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I was wondering if a rational function (quotient between polynomials) $f(x,a)$ exists such that $\int_0^1 f(x,a) dx = \frac{\ln(a)}{a}.$

For example, I could find, while playing with wolfram, a not rational $f(x,a) =\frac{ \ln(ax)+1}{a}$ that integrated on $x$ from $0$ to $1$ yelds $\frac{\ln a}{a}$. But the function not being rational does not really help me. How to proceed on a problem like this?

Pinteco
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  • This is equivalent to finding an $F(x, a)$ such that $F(1,a) - F(0,a) = \ln(a)/a$. Where F' is rational. – Kaynex Jul 15 '18 at 01:46
  • @Kaynex What does "$F'$ rational" mean in this multivariable case? $\frac{\partial F}{\partial x}$? – Randall Jul 15 '18 at 02:08

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Doesn't $f(x,a) = \frac{a-1}{a} \ \frac{1}{(a-1)x+1}$ work? Assume $a >0$ throughout (otherwise your log term makes no sense).

Let $g(x)= (a-1)x+1$. This is clearly monotone increasing or decreasing as long as $a \neq 1$. As $g(0)=1$ and $g(1) = a$, $g(x)$ is never $0$ on $[0,1]$ by the IVT. If $a=1$ then $g(x) \equiv 1$ and again $g$ is never $0$. So, either way, $f(x,a) = \frac{a-1}{a} \ \frac{1}{(a-1)x+1}$ has no discontinuities w/r/t $x$ over $[0,1]$.

You can now check---unless I did something really wrong---that $$\frac{a-1}{a} \int_0^1 \frac{1}{(a-1)x+1}\ dx = \frac{\ln a}{a}.$$

To answer your question how to proceed on a problem like this?, I don't know. I just experimented and made adjustments as I went, starting with $\int_1^a \frac{1}{x} \ dx = \ln (a)$, then dividing by $a$, then scooting the interval down to $[0,1]$, etc. Nothing clever.

Randall
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