# Simplifying a ratio of integrals

I'm doing theoretical economics, and after a few computation I end up with a ratio of two integrals in one of my model. The function being integrated is the same, but the bounds are different. It goes as the following

$$\frac{\int_{-\infty}^{x}f(t)dt}{\int_{-\infty}^{x(1 + a)}f(t)dt}=\frac{A}{B}$$

with $\forall t$ $f(t)>0$, and either both $x$ and $a$ positive or both of them negative. I don't think it should matter, but in my model the function $f$ is actually the density of a normal random variable.

Any idea if I can simplifies this quantity? In particular I'm interested in expressing x as a function of the remaining. This may not be feasible, but if I could at least simplify this ratio it'd help me for some proofs.

Any help appreciated, Thanks.

• Can we assume $x,a>0$?
– mlc
Mar 22, 2017 at 12:44
• @mlc actually in my model both are negative, but I guess you could treat both of them as positive, the point being that the whole ratio is lower than 1 (because $\forall t$ $f(t)>0$, I edited the question with those precisions). Mar 22, 2017 at 12:54

Your denominator is $$\int_{-\infty}^xf(t)dt+\int_x^{ax}f(t)dt$$ and has no particular relation to the numerator (except that it is larger). The fraction can take any value in $(0,1)$ and no simplification is possible.
If $f$ is a normal pdf, the integrals can be expressed in terms of the error function, but that doesn't yield a simple expression.