This is a question that originates out of a comment I made to Is there an integral for $\frac{1}{\zeta(3)} $?

After some playing around I found that the underlying conjecture appears to be

$$\int_0^a f(x)\, dx \, . \int_a^\infty \frac{f(x)}{\left(\int f(x)\, dx\right)^2}\, dx =1\tag{1}$$

where $$\int_a^\infty \frac{f(x)}{\left(\int f(x)\, dx\right)^2}\, dx=\left[- \frac{1}{\int f(x)\, dx} \right]_a^\infty$$

(1) appears to hold true if both definite integrals exist and $a$ is a real number, $a>0$.

The formula can be easily proved for simple functions e.g. $f(x)=x^n$, where $n>-1$, then $$\int_0^a f(x)\, dx=\frac{a^{(1 + n)}}{(1 + n)}$$ and $$\int_a^\infty \frac{f(x)}{\left(\int f(x)\, dx\right)^2}\, dx =\frac{(1 + n)}{a^{(1 + n)}}$$

with $\frac{a^{(1 + n)}}{(1 + n)} . \frac{(1 + n)}{a^{(1 + n)}}=1$

I have found an example involving the Euler-Mascheroni Constant $\gamma$ where a constant of integration ($-i\pi$) is required.

$$\gamma = -\int_0^1 \log \left(\log \left(\frac{1}{x}\right)\right) \, dx$$ with the inverse being (with the help of Mathematica) $$\frac{1}{\gamma}=\int_1^{\infty } -\frac{\log \left(\log \left(\frac{1}{x}\right)\right)}{\left(-\text{Ei}\left(-\log \left(\frac{1}{x}\right)\right)+x \log \left(\log \left(\frac{1}{x}\right)\right)-i \pi \right)^2} \, dx$$

where $$\int \log \left(\log \left(\frac{1}{x}\right)\right) \, dx= x \log \left(\log \left(\frac{1}{x}\right)\right)-\text{Ei}\left(-\log \left(\frac{1}{x}\right)\right)$$

and $Ei(x)$ is the exponential integral.

(Incidentally integrating the fractional harmonic number to find an integral for $\frac{1}{\gamma}$ does not need a constant of integration to work.)

Is it possible to develop a more general proof of formula (1) and discover the exact conditions under which it will fail to work?

  • $\begingroup$ Your second displayed equation: That should be $$\left[ \frac{-1}{\int f(x)\, dx} \right]_a^\infty$$ on the right. $\endgroup$ – zhw. Nov 9 '18 at 17:51
  • $\begingroup$ @zhw : Thanks fixed that. $\endgroup$ – James Arathoon Nov 9 '18 at 20:20

Suppose $f$ is continuous on $[0,\infty).$ Assume that $F(x)=\int_0^x f(t)\,dt >0$ for all $x>0.$ Then

$$\int_0^a f(x)\, dx \, \int_a^\infty \frac{f(x)}{F(x)^2}\, dx =1\tag{1}\,\, \text {for all } a>0$$

iff $\int_0^\infty f(x)\,dx = \infty.$

Proof: This is quite simple: Take $b>a$. Then the left side of $(1),$ with $b$ in place of $\infty,$ equals

$$F(a)\cdot \left (\frac{1}{F(a)} - \frac{1}{F(b)}\right ).$$

The limit of this as $b\to \infty$ equals $1$ iff $F(b)\to \infty,$ which is the same as saying $\int_0^\infty f = \infty.$


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