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.

  • $\begingroup$ Also I would like to say that the remark in the solution of Dr. MV answer is very nice. $\endgroup$ – user243301 Jun 23 '16 at 16:22
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    $\begingroup$ Also very thanks much to Cantarini by his claims. $\endgroup$ – user243301 Jun 23 '16 at 17:01

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:

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

  • $\begingroup$ Very thanks much, now I read your answer. $\endgroup$ – user243301 Jun 23 '16 at 16:57
  • $\begingroup$ @user243301: updated answer to deal also with your second question. $\endgroup$ – Jack D'Aurizio Jun 23 '16 at 17:06
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    $\begingroup$ There appears to be a sign error in the result in $(7)$. $\endgroup$ – Mark Viola Jun 23 '16 at 17:14
  • $\begingroup$ @Dr.MV: I noticed it. Now fixed, thanks ;) $\endgroup$ – Jack D'Aurizio Jun 23 '16 at 17:14
  • $\begingroup$ Very thanks much, tomorrow I study in details your answer and @Dr.MV Very thanks much both, now it is a great reference of nice examples of this theorem from your hands. $\endgroup$ – user243301 Jun 23 '16 at 19:21


The analysis succeeding Question $1$ in the OP is fine.

For $b=\frac12+iB$, $B>0$, we have

$$\begin{align} \log\left(\frac{b}{\log(2)}\right)&=\log\left(\frac{|b|}{\log(2)}\right)+i\arg(b)\\\\ &=\log\left(\frac{\sqrt{B^2+1/4}}{\log(2)}\right)+i\arctan\left(2B\right) \tag 1 \end{align}$$

Since $B>0$, $0<\arctan\left(2B\right)<\pi/2$.


Replacing $B>0$ with $t>0$ in the result given in $(1)$ reveals

$$\begin{align} \Phi(t)&=\int_0^\infty \frac{e^{-\log(2) x}-e^{-b(t)\,x}}{x}\,dx\\\\ &=\log\left(\frac{\sqrt{t^2+1/4}}{\log(2)}\right)+i\arctan\left(2t\right) \end{align}$$

Therefore, we find that

$$\begin{align} \Phi'(t)&=\frac{t}{t^2+1/4}+i \frac{1/2}{t^2+1/4}\\\\ &=\frac{i}{\frac12 +it}\\\\ &=\frac{b'(t)}{b(t)} \end{align}$$

If we can legitimize differentiating under the integral sign, then we can write

$$\begin{align} \Phi'(t)&=\int_0^\infty \frac{\partial }{\partial t}\left(\frac{e^{-\log(2)x}-e^{-b(t)\,x}}{x}\right)\,dx\\\\ &=\int_0^\infty b'(t)e^{-b(t)\,x}\,dx\\\\ &=\frac{b'(t)}{b(t)} \end{align}$$

as expected!

  • $\begingroup$ Very thanks much also for this new contribution in this kind of statements, now I read your answer. $\endgroup$ – user243301 Jun 23 '16 at 16:58
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$ – Mark Viola Jun 23 '16 at 19:26
  • $\begingroup$ You and other users with experience are the heart of this site, truly thanks. $\endgroup$ – user243301 Jun 23 '16 at 19:27
  • $\begingroup$ You're really quite welcome. I enjoy helping others. -Mark $\endgroup$ – Mark Viola Jun 23 '16 at 19:40

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