Firs, I am struggling with a problem in physics, but the issue lies with the math, so I think it fits here.
Now, let me explain, I have some normal/Gaussian distributions (of density but that's not relevant to the math) with known $\sigma$ and known height $a$ like: sketch1 they can be described with : $f(x) = a\exp(-\frac{x}{2\sigma^2})$
Integrating $\int f(x) dx = a\sqrt{2\pi\sigma^2}$
What this means for me is that for known area $A$ under the Gaussian (my measurement is proportional to) and known $\sigma$ I can calculate the height $a$. This is still fine, now my issue begins, the distribution is evolving in 2nd dimension (let's call it $z$) this evolution is an exponential decay, like so sketch2 The decay can be described with $\exp(-z*c)$ with $c$ some positive constant.
So, I think the function can be described as $f(x,z) = a\exp(-\frac{x}{2\sigma^2})\exp(-z*c)$, now when I "measure" the area $A$ under such curve with constant $z$ I can easily get the height $a$ for given z with $a = \sqrt{2\pi\sigma^2}/A$. The problem is that the "measurement" does not integrate along constant $z$ but rather at an angle $\alpha$ like: sketch3 red lines are the integration paths.
Now, I don't know anymore how to relate the $A$ under such a curve to the central height $a(x = 0, z = z)$. The angle $\alpha$ between the $z$ axis and the integration path is known, I would like to express the result in terms of the $\alpha$
Any help would be appricieted.
If I understand correctly, you want to integrate
$$ f(x,z)=a \exp \left(-\frac{x^2}{2 \sigma}-cz \right) $$
Over some given path $z(x)=\alpha + \beta x$. Along the path, substitute $z(x)$ into $f$, and note the line element is $dl=dx \sqrt{1+\beta^2}$. All together, we have a Gaussian integral
$$ \int\limits_{-\infty}^\infty dx \ \sqrt{1+\beta^2} a \exp\left(-\frac{x^2}{2 \sigma}-c \alpha-c \beta x \right)= a \sqrt{2 \pi \sigma (1+\beta^2)} \exp \left(-c \alpha +\frac{c^2 \beta^2 \sigma}{2} \right) $$
Note my $\alpha$ is not the path angle, but you should be able to relate $\alpha$ and $\beta$ to it.