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?
