Working on a problem and have reduced it to finding $ f ( x ) $ such that $$ \int _ 0 ^ 1 f ( x ) c ^ x \ \mathrm d x = c \text . $$ However, here's where it gets interesting. The function should satisfy this equation for all values of $ c $ in $ ( 0 , 1 ) $.
How would I solve this? I thought about Mellin transform, but I don't know if that would work.
You can show that there is no such integrable function $ f : [ 0 , 1 ] \to \mathbb R $. If you want to consider the case of generalized functions (like measures or distributions), then you can see that $ f $ must be Dirac's delta concentrated at $ 1 $.
One way of looking at the problem is using Laplace transform. Define $ s = - \log c $, and the given equation becomes $$ \int _ 0 ^ 1 f ( x ) \exp ( - s x ) \ \mathrm d x = \exp ( - s ) $$ for all $ s \in ( 0 , + \infty ) $. Extending the domain of $ f $ to be $ [ 0 , + \infty ) $ with $ f ( x ) = 0 $ for $ x > 1 $ (so that we deal with Laplace transform in the more traditional sense), the equation is in fact equivalent to $$ \mathcal L \{ f \} = \mathcal L \{ g \} \text , $$ where $ g ( x ) = \delta ( x - 1 ) $. So $ f $ must be equal to $ g $ (in the sense of almost everywhere extended to the realm of generalized functions), as Laplace transform is invertible.
You can also try to solve the problem directly like this: $$ \int _ 0 ^ 1 f ( x ) \exp ( - s x ) \ \mathrm d x = \int _ 0 ^ 1 f ( x ) \sum _ { n = 0 } ^ \infty \frac { ( - 1 ) ^ n x ^ n s ^ n } { n ! } \ \mathrm d x = \sum _ { n = 0 } ^ \infty \left( \frac { ( - 1 ) ^ n } { n ! } \int _ 0 ^ 1 f ( x ) x ^ n \ \mathrm d x \right) s ^ n $$ $$ \exp ( - s ) = \sum _ { n = 0 } ^ \infty \frac { ( - 1 ) ^ n } { n ! } s ^ n $$ Therefore, for every nonnegative integer $ n $, the coefficients of $ s ^ n $ in the two above series must be equal; i.e. we must have $ \int _ 0 ^ 1 f ( x ) x ^ n \ \mathrm d x = 1 $. This shows that in the sense of integrable functions, there is no solution, as $$ 1 = \lim _ { n \to \infty } \int _ 0 ^ 1 f ( x ) x ^ n \ \mathrm d x = \int _ 0 ^ 1 f ( x ) \left( \lim _ { n \to \infty } x ^ n \right) \ \mathrm d x = 0 \text , $$ which is a contradiction. In the case of generalized functions, the above limit under the integral will be equal to $ g ( x ) $ mentioned before. The justification of validity of some of the above steps (like interchanging sum and integral and interchanging limit and integral) will be different in cases of integrable functions and generalized functions, but are rather standard and straightforward to check.