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I've stumbled across the following identity $$ x^{+} = \frac{1}{2 \pi} \int_{-\infty}^\infty e^{(m + i z )x} \frac{1}{(m+i z )^2} d z, $$ supposed to hold for $x\in \mathbb{R}$ and $m >0$. I cannot seem to find a reference nor find out how to prove it. I hope someone will help with either a proof or idea.

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If you take the second derivative over x:

$$f''(x)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{(m+iz)x}{\rm d}z=e^{mx}\delta(x)=\delta(x)$$ I recognized the Fourier transform of the delta function. Now integrate twice. First integral is the Heaviside step function:

$$f'(x)=\int_{-\infty}^x \delta(t){\rm d}t=H(x)$$

Integrate again, and you get $$f(x)=\int_{-\infty}^x H(t){\rm d}t=x^+$$

The integrals are only defined in terms of distributions, but you can regularize the procedure, for instance, by including a dissipative term $e^{-\epsilon|z|}$ and sending $\epsilon\to 0$.

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