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Question

Find the Fourier sine transform of the function in the indicated problem, and write f(x) as a Fourier integral. Verify that the sine integral for f(x) is the same as the exponential integral found previously.

f(x)={1,π<x<01,0<x<π0,x>πf(x)=\left\{\begin{array}{rr} -1, & -\pi<x<0 \\ 1, & 0<x<\pi \\ 0, & |x|>\pi \end{array}\right.

Solution

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The Fourier sine transform of the function is equal to:

g(α)=2π0πsin(αx)dx=2πcos(αx)α=2π1cos(αx)α\begin{align*} g(\alpha)=&\sqrt{\dfrac{2}{\pi}}\int_0^{\pi}\sin(\alpha x)dx=-\sqrt{\dfrac{2}{\pi}}\dfrac{\cos(\alpha x)}{\alpha} =\sqrt{\dfrac{2}{\pi}}\dfrac{1-\cos(\alpha x)}{\alpha}\\ \end{align*}

The function is then:

f(x)=2π01cos(απ)αsin(αx)dα\begin{gather*} \boxed{f(x)=\dfrac{2}{\pi}\int_0^{\infty}\dfrac{1-\cos(\alpha\pi)}{\alpha}\sin(\alpha x)d\alpha}\\ \end{gather*}

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