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I'm working on a problem, and stuck with the following integral

$$\int_{-\infty}^\infty \dfrac{e^{-izx}}{x^2+y^2}dx$$ How can we compute this integral?

JJ Beck
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  • Are $z$ and $y$ constant? – tylerc0816 Dec 18 '13 at 16:08
  • @tylerc0816 Yes – JJ Beck Dec 18 '13 at 16:13
  • Is z=x+iy or the variable of FT ? @JJBeck – AHH Dec 18 '13 at 17:09
  • This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. – Carl Mummert Dec 18 '13 at 22:16

2 Answers2

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Such integrals can be nicely computed using complex methods, the residue theorem in particular.

If $z < 0$, the function

$$h(w) = \frac{e^{-izw}}{w^2+y^2}$$

is small in the upper half plane, and the residue theorem yields

$$\int_{-\infty}^\infty \frac{e^{-izx}}{x^2+y^2}\,dx = 2\pi i \operatorname{Res} \left(\frac{e^{-izw}}{w^2+y^2}; i\lvert y\rvert\right) = 2\pi i \left(\frac{e^{z\lvert y\rvert}}{2i\lvert y\rvert}\right) = \pi \frac{e^{z\lvert y\rvert}}{\lvert y\rvert}.$$

For $z > 0$, you can get the result either by the residue theorem again, or by symmetry considerations, the result is

$$\int_{-\infty}^\infty \frac{e^{-izx}}{x^2+y^2}\,dx = \pi\frac{e^{-\lvert zy\rvert}}{\lvert y\rvert}.$$

Daniel Fischer
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Substitute $x = y u.$ This will change your transform variable to $z y,$ and the transformed function to $\frac{1}{y} \frac{1}{1+x^2}.$ The transform of the last thing can be expressed in terms of Heaviside functions (the method is described here)

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Igor Rivin
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