difficult integral

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# difficult integral

Submitted by slashfan on Thu, 12/08/2005 - 21:23

Forums:

hello,

is is possible to show that integral of sinx.x^(-1/4) on [0,pi] is

less than pi^(3/4) and is it possible to find f that is square integrable such that (integral of (f-sinx)^2) <= 1 and

(integral of (f-cosx)^2) <= 4/9 on [0,pi]

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## Re: difficult integral

Part 1:

To begin, you might want to do the integral numerically and see how thevalue compares with the value you would like it to be. If so, you could look for some simple-to-integrate function like a piecewise linear function which is larger than your function but whose integral is less than your bound.

Part 2:

Think of this in terms of function space. sin and cos are two points in function space. Find the distance betwen these two points. You are looking to see if it is possible to find a point which is at most a certain distance from one point and some other distance from another point. This will possible if andonly if the triangle inequality is satisfied.

## Re: difficult integral

Part I:

integral of (sinx.x^(-1/4)) on [0,pi]=integral of (sinx.x^(-1/4)) on [0,pi/2] + integral of (sinx.x^(-1/4)) on [pi/2,pi].

but:

integral of (sinx.x^(-1/4)) on [0,pi/2] <= integral of (x.x^(-1/4)) on [0,pi/2] = (4/7)(pi/2)^(7/4),

and:

integral of (sinx.x^(-1/4)) on [pi/2,pi] <= integral of (sinx.(pi/2)^(-1/4)) on [pi/2,pi] = (pi/2)^(-1/4).

so:

integral of (sinx.x^(-1/4)) on [0,pi] <= (4/7)(pi/2)^(7/4) + (pi/2)^(-1/4) < (pi)^(3/4).

Part II:

nothing to add to what rspuzio mentioned. just note that because on [0,pi]: ||sinx-cosx||2 = (pi)^(1/2) > 1 + 2/3, the triangle inequality does not hold and so you can't find such a function.

good luck