Is $\int_{0}^\infty \frac{\sin(nx)}x \,dx$ is equal to $\pi/2$?

Question

Is $\int_{0}^\infty \frac{\sin(nx)}x \,dx$ is equal to $\pi/2$ for positive real $n$?

I've come to this answer by inverse Fourier transform. But since there is n, I am quite confused that I didn't get n in the answer. Is this answer incorrect? Thank you

Answer 1

Change of variable:

Let $u=nx$ Then $\frac{du}{u}=\frac{n\,dx}{nx} = \frac{dx}{x}$. So:

$$\int_{0}^{\infty} \frac{\sin nx}{x}\,dx = \int_0^{\infty}\frac{\sin u}{u}\,du$$

So this is true for any positive $n$. For negative $n$, you get the result:

$$\int_{0}^{\infty} \frac{\sin nx}{x}\,dx = -\int_0^{\infty}\frac{\sin u}{u}\,du$$

Answer 2

Make the change of variables $u = nx$. Then $$\int_0^\infty \frac{\sin(nx)}{x} dx = \int_0^\infty \frac{\sin(u)}{(u/n)} d(u/n) = \int_0^\infty \frac{\sin(u)}{u} du.$$ So indeed, the integral does not depend on $n$ (and your value is correct).