急~高等微積分

Question

To numerically integerate an imporoper integral ∫1~∞ f(x)dx,we must
apporximate it by the truncated integral ∫1~r f(x)dx. then we can use
numerical methods on ∫1~r f(x)dx.

b)since the choice of r that satisfies the tolerance of part a) is quite large,
we must see if there is another way to deal with the integral before
truncation.Make the change of variable x=t^p transforming the integral
into a form ∫1~∞g(t)dt. choose p so that
∫10~∞ [g(t)]dt小於等於10^-6


a)∫1~∞ cosx/(1+x^2) dx

How large must we choose r so that the error


[∫1~∞ cosx/(1+x^2) dx -∫1~r cosx/(1+x^2) dx]小於等於10^-6

[ ]是絕對值ㄉ意思

回答b小提即可
謝

Answer 1

令x= tp => dx = ptp-1dt , 代入原積分式,得
∫1∞ cosx/(1+x2) dx = ∫1∞ ptp-1costp/(1+t2p) dt
欲使 |∫10∞ ptp-1costp/(1+t2p) dt | < 10-6
取 p= 6即可, 因
|∫10∞ ptp-1costp/(1+t2p) dt | < ∫10∞ ptp-1/(1+t2p) dt (同除以 t2p )
= ∫10∞ pt-p-1/(1+t-2p) dt
< ∫10∞ pt-p-1dt = - t-p 代 t=10~∞ = 10-p
(還原至原積分 p=6 )
∫10∞ ptp-1costp/(1+t2p) dt =∫10∞ 6t6-1cost6/(1+t12) dt
令x=t6 ,得
∫(x=101/6~∞) cosx/(1+x2) dx
故求積分∫1r cosx/(1+x2) dx , r=101/6, 即可使誤差< 10-6

Ans: r=101/6