Complex differentiation under the integral sign (Ahlfors) In Ahlfors' Complex Analysis text, page 202, he claims that in $\{ \Re z>0 \} $ $$\frac{d}{dz} \int_0^\infty \frac{2 \eta}{\eta^2+z^2} \frac{\mathrm d \eta}{e^{2 \pi \eta}-1}=- \int_0^\infty \frac{4 \eta z}{(\eta^2+z^2)^2} \frac{ \mathrm{d} \eta}{e^{2 \pi \eta}-1} $$
"because the integral on the RHS converges uniformly when $z$ is restricted to any compact set in the half plane $x > 0$."
I can't see why is that the case. I tried forming the quotient $\frac{F(z+\Delta z)-F(z)}{\Delta z}$, but I can't see where does his remark kicks in.
Why is his reasoning valid?
 A: Without using Lebesgue theory, the step
$$\frac{\partial}{\partial z} \int_0^t F(z,\eta) \, d\eta = \int_0^t \frac{\partial}{\partial z} F(z,\eta) \, d\eta$$
requires continuity jointly in $z$ and $\eta$ of $F$ and its partial derivative, without the need for any kind of uniform convergence.
However, the step
$$\frac{\partial}{\partial z}\lim_{t \rightarrow \infty} \int_0^t F(z,\eta) \, d\eta = \lim_{t \rightarrow \infty} \frac{\partial}{\partial z}\int_0^t F(z,\eta) \, d\eta$$
is justified by the uniform convergence as $t \rightarrow \infty$.
Added later to elaborate:
Set
$$G(z,t) = \int_0^t F(z,\eta) \, d\eta.$$
We must justify the statement:
$$\frac{\partial}{\partial z} \lim_{t \rightarrow \infty} G(z,t) = \lim_{t \rightarrow \infty} \frac{\partial}{\partial z} G(z,t).$$
Now we apply a standard theorem on exchanging limits and derivatives, using
(1) $G(z,t)$ converges pointwise (i.e. for each $z$) in the right-half plane to $\int_0^\infty F(z,\eta) \, d\eta$,
(2) all the $G(z,t)$ are differentiable in $z$, and
(3) the derivatives $\frac{\partial}{\partial z} G(z,t) = \int_0^t \frac{\partial}{\partial z} F(z,\eta) \, d\eta$ converge uniformly to $\int_0^\infty \frac{\partial}{\partial z} F(z,\eta) \, d\eta.$
A: I convinced myself of Ahlfor's comment this morning by first showing the claim below.
Claim: Let $g(z, \eta)$ be analytic in $z$ for fixed $\eta$, jointly continuous in $(z, \eta)$, and let $\int _0 ^{\infty} g(z, \eta) d \eta$ converge uniformly on compact subsets of the open right half plane. For each $\eta$, define an anti-derivative of $g(z, \eta)$ by
$$
G(z, \eta) = \int _1 ^ z g(\zeta, \eta) d \zeta
$$
Then
$$
\frac{d}{dz} \int _0 ^{\infty} G(z, \eta) d \eta = \int _0 ^{\infty} g(z, \eta) d \eta 
$$
Once you have this you can verify Ahlfors' claim by using $g(z, \eta) = \frac{2z}{(\eta^2 + z^2)^2}$ and then noticing that
$$
\frac{1}{\eta^2 + z^2} = G(z, \eta) + \frac{1}{\eta^2 + 1}
$$
Proof of claim: Define
$$f_t(z) = \int _0 ^{t} g(z, \eta) d \eta$$
and
$$f(z) = \int _0 ^{\infty} g(z, \eta) d \eta$$
By integrating around closed paths, using Fubini's theorem and then Morera's theorem, show each $f_t$ is analytic in right half plane.
$f_t \to f$ uniformly on compact subsets of open right half plane, so $f$ is also analytic there.
Since $f_t \to f$ uniformly on compacta, get $\int _1 ^z f_t \to \int _1 ^z f$ as $t \to \infty$. So we get
\begin{align}
\int _1 ^z \int _0 ^{\infty} g(\zeta, \eta) \ d \eta \ d \zeta
  &= \int _1 ^z f(\zeta, \eta) \ d \zeta \\
  &= \lim _{t\to \infty} \int _1 ^z f_t(\zeta, \eta) \ d \zeta \\
  &= \lim _{t\to \infty} \int _1 ^z \int _0 ^{t} g(\zeta, \eta) \ d \eta \ d \zeta \\
  &= \lim _{t\to \infty} \int _0 ^{t}  \int _1 ^z g(\zeta, \eta) \ d \zeta \ d \eta  \\
  &= \lim _{t\to \infty} \int _0 ^{t}  G(z, \eta) \ d \eta \\
  &= \int _0 ^{\infty}  G(z, \eta) \ d \eta
\end{align}
