1
$\begingroup$

so I'm revising contour integration for an upcoming complex analysis exam. I have been asked to integrate $$\int_0^{2\pi}\frac{\sin^2x}{a+b \cos x}dx$$

I thought the sensible thing to do here would be to substitute in $z=e^{ix}$ and take a contour integral around the unit circle, call this path $\ast$ so that my integral becomes $$\frac{1}{2i}Re(\int_\ast\frac{1-z^2}{az+bz^2}dz)$$

Then, letting $f(z)=\frac{1-z^2}{az+bz^2}$, I thought the function had simple poles at $z=0$ with residue $\frac{1}{a}$ and another simple pole at $z=\frac{-a}{b}$ with residue $\frac{a}{b^2}-\frac{1}{a}$ and thus I get the that $$Re(\int_\ast\frac{1-z^2}{az+bz^2}dz)=2i(2\pi i)\frac{a}{b^2}=-4\pi(\frac{a}{b^2})$$ which is not the answer given, that is: $=\frac{2\pi}{b^2}[a-\sqrt{a^2-b^2}]$,but I can't work out why.

Any help appreciated, thank you in advance.

$\endgroup$
4
  • $\begingroup$ I got $$\frac{-i}{2}\int_*\frac{(1-z^2)^2}{z^2(bz^2+2az+b)}\;dz$$ $\endgroup$
    – GEdgar
    May 22 '19 at 17:30
  • $\begingroup$ You mistake is that $Re(x)/Re(y)\neq Re(x/y)$ $\endgroup$ May 22 '19 at 17:37
  • 1
    $\begingroup$ Is there some assumption about the values of $a$ and $b$? $\endgroup$ May 22 '19 at 17:40
  • $\begingroup$ Ah I see, thank you so much. Assumption on a and b is that a>b>0. $\endgroup$
    – xyz12345
    May 22 '19 at 17:51
1
$\begingroup$

$$ \begin{align} \int_0^{2\pi}\frac{\color{#C00}{\sin^2(x)}}{\color{#090}{a+b\cos(x)}}\,\color{#00F}{\mathrm{d}x} &=\oint\frac{\color{#C00}{-\frac{z^4-2z^2+1}{4z^2}}}{\color{#090}{\frac{bz^2+2az+b}{2z}}}\color{#00F}{\frac{\mathrm{d}z}{iz}}\\ &=-\frac1{2i}\oint\frac{z^4-2z^2+1}{bz^2+2az+b}\,\frac{\mathrm{d}z}{z^2} \end{align} $$ The residue of $\frac{z^4-2z^2+1}{bz^2+2az+b}\frac1{z^2}$ at $z=\frac{-a\pm\sqrt{a^2-b^2}}b$ (simple poles) is $\pm\frac{2\sqrt{a^2-b^2}}{b^2}$ and the residue at $z=0$ (degree $2$ pole) is $-\frac{2a}{b^2}$. Assuming $a\gt0$, we get $2\pi i$ times the sum of the residues inside the unit circle, $\frac{-a+\sqrt{a^2-b^2}}b$ and $0$, to be $$ \int_0^{2\pi}\frac{\sin^2(x)}{a+b\cos(x)}\,\mathrm{d}x=2\pi\frac{a-\sqrt{a^2-b^2}}{b^2} $$

$\endgroup$
0
$\begingroup$

If you substitute $z = e^{ix}$, then $\sin(x) = (z - \frac{1}{z})/(2i) $ and $\sin^2(x) =(z-\frac{1}{z})^2/(-4) = \frac{(z^2-1)^2}{-4z^2} $

$dz = ie^{ix}dx = izdx$ or $dx = \frac{dz}{zi}$

$a+b\cos x = a + b\frac{z+\frac{1}{z}}{2} = \frac{2az + bz^2 +1}{2z}$

So,

$$I = \int_*\frac{\frac{(z^2-1)^2}{-4z^2}dz}{zi\frac{2az + bz^2 +1}{2z}}$$

$$I = \int_*\frac{i(z^2-1)^2dz}{2z^2(2az + bz^2 +1)}$$

$\endgroup$
2
  • $\begingroup$ Quickly typed, but please check sign and that one in the denominator, then how the $z$-s cancel in the denominator of the first $I$, and the $z^{-2}$ from the numerator survives... $\endgroup$
    – dan_fulea
    May 22 '19 at 17:38
  • $\begingroup$ Thanks, adding it! $\endgroup$
    – Ak.
    May 22 '19 at 17:39
0
$\begingroup$

We suppose $b\ne 0$, and get rid of it by force, use rather instead of $a,b$ only the variable $c=a/b$.

(After some edits in the OP we have indeed $0<b<a$, thus $1<c$.)

We will assume $c$ real, $|c|>1$, so that there is no zero in the denominator. (And the integral does not diverge.)

Sooner or later we will have to fight against the two roots of $z^2+2c+1$, we denote them by $U,V$, their product is $UV=1$, and thus we may and do assume $|U|<1$, $|V|>1$.


Let $C$ be the unit circle centered in the origin of the complex plane, then formally using $z=e^{ix}$, $\frac 1{iz}\; dz=dx$ we have: $$ \begin{aligned} J &= \int_0^{2\pi}\frac{\sin^2x}{a+b\cos x}\;dx \\ &= \frac 1b\int_0^{2\pi}\frac{\sin^2x}{\cos x+c}\;dx \\ &= \frac 1b \int_C \frac{ \left(\frac 1{2i}\left(z-\frac 1z\right)\right)^2} {\frac 12\left(z+\frac 1z\right)+c}\;\frac 1{iz}\;dz \\ &= \frac i{2b} \int_C \frac{ (z^2-1)^2} {z^2(z^2+2cz+1)}\;dz\ . \\[3mm] &\qquad\text{ The partial fraction decomposition is:} \\ \frac{ (z^2-1)^2} {z^2(z^2+2cz+1)} &= \frac{ (z^2-1)^2} {z^2(z-U)(z-V)} \\ &= 1+\frac 1{z^2}+\frac{U+V}z -\frac {U-V}{z-V} -\frac {V-U}{z-U} \ . \\[3mm] &\qquad\text{ Only the residues in $0,U$ contribute, so...} \\ J&=2\pi i\cdot \frac i{2b} \cdot[(U+V)+(U-V)] \\ &=2\pi \frac {-U}b =2\pi \frac {c-\sqrt{c^2-1}}b \ . \end{aligned} $$ The last expression corresponds to the expected answer from the OP, recalling that $c=a/b$.


Sage check for the partial fraction decomposition:

sage: var('U,z');
sage: V = 1/U
sage: EE = (z^2-1)^2 / z^2 / (z-U) / (z-V)
sage: EE.partial_fraction(z)
-(U^2 - 1)/(U*z - 1) - (U^2 - 1)/((U - z)*U) + (U^2 + 1)/(U*z) + 1/z^2 + 1
sage: bool( _ == (1 + 1/z^2 + (U+V)/z - (V-U)/(z-U) - (U-V)/(z-V) ) )
True
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.