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

• A
• Anixx
In summary, the Delta function can be written in various forms, such as an integral of an exponential function or a combination of sine and cosine functions. However, when dealing with complex arguments, there can be discrepancies in the results, such as the integral depending on the sign of the complex number. This can be resolved by carefully considering the properties of the Delta function in complex analysis.

#### 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)##.

$$\int_{-\infty}^\infty \delta(t+bi)f(t)dt=f(-bi)$$

which depends on the sign of ##b##.

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)##.

$$\int_{-\infty}^\infty \delta(t+bi)f(t)dt=f(-bi)$$