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Im trying to find the Fourier Transform of $ f(x) = \sin(x) \cdot e^{-|x|}$

I applied the standard formula and got to this point:

$$\tilde{f(t)}= \frac{1}{\sqrt{2\pi}}\cdot \int_{-\infty}^{+\infty}\sin(x)e^{-|x|}e^{-itx} dx $$

= $$\frac{1}{\sqrt{2\pi}}\cdot \int_{-\infty}^{0}\sin(x)e^{x}e^{-itx} dx+\frac{1}{\sqrt{2\pi}}\cdot \int_{0}^{+\infty}\sin(x)e^{-x}e^{-itx} dx $$

How can I integrate this - I tried integration by parts and did not get anywhere because there will always be alternation between sin and cos and also there is an e function in the integrand.

Now Im trying to think about integrating by the help of substitution but I couldn't think of anything to substitute it with. Can you please help :) ?

Thanks a lot!

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    $\begingroup$ use that $sin(x) = \frac {e^{i\psi t} - e^{-i \psi t}}2$ $\endgroup$ – mike Apr 30 '12 at 20:03
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In general, if you use integration by parts twice, you can solve it like this: $$I=\int{\sin(x)e^{ax}dx}=-\cos(x)e^{ax} + a\int{\cos(x)e^{ax}dx} = -\cos(x)e^{ax} + a\sin(x)e^{ax} - a^2\int{\sin(x)e^{ax}dx}= -\cos(x)e^{ax} + a\sin(x)e^{ax} -a^2I$$

Leading to:

$$ I = \frac{-\cos(x)e^{ax} + a\sin(x)e^{ax}}{1+a^2} $$

As mike suggested, you can use the complex representation as well.

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