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

1 Answer
George C.
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)#

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