Estimating the integral of a decreasing trigonometric function

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TehAdzMan
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Hi,

So this is part of an assignment for my numerical analysis class.

The integral is this:

[tex] <br /> \int_0^{\infty} e^{-x} \cos^2 (x^2) dx<br /> [/tex]

We are instructed to evaluate the integral from 0 to some large A using numerical methods (which I'm fine with), and then estimate the tail, ie the integral from A to infinity.

Basically we need to come up with some method to estimate and bound the remainder in the tail.
My idea was to substitute [tex]u = x^2[/tex] which gives [tex]\int_A^{\infty} \frac{e^{- \sqrt{u}} \cos^2(u)}{\sqrt{u}} du[/tex]
and then I guess take the maximum of the cos function as 1 and just remove it, and use the remaining function, which I think would be quite easy to integrate and would give a maximum bound.
Any thoughts greatly appreciated.

Regards,
Adam
 
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TehAdzMan said:
Hi,

So this is part of an assignment for my numerical analysis class.

The integral is this:

[tex] <br /> \int_0^{\infty} e^{-x} \cos^2 (x^2) dx<br /> [/tex]

We are instructed to evaluate the integral from 0 to some large A using numerical methods (which I'm fine with), and then estimate the tail, ie the integral from A to infinity.

Basically we need to come up with some method to estimate and bound the remainder in the tail.
My idea was to substitute [tex]u = x^2[/tex] which gives [tex]\int_A^{\infty} \frac{e^{- \sqrt{u}} \cos^2(u)}{\sqrt{u}} du[/tex]
and then I guess take the maximum of the cos function as 1 and just remove it, and use the remaining function, which I think would be quite easy to integrate and would give a maximum bound.
Any thoughts greatly appreciated.

Regards,
Adam

Homework Statement


Homework Equations


The Attempt at a Solution


I don't see why you just don't use cos(x^2)^2<=1 to get a bound on the tail. But sure, you can do it that way too.
 
Last edited:
Yeah I just realized that my way is retarded and ends up with the exact same result.

Thanks!