show that
∫∞0sin3(x)x3dx=3π8
using different ways
thanks for all
Answer
Let f(y)=∫∞0sin3yxx3dx
Then,
f′(y)=3∫∞0sin2yxcosyxx2dx=34∫∞0cosyx−cos3yxx2dx
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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