Question

If it is known that \( \int_0^{\infty} \frac{e^{ix}}{x} dx = \frac{\pi i}{2}, \) then compute: \( \int_0^{\infty} \frac{\sin 5x}{x} dx = ? \)

• A. \( \frac{\pi}{10} \)
• B. \( \frac{5\pi}{2} \)
• C. \( \frac{\pi}{2} \)
• D. \( \frac{\pi}{2} + 5 \)

Hint:

The integral \( \int_0^{\infty} \frac{\sin(ax)}{x} dx = \frac{\pi}{2} \) is a standard result and is independent of \( a \), for \( a>0 \).

Answer 1

The Correct Option is C

To solve the problem, we need to compute the integral: \[\int_0^{\infty} \frac{\sin 5x}{x} dx\] We can use the following method involving Euler's Formula: \[e^{ix} = \cos x + i\sin x\] Given: \[\int_0^{\infty} \frac{e^{ix}}{x} dx = \frac{\pi i}{2}\] Consider the integral of the imaginary part: \[\int_0^{\infty} \frac{\sin x}{x} dx\] Given that \[\int_0^{\infty} \frac{e^{ix}}{x} dx = \int_0^{\infty} \left(\frac{\cos x}{x} + i\frac{\sin x}{x}\right) dx\] We focus on the imaginary part: \[\int_0^{\infty} \frac{\sin x}{x} dx = \frac{\pi}{2}\] Now apply the substitution \(u = 5x\), \(du = 5dx\), thus \(dx = \frac{du}{5}\). 
The integral becomes: \[\int_0^{\infty} \frac{\sin 5x}{x} dx = \int_0^{\infty} \frac{\sin u}{u/5} \cdot \frac{du}{5} = 5\int_0^{\infty} \frac{\sin u}{u} \cdot \frac{du}{5}\] Simplifying: \[\int_0^{\infty} \frac{\sin u}{u} du = \frac{\pi}{2}\] Therefore, multiplying by 1 due to conversion factor: \[5 \times \frac{\pi}{10} = \frac{\pi}{2}\] Hence, the value of the integral \(\int_0^{\infty} \frac{\sin 5x}{x} dx\) is \(\frac{\pi}{2}\).