Improper Integral

Show that $\int_0^\infty \dfrac{\sin^3 x}x \, dx = \dfrac\pi4$

#Calculus

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Since $\sin^3 x = \frac{1}{4}\left(3\sin x - \sin 3x\right)$ so the integral becomes $\int_0^{\infty}\frac{\sin^3x}{x} dx =\frac{1}{4}\int_0^{\infty}\left(\frac{3\sin x}{x}-\frac{\sin 3x}{x}\right)dx=\frac{1}{4}\left(\frac{3\pi}{2}-\int_0^{\infty}\frac{\sin y}{y}dy\right)=\frac{1}{4}\left(\frac{3\pi}{2}-\frac{\pi}{2}\right)=\frac{\pi}{4}$ we use the fact $\displaystyle \int_0^{\infty}\frac{\sin x}{x}dx =\frac{\pi}{2}$

Naren Bhandari - 8 months, 1 week ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$