Is function f defined by $f(z) = \int_0^{\infy} |e^{zt}| / (t+1) dt$ analytical on the left plane D: Re(z) < 0
user
Sure, I'd be happy to provide a response.

The integral in the definition of $f(z)$ appears to diverge for any finite upper limit due to the presence of the exponential function. This divergence can be seen by considering the behavior of $e^{zt}$ as $t \to \infty$ for fixed $z$ with real part less than zero. In particular, if $z = s + iy$, where $s < 0$, then as $t \to \infty$, we have

$$
e^{zt} \sim \left(\frac{-1}{t}\right)^s.
$$

So, the denominator, $t + 1$, does not prevent the integral from diverging if $|s| > 1$ because $(t+1) \sim t$ as $t \to \infty$. Therefore, the integral in the definition of $f(z)$ does not converge if the real part of $z$ is negative and greater than $-1$.

In summary, if $z = s + is$ with $s < 0$ and $|s| > 1$, the function $f(z)$ is not analytic on the left plane $D$. If $s \geq -1$, it is possible that $f(z)$ is analytic, but further analysis is needed to check the convergence of the infinite integral in its definition.

I hope this helps! Let me know if you have any further questions.
assistant
Is function f defined by $f(z) = \int_0^{\infy} |e^{zt}| / (t+1) dt$ analytical on the left plane D: Re(z) < 0
user
To determine if the function f(z) is analytic on the left plane D: Re(z) < 0, we can use the Morera's Theorem, which states that if a function is continuous in a domain D and the integral of the function over any closed contour in D is zero, then the function is analytic in D.

In this case, the function f(z) is defined as an infinite integral, and it's not immediately clear whether this integral converges or not. However, we can make an educated guess by looking at the behavior of the integrand.

The integrand in this case is $|e^{zt}| / (t+1)$. Since we are considering the left plane Re(z) < 0, we know that the real part of z is negative. Therefore, the exponential term $e^{zt}$ will decay to zero as t goes to infinity, because the negative real part of z will make the exponent negative for large t.

The factor of $(t+1)$ in the denominator is always positive and grows without bound as t goes to infinity. Therefore, it will not cause the integral to diverge.

Thus, we can expect that the integral defining f(z) will converge for any z in the left plane. This means that f(z) is well-defined and finite in the left plane, and it's likely to be continuous and analytic there as well.

However, this is only a heuristic argument and not a rigorous proof. A full proof would require showing that the integral converges uniformly on compact subsets of the left plane and that f(z) satisfies the Cauchy-Riemann equations.

In conclusion, it's likely that f(z) is analytic on the left plane D: Re(z) < 0, but a rigorous proof would require more detailed analysis.
assistant