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Fourier sine transform of $\displaystyle \frac{1}{x}$ is ..(fill in the blanks)..

My thoughts: By Definition, $\displaystyle F_s(s)=\int_{0}^{\infty}f(x)\sin sxdx $

$\displaystyle F_s(s)=\int_{0}^{\infty}\frac{\sin sx}{x}dx $

How do I integrate this? This just keeps blowing up if I integrate by parts.

An online search says this is a special function called sine integral.

$\displaystyle \int\frac{\sin ax}{x} \mathrm{d}x = \sum_{n=0}^\infty (-1)^n\frac{(ax)^{2n+1}}{(2n+1)\cdot (2n+1)!} +C$

This does not seem like the right path.... How do I solve this?

The given answer is $s^2/2$

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    $\begingroup$ Rescale the integration variable to get the parameter out of the sine function. $\endgroup$
    – David H
    Jul 26, 2014 at 15:26
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    $\begingroup$ The given answer has the disadvantage of being wrong. For real $s$, the value depends only on the sign of $s$. $\endgroup$ Jul 26, 2014 at 15:29
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    $\begingroup$ See this. $\endgroup$ Jul 26, 2014 at 15:36

1 Answer 1

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Notice that if $s>0$, $y=sx$ changes the integral $\int_0^\infty \frac{\mathbb{sin}x}{x} \mathbb{d}x$ ($s$ vanishes). There are two possibilities: one is to use contour integration in the complex plane. I am going to assume that you do not master such techniques yet, so let us examine a more elementary approach.

Consider the function $f(t)=\int_0^\infty \mathbb{e}^{-tx} \frac{\mathbb{sin}x}{x} \mathbb{d}x$ for $t \geq 0$. Notice that $f'(t)=-\int_0^\infty \mathbb{e}^{-tx} \mathbb{sin}x \mathbb{d}x$. Integrate this twice by parts and you will get a linear equation in $f'(t)$ which will give you $f'(t)=-\frac{1}{1+t^2}$, hence $f(t)=-\mathbb{arctan}t+C$, with $C$ a real constant. Notice that for $t \rightarrow \infty$ you get $f(t) \rightarrow 0$ because of the exponential, thus $C=\frac{\pi}{2}$. Then $f(0)=\int_0^\infty \frac{\mathbb{sin}x}{x} \mathbb{d}x = C = \frac{\pi}{2}$.

The final answer is $\frac{\pi}{2} \mathbb{sign}x$, not the one that you claim.

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