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

Don't add conditions in the comments, add them to the question. All details should be in the body of the question, not in comments. @NikhanbayevNursultan — Thomas Andrews, Aug 22 '15 at 13:29
Sorry, I am very beginner here. I will take it into account next time. @ThomasAndrews — Nikhanbayev Nursultan, Aug 22 '15 at 13:34

score 1 · Answer 1 · answered Aug 22 '15 at 13:26

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:

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

score 1 · Answer 2 · answered Aug 22 '15 at 13:27

1

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).

answered Aug 22 '15 at 13:27

hunter

29,847

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

2 Answers2