Friday, April 21, 2017

calculus - show that intinfty0fracsin3(x)x3dx=frac3pi8



show that


0sin3(x)x3dx=3π8


using different ways


thanks for all


Answer



Let f(y)=0sin3yxx3dx Then, f(y)=30sin2yxcosyxx2dx=340cosyxcos3yxx2dx f Therefore, f''(y) = \frac{9}{4} \int_{0}^{\infty} \frac{\sin{3yx}}{x} \mathrm{d}x - \frac{3}{4} \int_{0}^{\infty} \frac{\sin{yx}}{x} \mathrm{d}x


Now, it is quite easy to prove that \int_{0}^{\infty} \frac{\sin{ax}}{x} \mathrm{d}x = \frac{\pi}{2}\mathop{\mathrm{signum}}{a}


Therefore, f''(y) = \frac{9\pi}{8} \mathop{\mathrm{signum}}{y} - \frac{3\pi}{8} \mathop{\mathrm{signum}}{y} = \frac{3\pi}{4}\mathop{\mathrm{signum}}{y} Then, f'(y) = \frac{3\pi}{4} |y| + C Note that, f'(0) = 0, therefore, C = 0. f(y) = \frac{3\pi}{8} y^2 \mathop{\mathrm{signum}}{y} + D Again, f(0) = 0, therefore, D = 0.


Hence, f(1) = \int_{0}^{\infty} \frac{\sin^3{x}}{x^3} = \frac{3\pi}{8}


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