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I couldn't calculate $$\int \limits_{-\infty}^{+\infty} \frac{e^{-itx}}{2\pi} \frac{1}{a^2+x^2} dx. $$

I can either turn this into something along the lines of $\large \int \limits_0^{\pi/2} \cos( t \cdot \tan x ) dx$ or $ \large \int \limits_0^{+\infty} \frac{\cos tx}{a + ix} dx$ neither of which I can solve.

I've been told, that some tools of complex calculus could simplify this, but my book hasn't covered any before giving the exercise, so I wonder if there is a way without it.


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Its better we use duality property of Fourier transform. It says that if $$f(t)\Longleftrightarrow F(ix)$$ then $$F(it)\Longleftrightarrow 2\pi f(-x)$$Thus by knowing that $$\dfrac{1}{2a}e^{-a|t|}\Longleftrightarrow \dfrac{1}{a^2+x^2}$$, by invoking duality, we get $$\dfrac{1}{a^2+t^2}\Longleftrightarrow \dfrac{\pi}{a}e^{-a|x|}$$Hope this helps.


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Hint: you may find the initial Fourier transform in any table.

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