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This question is related to this other one, but this other question does not give the answer to this one.

Find all function $f\colon \mathbb R\to \mathbb R\in C^\infty(]0,+\infty[)$ such that $f'=0$ has at most one zero, and $$\int_0^1 f(x)\mathrm d x=\sum_{n\geqslant 1} f(n).$$

We can prove that $f(x)=\frac 1{x^x}$ will work:

  • $f'(x)=-x^{-x}(1+\log x)$ has only one zero;

  • using $$x^x=e^{x\log x}=\sum_{n\in \mathbb{N}} \frac{(x\log x)^n}{n!}$$

and an inversion $\sum-\int$ (and the value of $\int_0^1 (x\log x)^n\mathrm d x$);

we can show that $$\int_0^1 \frac 1{x^x}\mathrm d x=\sum_{n\geqslant 1} \frac 1{n^n}.$$

Is there any other function $f$ satisfying those conditions?

E. Joseph
  • 14,843

1 Answers1

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There are millions of them.

Take any function $g(x)$ such that $g'(x)e^x$ is monotonic* and such that the integral and the sum converge. Then let

$$f(x):=g(x)+ae^{-x}$$ and write

$$\int_0^1 f(x)\,dx=\int_0^1 g(x)\,dx+a(1-e^{-1})=I+a(1-e^{-1}),$$ $$\sum_{n=1}^\infty f(n)=\sum_{n=1}^\infty g(n)+\frac{ae}{1-e^{-1}}=S+\frac{ae}{1-e^{-1}},$$ which you can solve for $a$ after equating.

*This requirement ensures a single root of $f'(x)$.

  • About the same, but maybe a little easier: $f(x) = e^{-ax}$ for a suitable $a>0$ will work. This one has $f'(x) < 0$ everywhere. – zhw. Jun 07 '17 at 13:24
  • @zhw: in what way is it easier ? [And by the way, this $f$ works for a few $a$ only.] –  Jun 07 '17 at 13:33