1
$\begingroup$

Calculate with residue theorem: $\int_0^{\pi/2} \frac{1}{a+\sin^2(z)} dz$

I tried to use a contour as follow (without the blue circle) :

http://i.stack.imgur.com/P48XL.png

but it didn't work well. Any ideas?

$\endgroup$
1
  • $\begingroup$ @ nogalis :your contour looks strange because the parameter $a$ do not appear and $z=0$ is not a pole. $\endgroup$
    – JJacquelin
    May 31 '15 at 8:17
1
$\begingroup$

First, as the integrand $f(z)=\frac{1}{a+\sin^2 z}$ is even we have

$$I:=\int_0^{\pi/2}\frac{1}{a+\sin^2z}dz=\frac{1}{2} \int_{-\pi/2}^{\pi/2} \frac{1}{a+\sin^2 z}dz.$$

Second, observe that $f(z+\pi)=f(z)$ as well, so

$$I=\frac{1}{4} \int_{-\pi/2}^{-3 \pi/2} \frac{1}{a+\sin^2 z}dz. $$

Now write $\zeta=e^{iz}$; as $z$ traces the interval $[-\pi/2,3\pi/2]$, $\zeta$ traces the unit circle in the complex plane CCW. Also $$\sin^2 z=\left(\frac{e^{i z}-e^{-iz}}{2i} \right)^2=\left(\frac{\zeta-1/\zeta}{2i} \right)^2,$$ and $d \zeta=i e^{iz} dz=i \zeta dz$. Overall, we find that

$$I=\frac{1}{4}\oint_{|\zeta|=1} \frac{1}{a+\frac{\zeta^2-2+1/\zeta^2}{-4}} \frac{d \zeta}{i \zeta} =\frac{1}{i} \oint_{|\zeta|=1} \frac{\zeta}{4a \zeta^2-\zeta^4+2\zeta^2-1} d\zeta.$$

The denominator here is a biquadratic function, which helps you find the poles.

$\endgroup$
0
$\begingroup$

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

Let: $z=e^{iX}$ then $dz=iz \:dX$ and with $\cos(X)=\frac{1}{2}\left( e^{iX}+e^{-iX}\right)=\frac{1}{2}(z+z^{-1})$ $$I=\int \frac{1}{2a+1-\frac{1}{2}(z+z^{-1})}\frac{1}{iz} dz = 2i\int \frac{1}{z^2-(4a+2)z+1} dz $$ The roots of $z^2-(4a+2)z+1=0$ are $2a+1\pm 2\sqrt{a(a+1)}$

if $a>0$ there are two real poles

if $a=0$ there is one pole $z=1$

if $-1< a < 0$ there are two complex poles.

if $a=-1$ there is one pole $z=-1$

if $a<-1$ there are two real poles

So, depending of the case, one have to chose a different contour.

With this hit, I suppose that you can continue.

$\endgroup$
0
$\begingroup$

A similar approach to JJaquelin.

Note that by symmetry we have $$ I = \int_0^{\pi/2}\frac{1}{a+\sin^2(x)}dx = \int_{-\pi/2}^0 \frac{1}{a+\sin^2(x)}dx = \frac{1}{2}\int_{-\pi/2}^{\pi/2}\frac{1}{a+\sin^2(x)}dx.$$ It follows (using double angle formula) that $$ I = \int_{-\pi}^\pi\frac{1}{4a+2 - 2cos(\theta)}d\theta$$ We're going to think about this as a contour integral around the boundary of the unit disk, $\partial D$. We let $z= r\exp i\theta$, and hence $dz = i z d\theta $. We can note $2\cos \theta = z + z^{-1}$ on the boundary of the unit disk $\partial D$. Hence we have $$ I = i\int_{\partial D} \frac{1}{z^2 - (4a + 2)z + 1}dz$$ Since the denominator is a quadratic we have $$ I = i\int_{\partial D} \frac{1}{(z-z_0)(z-z_1)}dz$$ Where $z_0,z_1$ are the roots of the quadratic found in JJaquelins answer. We're going to exclude the cases where $a=0$ and $a=-1$. In both of these cases, one of the zeros $z_0,z_1$ is located on $\partial D$ and the integral doesn't even converge in principal value. We must also exclude the case where $a = -1/2$, as this gives solutions at $z_0 = i$ and $z_1 = \bar{z}_0$. For all other values of $a$, we can use the residue theorem to work the value of $I$.

I will leave it to you to finish the problem. Consider splitting the zeros into two cases; two real solutions, and a pair of conjugate solutions. Find out when there are no, one or two poles inside the unit disk, and then evaluate the residues.

$\endgroup$

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.