A Is ##\delta\left(a+bi\right)=\delta\left(a-bi\right)##?

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We can write Delta function as

$$\delta(z) = \frac{1}{2\pi}\int_{-\infty}^\infty e^{itz}\, dt=\delta\left(a+bi\right)=\frac1{2\pi}\int_{-\infty}^{+\infty}e^{-bx}\cos ax\, dx+\frac{i}{2\pi}\int_{-\infty}^{+\infty}e^{-bx}\sin ax\, dx.$$

The second integral is always zero (via Abel regularization, Laplace transform), the first integral does not depend on the sign of ##b##. So, ##\delta\left(a+bi\right)## should be equal to ##\delta\left(a-bi\right)##.

But this contradicts https://math.stackexchange.com/a/4045521/2513
$$\int_{-\infty}^\infty \delta(t+bi)f(t)dt=f(-bi)$$

which depends on the sign of ##b##.
 
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Fix the integrals! Off hand they look like divergent.
 
I am guessing that the following leap $$\int_{-\infty}^\infty e^{-bx}(\cos(ax)+i\sin(ax))\,dt=\int_{-\infty}^\infty e^{-bx}\cos(ax)\,dz+i\int_{-\infty}^\infty e^{-bx}\sin(ax)\,dt$$ is not justified due to failure of absolute convergence (in some generalized sense that applies to distributions).
 
Anixx said:
We can write Delta function as

$$\delta(z) = \frac{1}{2\pi}\int_{-\infty}^\infty e^{itz}\, dt=\delta\left(a+bi\right)=\frac1{2\pi}\int_{-\infty}^{+\infty}e^{-bx}\cos ax\, dx+\frac{i}{2\pi}\int_{-\infty}^{+\infty}e^{-bx}\sin ax\, dx.$$

The second integral is always zero (via Abel regularization, Laplace transform), the first integral does not depend on the sign of ##b##. So, ##\delta\left(a+bi\right)## should be equal to ##\delta\left(a-bi\right)##.

But this contradicts https://math.stackexchange.com/a/4045521/2513
$$\int_{-\infty}^\infty \delta(t+bi)f(t)dt=f(-bi)$$

which depends on the sign of ##b##.
You are confusing complex and real arguments here. See:

https://mathoverflow.net/questions/118101/dirac-delta-function-with-a-complex-argument
 

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