# Understanding when the Multiplication of two Definite Integrals gives Unity

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?

• Your second displayed equation: That should be $$\left[ \frac{-1}{\int f(x)\, dx} \right]_a^\infty$$ on the right. – zhw. Nov 9 '18 at 17:51
• @zhw : Thanks fixed that. – 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 $$\int_0^\infty f = \infty.$$