# Evaluate $\int_{x=0}^{2 \pi}\frac{dx}{(1+a \cos{x})^{b^2}}$

Evaluate: $$\int_{0}^{2 \pi}\frac{dx}{(l^2+r^2+2 l r \cos{x})^{b^2}}$$ where $b^2$ is a real number, $r>0$, and $l \geq 0$.

I can just simplify it to $$c\int_{0}^{2 \pi}\frac{dx}{(1+a \cos{x})^{b^2}}$$ where $a>0$.

Any ideas? simplifications? results?

• Do you have access to the residue theorem? – user217285 Apr 13 '17 at 16:33
• @Nitin Thanks. That's a good point. I am reading it on Wiki and I know it will take a while for me to understand and use it. Can you help how I can apply that to this problem? – Susan_Math123 Apr 13 '17 at 16:46
• is $b^2$ integer? is $a>-1$? – tired Apr 13 '17 at 16:48
• @tired $b^2$ is not integer, it is a real number, and $a>0$. – Susan_Math123 Apr 13 '17 at 16:52
• @tired, I think that could be proved using the integral representation of the hypergeometric function. But we must convert the interval $(0,\pi/2)$ and apply the substitution $t =\cos (x)$ – Zaid Alyafeai Apr 13 '17 at 17:31

We want to calculate the integral (set $x\rightarrow 2\pi-x$)

$$I(a,\beta)=\int_0^{2\pi}\frac{1}{(1+a \cos(x))^{\beta}}=2\int_0^{\pi}\frac{1}{(1+a \cos(x))^{\beta}}=2\int_0^{\pi}\frac{1}{(1+a-2a \sin^2(x/2))^{\beta}}$$

where $|a|<1$ and $\beta\in \mathbb{R}_+$. Let us start with a subsitution $y=\sin(x/2)$. This yields

$$I(a,\beta)=\frac{4}{(1+a)^\beta}\int_{0}^{1}\frac{1}{\sqrt{1-y^2}}\frac{1}{(1-\frac{2a}{1+a} y^2)^{\beta}}$$

setting $y=t^{1/2}$ we get

$$I(a,b)=\frac{2}{(1+a)^\beta}\int_{0}^{1}\frac{t^{-1/2}}{\sqrt{1-t}}\frac{1}{(1-\frac{2a}{1+a} t)^{\beta}}$$...

which equals by Euler's formula (we use $B(1/2,1)=\Gamma(1/2)^2/\Gamma(1)=\pi$ in the second step)

$$I(a,\beta)=\frac{2}{(1+a)^\beta}B(1/2,1){_2F_1}\left(\beta,\frac{1}{2};1;\frac{2a}{1+a}\right)=\\\frac{2\pi}{(1+a)^\beta}{_2F_1}\left(\beta,\frac{1}{2};1;\frac{2a}{1+a}\right) \quad$$

This reproduces Mathematicas result from the comments by the means of Pfaff's transformation Set: $\frac{2a}{1+a}\rightarrow\frac{2a}{a-1}$ and use that $1-\frac{1}{2}=\frac{1}{2}$ :-)

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edit: To make the last step more clear:

Due to Pfaff it holds that

$$_2F_1(a,b;c,z)=\frac{1}{(1-z)^b}{_2F_1}\left(b,c-a;c,\frac{z}{z-1}\right)$$

now we set $z=\frac{2a}{a-1}$,$a=\frac{1}{2}$, $b=\beta$ and $c=1$. We find

$$\frac{1}{(1-a)^{\beta}}{_2F_1}\left(\frac{1}{2},\beta;1,\frac{2a}{a-1}\right)=\frac{1}{(1+a)^{\beta}}{_2F_1}\left(\beta,\frac{1}{2};1,\frac{2a}{a+1}\right)$$

which also means that

$$I(a,\beta)=\frac{2\pi}{(1-a)^{\beta}}{_2F_1}\left(\frac{1}{2},\beta;1,\frac{2a}{a-1}\right)$$

which is exactly Mathematicas claim

• +1 maybe you can write some special cases exploiting known values of the Hypergeometric function. – Zaid Alyafeai Apr 13 '17 at 19:50
• @ZaidAlyafeai i'm very bad with hypergeometrics..the case $\beta=1/4$ seems to reduce to an ellipitic integral which i am too stupid to show – tired Apr 13 '17 at 20:22
• $$I(-1,\beta)=\frac{2\pi}{2^{\beta}}{_2F_1}\left(\frac{1}{2},\beta;1,1\right) = \frac{\sqrt{\pi}}{2^{\beta -1}}\frac{ \Gamma\left(\frac{1}{2} - \beta\right)}{\Gamma(1 - \beta)}$$ – Zaid Alyafeai Apr 13 '17 at 21:35