With abs c gt absa+absb calculate lim_(x->oo)1/x int_0^x (dt)/(a sint+ bcost+c)?

1 Answer
Jun 4, 2017

1/sqrt(c^2-a^2-b^2)

Explanation:

Note that sin t and cos t are both periodic with period 2pi.

Hence the limit boils down as:

lim_(x->oo) 1/x int_0^x dt/(a sin t + b cos t + c) = 1/(2pi) int_-pi^pi dt/(a sin t + b cos t + c)

Note also that:

a sin t + b cos t = sqrt(a^2+b^2) cos (t + theta)

where theta = -tan^(-1)(a/b)

which is also periodic with period 2pi. So our integral boils down to:

1/(2pi) int_-pi^pi dt/(sqrt(a^2+b^2)cos t+c)

Next use a Weierstrass substitution, using:

cos t = (1-tau^2)/(1+tau^2)

where:

tau = tan (t/2)

and:

dt = 2/(1+tau^2) d tau

So:

1/(2pi) int_-pi^pi dt/(sqrt(a^2+b^2)cos t+c)

=1/(2pi) int_-oo^oo 1/(sqrt(a^2+b^2)(1-tau^2)/(1+tau^2)+c)*(2d tau)/(1+tau^2)

=1/pi int_-oo^oo (d tau)/(sqrt(a^2+b^2)(1-tau^2)+c(1+tau^2))

=1/pi int_-oo^oo (d tau)/((c+sqrt(a^2+b^2))+(c-sqrt(a^2+b^2))tau^2)

=1/(pi sqrt(c^2-a^2-b^2)) int_-oo^oo ((c-sqrt(a^2+b^2))/sqrt(c^2-a^2+b^2)d tau)/(1+((c-sqrt(a^2+b^2))/sqrt(c^2-a^2-b^2)tau)^2)

=1/(pi sqrt(c^2-a^2-b^2)) int_-oo^oo (du)/(1+u^2)" " with u = (c-sqrt(a^2+b^2))/sqrt(c^2-a^2-b^2)tau

=1/(pi sqrt(c^2-a^2-b^2)) [tan^(-1) u]_-oo^oo

=1/(pi sqrt(c^2-a^2-b^2)) (pi/2-(-pi/2))

=1/sqrt(c^2-a^2-b^2)