I came across to the following question: let $f$ and $g$ be nice enough functions. I have
$$I(\theta)\triangleq \int_a^b f(x)g(\theta x) dx.$$
Is it possible to find $\theta$ such that $I(\theta)=0$ using analytical methods or posing a problem that $\theta$ has to solve? I do not know why, but I see this related to an (deterministic) optimal control problem, but I do not see how to adapt it.
Comment: $g$ is actually an exponential function, should it help.
I tried of course differentiating outside, inside, etc. with little success. Thanks a lot for any hints or help!
The fact that $g$ is the exponential function seems to be important to me. First, it implies that $g$ is always positive so $f$ has to be positive and negative as well, otherwise there is no solution. Unfortunately, the "nice enough" condition is quite weigh, so is my guessing below.
You can separate the positive and negative parts of $f$ leading to $$ \int_a^b f(x) I\{f(x) > 0\} e^{\theta x} dx = -\int_a^b f(x) I\{f(x) < 0\} e^{\theta x} dx $$ where $I\{\cdot\}$ denotes the 0/1-indicator function, which might allow you to exploit some structure of $f$ and simplify the problem.
You can also try applying integration by parts until the derivative becomes "manageable". For example: $$ 0 = \int_a^b f(x) e^{\theta x} dx = \int_a^b f(x) \big(e^{\theta x} / \theta\big)' dx \\ = \frac1\theta\Big(\big[f(x)e^{\theta x}\big]_a^b - \int_a^b f'(x) e^{\theta x} dx\Big) \\ = \frac1\theta\big[f(x) e^{\theta x}\big]_a^b - \frac1{\theta^2}\big[f'(x)e^{\theta x}\big]_a^b + \frac1{\theta^2} \int_a^b f''(x) e^{\theta x} dx \\ = \ldots $$ and continue this until the derivative of $f$ vanishes (if it does), which gives an equality you can solve by a root finding numerical method. For example if $f''(x) = 0$ for all $x \in [a,b]$, like when $f$ is linear or relu, then you get $$ 0 = \big[\big(\theta f(x) - f'(x)\big) e^{\theta x}\big]_a^b = \big(\theta f(b) - f'(b)\big) e^{\theta b} - \big(\theta f(a) - f'(a)\big) e^{\theta a} $$ which you can solve for $\theta$ numerically.