{\omega}x\sin(\omega x)\vert_0^{10} + \frac{2}{\omega^2}\cos(\omega x)\vert_0^{10}\right)\\ &= \sqrt{\frac{2}{\pi \omega^2}}\left(-100 \cos(100 \omega) + \frac{20}{\omega}\sin(20\omega) + \frac{2}{\omega^2}(\cos(10 \omega) - 1)\right) \end{aligned} \\ \textsf{Fourier cosine transform of}\,f(x)\\ \begin{aligned} f_c(\omega) &= \sqrt{\frac{2}{\pi}}\int_{-\infty}^{\infty} f(x) \cos(\omega x) \mathrm{d}x\\ &=\sqrt{\frac{2}{\pi}}\int_0^{10} x^2 \cos(\omega x) \mathrm{d}x + \sqrt{\frac{2}{\pi}}\int_{10}^{\infty} 0 \times \cos(\omega x) \mathrm{d}x\\ &= \frac{1}{\omega}\sqrt{\frac{2}{\pi}}\int_0^{10} x^2 \mathrm{d}(\sin(\omega x))\\ &= \sqrt{\frac{2}{\pi \omega^2}}\left(x^2 \sin(\omega x)\vert_0^{10} - 2\int_0^{10} x\sin(\omega x) \mathrm{d}x\right)\\ &= \sqrt{\frac{2}{\pi \omega^2}}\left(100 \sin(100\omega) + \frac{2}{\omega}\int_0^{10} x\mathrm{d}(\cos(\omega x))\right)\\ &= \sqrt{\frac{2}{\pi \omega^2}}\left(100 \sin(100 \omega) +\frac{2}{\omega} x\cos(\omega x)\vert_0^{10} - \frac{2}{\omega}\int_0^{10} \cos(\omega x) \mathrm{d}x\right)\\ &= \sqrt{\frac{2}{\pi \omega^2}}\left(100 \sin(100 \omega) +\frac{2}{\omega}x\cos(\omega x)\vert_0^{10} - \frac{2}{\omega^2}\sin(\omega x)\vert_0^{10}\right)\\ &= \sqrt{\frac{2}{\pi \omega^2}}\left(100 \sin(100 \omega) + \frac{20}{\omega}\cos(20\omega) - \frac{2}{\omega^2}\sin(10 \omega)\right) \end{aligned} \\ (ii)\\ \textsf{Fourier sine transform of}\,f(x) \\ \begin{aligned} f_c(\omega) &= \sqrt{\frac{2}{\pi}}\int_{-\infty}^{\infty} f(x) \cos(\omega x) \mathrm{d}x\\ &=\sqrt{\frac{2}{\pi}}\int_0^{1} x\cos(\omega x) \mathrm{d}x + \sqrt{\frac{2}{\pi}}\int_1^{2} (x + 1)\cos(\omega x) \mathrm{d}x \\&+ \sqrt{\frac{2}{\pi}}\int_{2}^{\infty} 0 \times \cos(\omega x) \mathrm{d}x \\ &= \frac{1}{\omega}\sqrt{\frac{2}{\pi}}\left(\int_0^{1} x\mathrm{d}(\sin(\omega x)) + \int_1^{2} (x + 1)\mathrm{d}(\sin(\omega x))\right)\\ &= \frac{1}{\omega}\sqrt{\frac{2}{\pi}}\left(x\sin(\omega x)\vert_0^{1} - \int_0^{1} \sin(\omega x) \mathrm{d}x + (x + 1)\sin(\omega x)\vert_1^{2}\right.\\&\left. - \int_1^{2} \sin(\omega x) \mathrm{d}x\right)\\ &= \frac{1}{\omega}\sqrt{\frac{2}{\pi}}\left(\sin(\omega) + \frac{\cos(\omega x)}{\omega}\vert_0^{1} +3\sin(2\omega) \right.\\&\left.- 2\sin(\omega) + \frac{\cos(\omega x)}{\omega}\vert_1^{2}\right)\\ &= \frac{1}{\omega}\sqrt{\frac{2}{\pi}}\left(\sin(\omega) + \frac{\cos(\omega)}{\omega} - \frac{1}{\omega} + 3\sin(2\omega) \right.\\&\left.