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

  • Thread starter buggykong
  • Start date
  • Tags
    Integration
In summary, the conversation discusses solving for A, which is a function of x, in the given integral equation with a constant b. Possible solutions include A(x) = \frac{\delta(x-x_0)}{\sin(x_0)\exp(-b\sin^2(x_0))}, A(x) = \frac{2}{\pi^2}\frac{\exp(b\sin^2(x))}{\mbox{sinc}(x)} with a slight problem at x=pi, and A(x) = \frac{1}{\pi} \frac{\exp(b\sin^2(x))}{\sin(x)}. It is also suggested to introduce a parameter y and solve for it using Fourier series/
  • #1
buggykong
4
0
I have an integral:

[tex]{\int}[/tex][tex]^{\pi}_{0}[/tex]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.
 
Last edited:
Physics news on Phys.org
  • #2
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
[tex]A(x) = \frac{\delta(x-x_0)}{\sin(x_0)\exp(-b\sin^2(x_0))}[/tex]
for any x_0 such that [itex]0 < x_0 < 2\pi[/itex]. I'm not sure if there are any non-trivial solutions.

I could also try
[tex]A(x) = \frac{2}{\pi^2}\frac{\exp(b\sin^2(x))}{\mbox{sinc}(x)}[/tex]
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

[tex]A(x) = \frac{1}{\pi} \frac{\exp(b\sin^2(x))}{\sin(x)}[/tex]
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

[tex]f(y) = \int_{0}^\pi dx A(x) \sin(x-y) \exp(-b\sin^2(x-y)).[/tex]
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.
 
Last edited:

1. What is the purpose of solving this integral?

The purpose of solving this integral is to find the value of A that makes the equation equal to 1. This can help us understand the behavior of the function and make predictions based on its properties.

2. What is the significance of the limits of integration in this integral?

The limits of integration, 0 and π, determine the range of values for which the integral is being evaluated. In this case, the integral is being evaluated over one full period of the sine function, from 0 to π.

3. What is the meaning of the symbols used in this integral?

The symbol A represents a constant, while b represents a coefficient. The integral sign, ∫, indicates that we are finding the area under the curve of the function. The sine function, sin(x), represents the relationship between the angle of a right triangle and the ratio of its opposite side to its hypotenuse. The exponential function, exp(-bsin(x)^2), represents a decaying or growing quantity over time.

4. How can this integral be solved?

This integral can be solved using various techniques, such as integration by parts, substitution, or trigonometric identities. The specific method used will depend on the complexity of the function and the skills of the person solving it.

5. What are the potential applications of solving this integral?

Solving this integral can have various applications in different fields, such as physics, engineering, and economics. It can help in understanding the behavior of a system or in predicting future outcomes based on the given function. It can also be used to find the maximum or minimum values of a function, which can be useful in optimization problems.

Similar threads

Replies
3
Views
1K
Replies
3
Views
937
Replies
2
Views
180
Replies
4
Views
192
Replies
3
Views
1K
Replies
2
Views
831
  • Calculus
Replies
6
Views
1K
Replies
2
Views
1K
Back
Top