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As stated in the title I should calculate the cosine and sine Fourier transform of:

$$f_1(x)=\exp{(-ax)}(1+bx)^{-1}$$

and

$$f_2(x)=\exp{(-ax)}(1+bx)^{-2}$$

That obviously means calculating:

$$\int_0^{\infty}\, f_i(x)\cos(\omega x)dx$$ and $$\int_0^{\infty}\, f_i(x)\sin(\omega x)dx$$

Are those definite integrals known?

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By considering the sine and cosine functions as the imaginary and real part of a complex exponential, the problem boils down to finding: $$ I_1(z) = \int_{0}^{+\infty}\frac{e^{-zx}}{1+x}\,dx, \qquad I_2(z) = \int_{0}^{+\infty}\frac{e^{-zx}}{(1+x)^2}\,dx,$$ (where $\text{Re}(z)>0$) both depending on the incomplete $\Gamma$ function: $$ I_1(z) = e^{z}\, \Gamma(0,z),\qquad I_2 = 1-z\,e^z\, \Gamma(0,z). $$

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  • $\begingroup$ Thanks. I still don't understand how to proceed from your comment. Could you expand it a bit? $\endgroup$ – JFNJr Feb 2 '15 at 18:11
  • $\begingroup$ $I_1(z)$ and $I_2(z)$ are the wanted Fourier series, so what else do you need? To get rid of the two parameters, just set $x=\frac{1}{b}x'$. $\endgroup$ – Jack D'Aurizio Feb 2 '15 at 18:14
  • $\begingroup$ Stupid me! I was fooled by a bad connection on my mobile! Thanks! $\endgroup$ – JFNJr Feb 10 '15 at 8:30

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