How Can You Approximate f(m) in Integration Using Bounds and Taylor Expansion?

[Togli]

It is basically an integration that cannot be properly solved, so I look for an approximation or maximum&minimum bounds of f1(m) and f2(m) such that f1(m) < f(m) < f2(m).

Here is the integral: f(m) = Integrate [ exp( -0.5* (sin(x)^2) *m) dx, x=0:pi/2] where m is a variable.

When I take sinx ~ x when x is close to 0, it becomes Integrate[exp(-0.5x^2)], but the solution to that is "error function" having a closed form.

So can anyone suggest a "good" approximation (or minimum/maximum bounds) to f(m)? Thank you.

 

[hunt_mat]

Two very simple approximations spring to mind, on the interval $$[0,\pi /2]$$, you know that $$0\leqslant\sin x\leqslant 1$$, but you need to couple this to the fact that $$\exp (-x)$$ is a strictly decreasing function, this will give you bounds but I don't know if they're the ones you want.

 

[hunt_mat]

Or failing that, you could calculate f'(m) on the interval of interest and treat it like a normal calculus problem.

 

[Togli]

Thank you! You are right, it is easy to show that

pi/2 * exp(-0.5*m) < f(m) < pi/2

by simply inserting sin(x) for the boundaries. However, these are a bit relaxed bounds, I prefer a real approximation.

I did not understand your statement about f'(m); if I take the derivative as you say

d/dm f(m) = f'(m) = Integrate [ -0.5* (sin(x)^2) * exp( -0.5* (sin(x)^2) *m) dx, x=0:pi/2]

I am not sure how I proceed then?

 

[hunt_mat]

It depends what you mean by real approximation I suppose, you are allowed to do that with the boundaries because on [0,pi/2], both sin^2(x) and e^-x are monotonic. It depends on what you want the approximation for. For small values of m, the approximation is pretty good. For large values of m, you have a well known expansion which works and you can integrate that to get an approximation, I don't think however that m has to be all that large. for the value to be fairly close to zero.

 

[Togli]

In fact, the range of m in my problem is : 0 < m < 2*pi

And yes, for m ~ 0, the approximation exp(-0.5*m) ~ 1

but for large m ~ 2pi, exp(-0.5*2*pi) = 0.04, hence it substantially deviates from 1.

I am not sure whether Taylor's expansion exp(-x)=$$\sum$$ (-x^n) / n! would work in this case however, as you suggest. I'll give a try though.

 

[hunt_mat]

I would do a Taylors expansion around m-pi/2.