With abs c gt absa+absb calculate lim_(x->oo)1/x int_0^x (dt)/(a sint+ bcost+c)?
1 Answer
Explanation:
Note that
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
which is also periodic with period
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)" " withu = (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)