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Define the signum function, $\text{sgn}(x)$, by

$$\text{sgn}(x)=\begin{cases} 1, & x>0\\0, & x=0\\-1, & x<0 \end{cases}$$

Establish the identity

$$\dfrac{2}{\pi}\int_0^ \infty \dfrac{\sin(xt)}{t}dt=\text{sgn}(x)$$

There is a hint that we can make a change of variables $\omega = xt$

Here is my work: Since it's an odd function, we have $A(\omega)=0$, and

$$B(\omega ) = \frac{2}{\pi} \int_{- \infty }^ \infty \sin( \omega t) \,dt$$

I stuck here since $B(\omega)$ does not converge. Can someone help me please? Thank you very much!

Edward Jiang
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jinha0001
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  • $$\frac2\pi\int_0^\infty\dfrac{sin(xt)}{t},dt=\frac{2x}{\pi}\int_0^\infty \frac{sin(u)}u , du=\dfrac{2x}\pi\times\frac \pi2=x$$ is what I got...does not equal sign(x) – Teoc Dec 23 '14 at 04:03
  • @MathNoob If $x<0$ then there's a - in the $du$. – Ian Dec 23 '14 at 04:22
  • It would still result in $f(x)=\pm x,$however, and not sign(x) – Teoc Dec 23 '14 at 04:25
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    $du=d(xt) = xdt$ so, the second term is missing a $1/x$ then we will get the answer 1. I wonder if I can just establish the identity on this way instead of by using the fourier integral theorem? – jinha0001 Dec 23 '14 at 04:40

1 Answers1

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You should have $t=\omega/x$, so $\sin(xt)/t = x \sin(\omega)/\omega$. This lets you factor out the $x$. Now try to rewrite the integral after the substitution in terms of the integral before the substitution.

Ian
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