Fourier sine transform of $e^{-x}$

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 $$\dfrac{k}{1+k^2}$$

• You need to describe your problem, especially since you have the answer. – herb steinberg Nov 8 '18 at 19:53
• do you want help computing the integral? – gt6989b Nov 8 '18 at 20:01

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