Trigonometric Integral Help: Solving for a<1 Using Complex Contour Method

von_biber
Messages
5
Reaction score
0
need some assistance with the following integral:

\int_0^{2\pi} cosx/(a-cosx), a-parameter (say a>0)

i've converted it into a complex contour integral over z=e^(ix):

~ \int_{|z|=1} dz (z^2+1)/[z(z^2-2az+1)]

which is easily evaluated for a>1. my question regards a<1 - i am not sure how to solve it in this case, because the the 2 poles

z_1=a+Sqrt[a^2-1], z_2=a-Sqrt[a^2-1]

are exactly on the unit circle and off the real axis. thanks for any suggestions!
 
Physics news on Phys.org
Hmm..is it really necessay to venture out in complexity here?
Rational expressions in the trig. functions are easily handled by the transformation u=tan(x/2).
 
it get's a bit complicated with this tan(x/2) substitution, with complex substitution it's very easily evaluated for a>1 and i'd like to find out what happens for a<1.
 
Well, for a<1, the integral will readily diverge, so no wonder if you get into some problems.
 
anyway, what is the principal value for a<1?
thanks for your help, btw. i really appreciate it.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top