复积分$I(z)=\int_{[0,z]}{(e^{-t²})}^{\text{erf(t)}}dt$的收敛性与整函数判定问询

Great questions about this complex integral! Let's break down each part clearly, starting with some simplification to make the analysis easier.

First, rewrite the integrand to simplify it:
$$
{(e^{-t²})}{\text{erf}(t)} = e^{-t2 \cdot \text{erf}(t)}
$$
Since $\text{erf}(z)$ is an entire function (holomorphic everywhere on the complex plane), the product $t^2 \cdot \text{erf}(t)$ is also entire. Composing this with the exponential function (which preserves holomorphy) gives us an entire integrand $f(t) = e^{-t2 \cdot \text{erf}(t)}$.

• Convergence: For any finite complex $z$, pick any path from $0$ to $z$ (the integral is path-independent because $f(t)$ is entire). On any compact subset of $\mathbb{C}$, $f(t)$ is bounded, and the path from $0$ to $z$ has finite length—so the integral $I(z)$ converges absolutely for all $z \in \mathbb{C}$.
• Analyticity: The derivative of $I(z)$ is just the integrand evaluated at $z$:
  $$
  I'(z) = f(z) = e^{-z2 \cdot \text{erf}(z)}
  $$
  Since $f(z)$ is entire, $I'(z)$ exists and is holomorphic everywhere on $\mathbb{C}$. This means $I(z)$ is analytic across the entire complex plane.

Absolutely! By definition, an entire function is one that's holomorphic everywhere on $\mathbb{C}$. We just showed $I(z)$ meets this criteria: its derivative exists everywhere, and the derivative itself is an entire function. The key here is that integrating an entire function over a path from $0$ to $z$ always produces another entire function.

Let's split this into real and complex cases, since "positivity" doesn't apply to complex numbers (they don't have a total order like real numbers):

Real $z$ #

• For $z > 0$: The integrand $f(t) = e^{-t2 \cdot \text{erf}(t)}$ is positive for all real $t ≥ 0$ (since $\text{erf}(t) > 0$ here, making the exponent negative and the exponential positive). Integrating a positive function from $0$ to $z > 0$ gives $I(z) > 0$.
• For $z < 0$: Substitute $t = -s$ (where $s > 0$). The integral becomes:
  $$
  I(z) = \int_0^{-|z|} f(t) dt = -\int_0^{|z|} e^{-s2 \cdot \text{erf}(-s)} ds = -\int_0^{|z|} e^{s2 \cdot \text{erf}(s)} ds
  $$
  The integrand here is positive, so $I(z) < 0$ for $z < 0$. On the real line, $I(z)$ is non-zero for all $z ≠ 0$, positive on the positive axis, negative on the negative axis.

Complex $z$ #

• Positivity: This term doesn't make sense for complex numbers—you can't call a non-real complex number "positive" or "negative". For example, take $z = iy$ (pure imaginary, $y ≠ 0$):
  $$
  I(iy) = i \int_0^y e^{i s^2 \cdot \text{erfi}(s)} ds
  $$
  where $\text{erfi}(s)$ is the imaginary error function (positive for $s > 0$). The integrand has modulus $1$, so $I(iy)$ is a non-real complex number, which can't be classified as positive or negative.
• Zeroes: $I(z)$ is a non-polynomial entire function (its derivative $I'(z) = e^{-z2 \cdot \text{erf}(z)}$ is not a polynomial). By Picard's Theorem, any non-polynomial entire function takes every complex value infinitely many times, except possibly one exceptional value. Since $I(0) = 0$, zero is not an exceptional value—so there are infinitely many non-zero complex $z$ where $I(z) = 0$. Note that all these zeroes are simple (since $I'(z) ≠ 0$ everywhere, as the exponential function never equals zero).

内容的提问来源于stack exchange，提问作者zeraoulia rafik