# 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. Nov 8 '18 at 19:53
• do you want help computing the integral? 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.
I would suggest not to use $$e^{-x} = coshx - sinhx$$
f(x) = $$e^{-x}$$
Now for Fourier sine transform, we know, $$F_s[f(x)] = \int_0^∞ e^{-x}sin(sx)dx$$ Now,applying integration by parts $$I = \int e^{-x}sin(sx)dx = e^{-x}\int sin{sx}dx + \int e^{-x}(\int sin{sx}dx)dx$$ $$= \frac{-e^{-x}(cos{sx})}{s} - \frac{\int e^{-x}(cos{sx})dx}{s}$$ $$= \frac{-e^{-x}(cos{sx})}{s} - \frac{[\frac{e^{-x}(sin{sx})}{s} + \frac{\int e^{-x}sin(sx)dx}{s}]}{s}$$ $$= \frac{-e^{-x}(cos{sx})}{s} - \frac{e^{-x}(sin{sx})}{s^2} - \frac{I}{s^2}$$ $$or, I(1 +\frac{1}{s^2}) = -\frac{-e^{-x}(scos{sx} + sin{sx})}{s^2}$$ $$or, I = -\frac{-e^{-x}(scos{sx} + sin{sx})}{s^2+1}$$ Now, putting the limits, $$F_s[f(x)] = [-\frac{-e^{-x}(scos{sx} + sin{sx})}{s^2+1}]_0^∞$$ $$= \frac {s}{s^2 + 1}$$ I hope it helps