Re: Help to calculate an integral

Matthew <xuzlmath@xxxxxxxxx> writes:

Hi, all, I need help finding an integration arising in my research
topic as follows:

\int_0^2pi cos(m x)/sqrt(a+b cos(x)) dx,

in which "a" and "b" are real, positive, and satisfy a>b, m=1,2,3,...

Please help me. Thank you in advance.

For simplicity take b=1 (we can always scale it).
We can write the integrand as

a^(-1/2) cos(m x) sum_{k=0}^infty (2k)! (-1/4)^k cos(x)^k (b/a)^k/(k!)^2

Now

int_0^{2 pi} cos(m x) cos(x)^k dx = (k choose (k-m)/2) 2^(1-k) pi
for 0 <= m <= k with k == m mod 2, 0 otherwise

so the integral becomes

a^(-1/2) sum_{j=0}^infty (2m+4j)! (-1/4)^(m+2j) (b/a)^(m+2j) (m+2j choose j)
2^(1-m-2j) pi/((m+2j)!)^2

which according to Maple is

2(2m)!(-1)^m pi hypergeom([m/2+3/4,m/2+1/4],[m+1],1/a^2) (b/a)^m
/ (a^(1/2)(m!)^2 8^m)
= 2 sqrt(2/b) (-1)^m LegendreQ(m-1/2, a/b)

LegendreQ(m-1/2,a/b) is a Legendre function of the second kind.
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.

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