integration - Frullani's theorem in complex context, other examples

One has as application of Frullani's theorem in complex context that

$$\int_0^\infty \frac{e^{-x\log 2}-e^{-xb}}{x}dx=\mathcal{Log} \left( \frac{1}{2\log 2}+i\frac{B}{\log 2} \right)$$
where I taken $a=\log 2+i\cdot 0$ and $b=\frac{1}{2}+iB$, that are complex numbers whose real parts are positive, here also I take $B>0$ as a fixed real number.

Question 1. Can you define and compute RHS of previous identity?

I would like to refresh the notion of complex logarith. My attempt for Question 1, is that I believe that I need write RHS should be

$$\frac{1}{2}\log \left( \frac{1}{4} +B^2\right) -\log\log 2+i\arctan 2B,$$
and saying that is required $-\pi\leq \arctan 2B<\pi$. Also I've tried write the previous imaginary part as $$\arcsin \left( \frac{2B}{\sqrt{4B^2+1}} \right).$$ Notice that it is possible take advantage from the computations of Dr.MV answer. Then since $0=\text{arg}(\log 2)\neq \text{arg}(\pm b)$ I did the following computations $$\log(|b/a|)=-\log\log 2+\frac{1}{2}(1/4+B^2),$$ $$(\Re(a\bar{b})-|a|^2)/\Im(a\bar{b})=(\log2-1/2)/B$$ and $$(|b|^2-\Re(a\bar{b}))/\Im(a\bar{b})=(1/4+B^2)(1/2-1/\log 2)/B.$$

Question 2. Denoting by $\Phi(t)$ previous deduction from Frullani's theorem with same (negative exponential function) $a=\log 2$ but now taking $b=\frac{1}{2}+it$, where $t>0$ is a real variable, can you provide us the statement that it is possible to write combining previous deduction and Lebesgue's Dominated Convergence theorem to compute the derivative $\Phi'(t)$? If you consider only hints to this second question, I believe that I can get the final statement with a clarification of the notation, as you see.

I am almost sure that I can justify the differentiation under the integral sign, but in my book the justification is explained in terms of $I(\cdot,t)$ and $I(x,\cdot)$, when $I(x,t)$ is denoting the function in the integrand, and now I am a few stuck with this notation. Also the computations of the derivative of the RHS could be tedious, without a good simplification in previous question. Thanks in advance all users.

Answer

With a change of variable, it is enough to study

$$J(z)=\int_{0}^{+\infty}\frac{e^{-x}-e^{-zx}}{x}\,dx \tag{1}$$
with $z=\sigma+it$ being a complex number in the right half-plane. The integral is convergent by Dirichlet's test, since $\frac{1}{x}$ is decreasing towards zero and $\int_{0}^{M}\left(e^{-x}-e^{-zx}\right)\,dx$ is bounded. We may study the real part and the imaginary part of $(1)$ as two separate integrals, computing them through the Laplace transform. Since $\mathcal{L}^{-1}\left(\frac{1}{x}\right)=1$,
$$\int_{0}^{+\infty}\frac{e^{-x}-e^{-\sigma x}\cos(tx)}{x}\,dx = \int_{0}^{+\infty}\left(\frac{1}{1+s}-\frac{s+\sigma}{t^2+(s+\sigma)^2}\right)\,ds\tag{2}$$
$$\int_{0}^{+\infty}\frac{-e^{-\sigma x}\sin(tx)}{x}\,dx = \int_{0}^{+\infty}\left(-\frac{t}{t^2+(s+\sigma)^2}\right)\,ds\tag{3}$$
and the integrand functions in the RHSs of $(2)$ and $(3)$ have elementary primitives, leading to:
$$\text{Re}\left(J(\sigma+it)\right) = \frac{1}{2}\log(t^2+\sigma^2),\tag{4}$$
$$\text{Im}\left(J(\sigma+it)\right) = \arctan\left(\frac{t}{\sigma}\right),\tag{5}$$
hence:

For any $z\in\mathbb{C}$ such that $\text{Re}(z)>0$,

$$J(z)=\int_{0}^{+\infty}\frac{e^{-x}-e^{-zx}}{x}\,dx = \color{red}{\text{Log}(z)} = \log(\|z\|)+ i\,\arctan\left(\frac{\text{Im}(z)}{\text{Re}(z)}\right). \tag{6}$$

If we assume now that $\sigma>0$ is fixed, we may study:

$$\frac{J(\sigma+it_1)-J(\sigma+it_2)}{t_1-t_2} = \int_{0}^{+\infty}\color{blue}{\left(\frac{e^{-it_1 x}-e^{-it_2 x}}{x(t_1-t_2)}\right)} e^{-\sigma x}\,dx$$
where the blue term is an entire function, bounded over the positive real axis. Since $\int_{0}^{+\infty}e^{-\sigma x}\,dx$ is finite, by the dominated convergence theorem ($t_2\to t_1$) we are allowed to state:

$$\frac{d}{dt}\,J(\sigma+it) = \int_{0}^{+\infty}i e^{-itx}e^{-\sigma x}\,dx =\frac{i}{\sigma+it}=\color{red}{\frac{i}{z}}.\tag{7}$$