Deriving a function from within an integral with a known solution

[Phylosopher]

Hello,I am not sure if these types of problems are Intermediate or advanced. I am not sure too whether they have a certain name or not.

I have a function inside a definite integral. The solution of this definite integral is known. What is the function that satisfy the known solution.

In mathematical terms: $\int_{-∞}^{∞} f(x) g(x) \, dx = h$

f(x) is the unknown function while g(x) is a known function and h is the known solution of the integral. The actual problem that I have is way harder than demonstrated, but the basic idea is the same. How can I find f(x) that satisfy the solution h.

(If its indefinite integral of course) If h ∝ x. I would have an indefinite integral and then differentiate both sides of the equation and finally have $f(x)=\frac {1} {g(x)}\frac {dh(x)} {dx}$. I think this is the right approach if h∝x with indefinite integrals. But its not, its independent of x.

 

[kuruman]

I don't believe it can be done with a definite integral in which case $h$ is just a number. I don't have a general proof, just a counterexample that $f(x)$ is not unique. Suppose it is given that $g(x)=\frac{1}{1+x^2}$ and $h=\pi/6.$
Then $f(x) = \frac{1}{x^2+4}$ and $f(x) = \frac{4}{9(x^2+1)^2}$ work equally well to give $\int_{-∞}^{∞} f(x) \frac{1}{1+x^2} \, dx = \pi/6.$

 

[Phylosopher]

I don't believe it can be done with a definite integral in which case $h$ is just a number. I don't have a general proof, just a counterexample that $f(x)$ is not unique. Suppose it is given that $g(x)=\frac{1}{1+x^2}$ and $h=\pi/6.$
Then $f(x) = \frac{1}{x^2+4}$ and $f(x) = \frac{4}{9(x^2+1)^2}$ work equally well to give $\int_{-∞}^{∞} f(x) \frac{1}{1+x^2} \, dx = \pi/6.$

I found a way to find f(x).

I basically took H(∞)-H(-∞)=h and then assumed H(-∞)=0, and so H(∞)=h which means H(x) must be a function with lim→∞ =h while lim→-∞ =0.

One can use any function that satisfies the two conditions and define H(x)=z(x)*h. z(x) satisfies the conditions lim→∞ =h while lim→-∞ =0. After that, we can reduce the definite integral to indefinite one since we know what is H(x) and use the method wrote in the earlier post:

$f(x)=\frac {1} {g(x)}\frac {dh(x)} {dx}$
$f(x)=\frac {1} {g(x)} h \frac {dz(x)} {dx}$

But as you said, the solution for f(x) is not unique. Anyway, for my problem I actually don't need a unique solution. In the contrary, I actually want this discrepancy.

Thanks for your reply sir.

 

[Svein]

This is a classic Hilbert space problem. Essentially your problem is: Given $(f, g)=h$ find f. First, observe that f is not uniquely defined. Any q such that $(q, g)=0$ can be added to any solution and the resulting f will still be a solution.

Now take any orthonormal basis in your function space and express g in that basis: $g=\sum_{n=0}^{\infty}g_{n}e_{n}$. Assume that a solution exists, then it can be expressed in the same basis as $f=\sum_{n=0}^{\infty}f_{n}e_{n}$. This means that $(f,g)=\sum_{n=0}^{\infty}f_{n}\cdot g_{n}$. This transforms your problem into $\sum_{n=0}^{\infty}f_{n}\cdot g_{n} = h$ which obviously is underdetermined.

 

[Phylosopher]

This is a classic Hilbert space problem. Essentially your problem is: Given $(f, g)=h$ find f. First, observe that f is not uniquely defined. Any q such that $(q, g)=0$ can be added to any solution and the resulting f will still be a solution.

Now take any orthonormal basis in your function space and express g in that basis: $g=\sum_{n=0}^{\infty}g_{n}e_{n}$. Assume that a solution exists, then it can be expressed in the same basis as $g=\sum_{n=0}^{\infty}f_{n}e_{n}$. This means that $(f,g)=\sum_{n=0}^{\infty}f_{n}\cdot g_{n}$. This transforms your problem into $\sum_{n=0}^{\infty}f_{n}\cdot g_{n} = h$ which obviously is underdetermined.

That is some advanced mathematics for me. I will read more on the subject of Hilbert spaces.

Thank you sir.

 

[Mark44]

I found a way to find f(x).

I basically took H(∞)-H(-∞)=h and then assumed H(-∞)=0, and so H(∞)=h which means H(x) must be a function with lim→∞ =h while lim→-∞ =0.

Since neither ∞ nor -∞ are considered actual numbers, it's not legitimate to evaluate H at either of these or otherwise do arithmetic with either one. If H is some function of x, you can however take the limit: $\lim_{x \to \infty}H(x)$.

 

[jasonRF]

I am not sure if these types of problems are Intermediate or advanced. I am not sure too whether they have a certain name or not.

They are called integral equations. If you have learned linear algebra then you have the background needed to learn how to solve (exactly or approximately) some of them.