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

[Anixx]

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 $b$.

 

[mathman]

Fix the integrals! Off hand they look like divergent.

 

[suremarc]

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).

 

[PeroK]

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