- 2\sin(\omega) + \frac{\cos(2\omega)}{\omega} - \cos(\omega)\right)\\ &= \frac{1}{\omega}\sqrt{\frac{2}{\pi}}\left(-\sin(\omega) + 3\sin(2\omega) + \frac{\cos(2\omega)}{\omega} - \frac{1}{\omega}\right) \end{aligned} \\ \textsf{Fourier cosine transform of}\,f(x) \\ \begin{aligned} f_c(\omega) &= \sqrt{\frac{2}{\pi}}\int_{-\infty}^{\infty} f(x) \cos(\omega x) \mathrm{d}x \\&=\sqrt{\frac{2}{\pi}}\int_0^{1} x\cos(\omega x) \mathrm{d}x + \sqrt{\frac{2}{\pi}}\int_1^{2} (x + 1)\cos(\omega x) \mathrm{d}x \\&+ \sqrt{\frac{2}{\pi}}\int_{2}^{\infty} 0 \times \cos(\omega x) \mathrm{d}x \\ &= \frac{1}{\omega}\sqrt{\frac{2}{\pi}}\left(\int_0^{1} x\mathrm{d}(\sin(\omega x)) + \int_1^{2} (x + 1)\mathrm{d}(\sin(\omega x))\right)\\ &= \frac{1}{\omega}\sqrt{\frac{2}{\pi}}\left(x\sin(\omega x)\vert_0^{1} - \int_0^{1} \sin(\omega x) \mathrm{d}x + \int (x + 1)\sin(\omega x)\vert_1^{2} \right.\\&\left.- \int_1^{2} \sin(\omega x) \mathrm{d}x\right)\\ &= \frac{1}{\omega}\sqrt{\frac{2}{\pi}}\left(\sin(\omega) + \cos(\omega x)\vert_0^{1} +3\sin(2\omega) \right.\\&\left.- 2\sin(\omega) + \cos(\omega x)\vert_1^{2}\right)\\ &= \frac{1}{\omega}\sqrt{\frac{2}{\pi}}\left(\sin(\omega) + \cos(\omega) - 1 + 3\sin(2\omega) \right.\\&\left.- 2\sin(\omega) + \cos(2\omega) - \cos(\omega)\right)\\ &= \frac{1}{\omega}\sqrt{\frac{2}{\pi}}\left(-\sin(\omega) + 3\sin(2\omega) + \cos(2\omega) - 1\right) \end{aligned} \\ \begin{aligned} f_s(\omega) &= \sqrt{\frac{2}{\pi}}\int_{-\infty}^{\infty} f(x) \sin(\omega x) \mathrm{d}x\\ &=\sqrt{\frac{2}{\pi}}\int_0^{1} x\sin(\omega x) \mathrm{d}x + \sqrt{\frac{2}{\pi}}\int_1^{2} (x + 1)\sin(\omega x) \mathrm{d}x \\&+ \sqrt{\frac{2}{\pi}}\int_{2}^{\infty} 0 \times \sin(\omega x) \mathrm{d}x \\ &= \frac{1}{\omega}\sqrt{\frac{2}{\pi}}\left(\int_0^{1} x\mathrm{d}(-\cos(\omega x)) + \int_1^{2} (x + 1)\mathrm{d}(-\cos(\omega x))\right)\\ &= \frac{1}{\omega}\sqrt{\frac{2}{\pi}}\left(-x\cos(\omega x)\vert_0^{1} + \int_0^{1} \cos(\omega x) \mathrm{d}x - (x + 1)\cos(\omega x)\vert_1^{2} \right.\\&\left.+ \int_1^{2} \cos(\omega x) \mathrm{d}x\right)\\ &= \frac{1}{\omega}\sqrt{\frac{2}{\pi}}\left(-\cos(\omega) + \frac{\sin(\omega x)}{\omega}\vert_0^{1} - 3\cos(2\omega) \right.\\&\left.+2\cos(\omega) + \frac{\sin(\omega x)}{\omega}\vert_1^{2}\right)\\ &= \frac{1}{\omega}\sqrt{\frac{2}{\pi}}\left(-\cos(\omega) + \frac{\sin(\omega)}{\omega} - 3\cos(2\omega) + 2\cos(\omega) \right.\\&\left.+ \frac{\sin(2\omega)}{\omega} - \frac{\sin(\omega)}{\omega} \right)\\ &= \frac{1}{\omega}\sqrt{\frac{2}{\pi}}\left(\cos(\omega) - 3\cos(2\omega) + \frac{\sin(2\omega)}{\omega}\right) \end{aligned}$