I've deduced and I know how justify an example of a function $f:[0,\infty)\to\mathbb{R}$ satisfying $$\int_0^\infty f(x)\log xdx=1.$$

Now I state question, and after I write hints to get my example.

Question. Imagine that a friend ask me about functions $$f:[0,\infty)\to\mathbb{R}$$ and its more relevant properties, that satisfy $$\int_0^\infty f(x)\log xdx=1.$$ Because it seems too broad, that is I believe that is difficut to determine all functions, impossible to state all relevant statements related to this kind of functions, I am asking here:

What should be the first step/s to investigate this question, that is what are examples of functions that satisfies our condition, and what is the obvious property that satisfy as consequence of it? Many thanks.

You are able to combine with techniques of real and complex analysis, integration methods ... to show me what should be the strategy and/or hints to study this problem.

The example was to combine the closed-form for $\int_0^\infty\frac{\log x}{e^{nx}}dx$, the Prime Number Theorem, and $(11)$ of this MathWorld's article for the Möbius function to deduce $$f(x)=\sum_{n=1}^\infty\frac{\mu(n)}{e^{nx}}$$ is a function that satisfies it, because is a consequence of Fubini's theorem:

$$\int_0^{\infty}\sum_{n=1}^{\infty}\left|f_n(x)\right|dx\leq\int_0^\infty\frac{x}{e^x-1}dx=\zeta(2)<\infty,$$ with $f_n(x)=\mu(n)\log(x)e^{-nx}$.

  • $\begingroup$ Your answer can to be from Riemann's approach for the integral, or well Lebesgue integration. My example is from the second. Many thanks all users. $\endgroup$ – user243301 Feb 14 '17 at 12:59
  • $\begingroup$ @mickep I accept your answer, because you provide me several examples and state the characterization and conditon of such functions. Any case, if in next future you or other users want do more contribution, feel free. Many thanks. $\endgroup$ – user243301 Feb 14 '17 at 21:07

I don't see the purpose of this question. For example, it is known that $$ \int_0^{+\infty}e^{-x}\log x\,dx=-\gamma, $$ so $$ f(x)=-\frac{1}{\gamma}e^{-x} $$ is one example. Since $$ \int_0^{+\infty}e^{-x^2}\log x\,dx=-\frac{(\gamma+2\sqrt{2})\sqrt{\pi}}{4}, $$ another example is $$ f(x)=-\frac{4}{(\gamma+2\sqrt{2})\sqrt{\pi}}e^{-x^2}. $$

In general, and in a similar manner, you can take any function $\tilde f$ so that the integral $$ I=\int_0^{+\infty}\tilde f(x)\log x\,dx $$ is convergent (and non-zero), and then define $f$ by $$ f(x)=\frac{1}{I}\tilde f(x). $$

I give you a final example, $$ f(x)=-\frac{8\sqrt{2}}{\pi^2(1+x^4)}. $$

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    $\begingroup$ My main purpose is explore questions about Möbius function and arithmetic. Well then I was asking about general function satisfying previous requeriment to know more about the situation. Many thanks for your attention and help, as companion of my example and a present for you, here you can state the following: Take $m_k=\frac{\mu(k)}{k}$ for $1\leq k\leq n-1$, and $m_n=\sum_{k=n}^\infty\frac{\mu(k)}{k}$ and with $a_k=\frac{1}{k}$ $\forall k\geq 1$, $f(x)=e^{-x}$ in FRU2 in page 3 of Jameson, The Frullani Integral $\endgroup$ – user243301 Feb 14 '17 at 15:53

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