If $k(x)=\frac{i(x)}{j(x)}$, is there a way to re-write $\frac{\int i(x)f(x)dx}{\int j(x)f(x)dx}$ without $i(x)$ and $j(x)$? If I have the following equation;
$$T=\frac{\int i(x)f(x)dx}{\int j(x)f(x)dx}$$
and $i(x)$ and $j(x)$ are related to each other as follows;
$$k(x)=\frac{i(x)}{j(x)}$$
then can I rewrite the first formula with the use of $k(x)$ and not with the use of $i(x)$ and $j(x)$, and how would I do that?
In order to give some context, I'm trying to calculate the transmission  $T$, of a glass sample from the spectral transmission of glass. Where $i(x)$ is the spectrum of some light source transmitted through glass, and $j(x)$ is the unaltered spectrum of the light source, then $k(x)$ is the spectral transmission function. In the first formula $f(x)$ is the luminous efficiency function (i.e. a weighting for the visual spectrum as perceived by the human eye). Intuitively I think there should be a way to express T with the use of the transmission function, i.e. without the use of the spectra used to determine the transmission function, but I don't know how.
 A: What you ask is (in general) impossible (see the comments and below), but it's possible to give a new "interpretation" to the integral.
I assume $j(x)>0$ and $f(x)>0$ and
$$\int j(x) f(x) =N \, ,$$
with $N>0$ and finite (I do not specify the integration domain, but the integrals are definite integrals). Let's rewrite your expression:
$$
T=\frac{\int i(x)f(x)dx}{\int j(x)f(x)dx} = 
\frac{\int k(x) j(x)f(x)dx}{\int j(x)f(x)dx} = 
\int k(x) p(x)dx = \bar{k}
$$
where $\bar{k}$ is the average of the function $k$ calculated via the PDF $p(x)$ defined as
$$
p(x) = \frac{1}{N} j(x) f(x)
$$
Here $N$ is the normalization factor, so that $p(x)$ is normalized:
$$
\int p(y) dy = 1
$$
This is not a mathematical solution to what you are asking, it is just a way to show that you can interpret $T$ as an average of your ratio $k(x)=i(x)/j(x)$.
Why it is impossible: an integral is, loosely speaking, a sum. If you have a fraction between numbers like
$$
\frac{i_1 f_1}{j_1 f_1} = k_1\frac{f_1}{f_1}
$$
everything works fine... but if you have
$$
\frac{i_1 f_1 + i_2 f_2}{j_1 f_1 + j_2 f_2} =
\frac{k_1 j_1 f_1 + k_2 j_2 f_2}{j_1 f_1 + j_2 f_2} 
$$
and there is nothing you can do to eliminate both the $i$s and the $j$s. Note, however, that again you ended up with a weighted arithmetic mean for the $k$s.
Practical approximation: assume that you plot $p(x) = j(x) f(x)$ and that it turns out that $p$ is sharply peaked around a value $x_p$, where $p(x_p)$ is maximum. Now plot $k(x)$: if around $x_p$ you find that $k(x)$ is fairly constant (i.e. $k(x)\approx k(x_p)$ for $x$ close to $x_p$), then
$$
T = \bar{k} \approx  k(x_p) \, .
$$
You can make this a bit more rigorous by Taylor expanding around $x_p$, and also estimate the error associated to the approximation.
