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I'm trying to find the Fourier sine transform of $e^{-x}$. I know that $e^{-x}=\cosh x-\sinh x$. Keeping in mind that $\cosh x$ is an even function so I have the following transformation: $$\sqrt{\frac{2}{\pi}}\int_0^\infty -\sin(kx)\sinh(x)dx$$

However I have problem calculating this integral or I'm doing something completely off. What I found out from Fourier sine transform table with exponential functions it that transform equals $\dfrac{k}{1+k^2}$

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    $\begingroup$ You need to describe your problem, especially since you have the answer. $\endgroup$ – herb steinberg Nov 8 '18 at 19:53
  • $\begingroup$ do you want help computing the integral? $\endgroup$ – gt6989b Nov 8 '18 at 20:01
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Hint: you can use complex numbers and write

$$e^{ikx}=\cos(kx)+i\sin(kx)$$

Calculating the integral of $e^{-x}e^{ikx}$ should be trivial. The real part of the answer is the cosine transform, the imaginary part is the sine transform.

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