Discussion
Given that Fourier integral representation for the function
\[ f(x) = \begin{cases} 0 & \text{if } |x| \gt 1 \\ 1 & \text{if } |x| \lt 1 \end{cases} \]
is \( f(x)=\frac{2}{\pi}\int_0^\infty\frac{\cos\omega x\sin\omega}{\omega}d\omega \)
Which of the following options is correct ?
Answer: A
The correct statement is the piecewise result: the integral equals \( \pi/2 \) for \( 0 \le x \lt 1 \), \( \pi/4 \) at \( x = 1 \) and 0 for \( x \gt 1 \).
Rearranging the given representation, the integral is \( \frac{\pi}{2} f(x) \) wherever \( f \) is continuous:
\[ \begin{aligned} &\int_0^\infty \frac{\cos\omega x \sin\omega}{\omega}\, d\omega \\ &= \frac{\pi}{2} f(x) \end{aligned} \]
So it is \( \pi/2 \) inside \( |x| \lt 1 \) and 0 outside. At the jump \( x = 1 \) a Fourier integral converges to the mean of the left and right limits, \( \frac{\pi}{2} \cdot \frac{1 + 0}{2} = \frac{\pi}{4} \).
Putting \( x = 0 \) gives the Dirichlet integral \( \int_{0}^{\infty} \frac{\sin \omega}{\omega} d\omega = \frac{\pi}{2} \), so the options giving \( \pi \) and \( \pi/4 \) are wrong, and the integral is 0 only for \( |x| \gt 1 \), not for every \( x \).
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