Solving Integral for A: {\int}^{\pi}_{0}A(x)sin(x) exp(-bsin(x)^{2})dx=1

[buggykong]

I have an integral:

$${\int}$$$$^{\pi}_{0}$$A(x)sin(x) exp(-bsin(x)^{2})dx=1

I want to solve for A which is a function of x, how do I go about doing it? b is a constant.

 

[Mute]

Well, what have you tried? This is a Fredholm integral equation. Have you looked those up? It might help you determine if there's a well-defined solution or not.

The obvious solution is
$$A(x) = \frac{\delta(x-x_0)}{\sin(x_0)\exp(-b\sin^2(x_0))}$$
for any x_0 such that $0 < x_0 < 2\pi$. I'm not sure if there are any non-trivial solutions.

I could also try
$$A(x) = \frac{2}{\pi^2}\frac{\exp(b\sin^2(x))}{\mbox{sinc}(x)}$$
where sinc(x) = sin(x)/x, x not 0 and 1 for x = 0. Slight problem at x = pi, but you can just remove that single point from the domain of integration and everything is fine.

Or I could try

$$A(x) = \frac{1}{\pi} \frac{\exp(b\sin^2(x))}{\sin(x)}$$
where x = 0 and pi are removed from the range of integration.

I think you need some boundary conditions for A(x) otherwise one could probably concoct as many solutions as desired for it.

To connect to the theory of fredholm equations of the first kind, you might try introducing another parameter into the equation, y, such that your equation becomes

$$f(y) = \int_{0}^\pi dx A(x) \sin(x-y) \exp(-b\sin^2(x-y)).$$
Picking f(y) such that f(0) = 1 causes this to reduce to your equation when y = 0. You might be able to solve this using Fourier series/transform methods.

However, note that f(y) is arbitrary aside from f(0) = 1, and the way I inserted y into the integrand was also arbitrary. I chose it such that it looked similar to the convolution form, which is a known method of solving the problem. As a result, unless by some miracle the solution turns out to be independent of the choices of f(y) and how y is placed in the integral, there's no unique solution to this problem.