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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. $\endgroup$ Commented Nov 8, 2018 at 19:53
  • $\begingroup$ do you want help computing the integral? $\endgroup$
    – gt6989b
    Commented Nov 8, 2018 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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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

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