I have the following equality :

$$I_1=-\frac{ab}{2\pi}\int_0^\pi \frac{\cos(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt=\frac{a}{a+b}-\frac{1}{2}$$ $$I_2=-\frac{ab}{2\pi}\int_0^\pi \frac{\sin(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt=0$$

where $0 <b \leq a$.

I used the residues but I could not prove this equality

  • $\begingroup$ Try $t=\pi/2 + u$ $\endgroup$ – Michael Galuza Jul 17 '15 at 17:10
  • $\begingroup$ For $I_2$ , let $\sin t = x $, then you can calculate the integral. Note that $I_1$ has the same form with $I_2$ if you let $t = \pi/2 +u$ as @MichaelGaluza 's recommendation. $\endgroup$ – Hoping_Blessing Jul 17 '15 at 17:22
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    $\begingroup$ Looks like you asked this a couple of months ago … math.stackexchange.com/questions/1306776/… . It is not nice when you ask the same question twice in this site, you can always manage to drag more attention to your first question... $\endgroup$ – Leo Sera Jul 17 '15 at 18:31
  • $\begingroup$ @LeoSera Evidently the OP did not receive a satisfactory answer. Perhaps, you can describe a way to "drag more attention" to the original post. I believe that the question is legitimate since the OP was given one wrong answer (the posted solution for $I_1$ is not correct) and thus should not be closed. $\endgroup$ – Mark Viola Jul 17 '15 at 19:07
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    $\begingroup$ @Dr.MV First of all there is no answers in the OP´s first question, just a hint, which I think may lead to a correct solution, I don't see why it is wrong. Nevertheless it is clear the OP wasn't satisfied with the hint. Second of all you can always edit your old question, adding for example the $I_2$ integral to put your question in the active ones and receive new visits …. So no need to re-ask practically the same question $\endgroup$ – Leo Sera Jul 17 '15 at 20:56

Let $$I=I_1+iI_2.$$ Then \begin{eqnarray} I&=&-\frac{ab}{2\pi}\int_0^\pi \frac{e^{2it}}{a^2\sin^2(t)+b^2\cos^2(t)}dt\\ &=&-\frac{ab}{4\pi}\int_0^{2\pi} \frac{e^{it}}{a^2\frac{1-\cos t}{2}+b^2\frac{1+\cos t}{2}}dt\\ &=&-\frac{ab}{2\pi}\int_0^{2\pi} \frac{e^{it}}{(a^2+b^2)+(b^2-a^2)\cos t}dt\\ &=&-\frac{ab}{2\pi i}\int_{|z|=1} \frac{1}{(a^2+b^2)+(b^2-a^2)\frac12(z+\frac{1}{z})}dz\\ &=&-\frac{ab}{\pi i}\int_{|z|=1} \frac{z}{2(a^2+b^2)z+(b^2-a^2)(z^2+1)}dz\\ &=&-\frac{ab}{\pi i}(2\pi i)\text{Res}\left(\frac{z}{2(a^2+b^2)z+(b^2-a^2)(z^2+1)},\frac{a-b}{a+b}\right)\\ &=&-\frac{ab}{\pi i}2\pi i\frac{a-b}{4ab(a+b)}\\ &=&\frac{b-a}{2(a+b)}\end{eqnarray} and hence $$ I_1=\frac{b-a}{2(a+b)}, I_2=0. $$

  • $\begingroup$ Way to go !! +1 . How did you managed to see that the only pole inside $|z|=1$ was in fact $z_0=\frac{a-b}{a+b}$ ? $\endgroup$ – Leo Sera Jul 17 '15 at 20:58
  • $\begingroup$ Note that $2(a^2+b^2)z+(b^2-a^2)(z^2+1)=0$ has two roots $z_1=\frac{a-b}{a+b},z_2=\frac{a+b}{a-b}$. It is easy to see $|z_1|<1,|z_2|>1$. Done. $\endgroup$ – xpaul Jul 17 '15 at 21:05
  • $\begingroup$ Thanks !! Also note that your answer agrees with the one of @Dr.MV, since $$ \frac{b-a}{2(a+b)} = \frac12 - \frac{a}{a+b} $$ So in fact it looks like the OP has the wrong answer by a factor of $-1$. $\endgroup$ – Leo Sera Jul 17 '15 at 21:07
  • $\begingroup$ @LeoSera, you are right. $\endgroup$ – xpaul Jul 17 '15 at 21:09
  • $\begingroup$ Looks like mine. ;-))) $\endgroup$ – Mark Viola Jul 17 '15 at 21:09

One trick to facilitate analysis is to write

$$\sin^2x=\frac{1-\cos 2x}{2}$$


$$\cos^2x=\frac{1+\cos 2x}{2}$$


$$a^2\sin^2(t)+b^2\cos^2(t)=\frac{b^2+a^2}{2}+\frac{b^2-a^2}{2}\cos 2t$$

For the integral of interest, we can write

$$\begin{align} I_1&=-\frac{ab}{2\pi}\int_0^{\pi} \frac{\cos(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt\\\\ &=-\frac{ab}{4\pi}\int_0^{2\pi}\frac{\cos t}{\frac{b^2+a^2}{2}+\frac{b^2-a^2}{2}\cos t}dt\\\\ &=-\frac{ab}{2\pi(b^2-a^2)}\int_0^{2\pi}\frac{\cos t}{\frac{b^2+a^2}{b^2-a^2}+\cos t}dt\\\\ &=-\frac{ab}{(b^2-a^2)}+\frac{ab(b^2+a^2)}{2\pi(b^2-a^2)^2}\int_0^{2\pi}\frac{1}{\frac{b^2+a^2}{b^2-a^2}+\cos t}dt \end{align}$$

Can you finish now?


Starting with the last term $I_1=-\frac{ab}{(b^2-a^2)}+\frac{ab(b^2+a^2)}{2\pi(b^2-a^2)^2}\int_0^{2\pi}\frac{1}{\frac{b^2+a^2}{b^2-a^2}+\cos t}dt$, we move to the complex plane by letting $z=e^{it}$. Then, $I_1$ becomes $$\begin{align}I_1&=-\frac{ab}{(b^2-a^2)}+\frac{ab(b^2+a^2)}{2\pi(b^2-a^2)^2}\oint_C \frac{-2i}{z^2+2\frac{b^2+a^2}{b^2-a^2}\,z+1}dz\\\\&=-\frac{ab}{(b^2-a^2)}+\frac{ab(b^2+a^2)}{2\pi(b^2-a^2)^2}\left(2\pi i \frac{-2i}{-2\frac{2ab}{a^2-b^2}}\right)\\\\&=-\frac{ab}{(b^2-a^2)}-\frac{a^2+b^2}{2(a^2-b^2)}\\\\&=\frac12 -\frac{a}{a+b}\end{align}$$which differs from the posted answer by a factor of $-1$. For $a=2$ and $b=1$, Wolfram Alpha confirms the value of the integral as $-1/6$ agreeing with the answer herein, whereas the posted answer is $+1/6$.

For $I_2$, enforce the substitution $t= \pi -x$. Then, we see

$$\begin{align} I_2&=\int_0^{\pi} \frac{\sin(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt\\\\ &=\int_{\pi}^{0} \frac{\sin 2(\pi-x)}{a^2\sin^2(\pi-x)+b^2\cos^2(\pi-x)}(-1)dx\\\\ &=\int_{0}^{\pi} \frac{-\sin (2x)}{a^2\sin^2(x)+b^2\cos^2(x)}dx\\\\ &=-I_2 \end{align}$$

Thus, we have $I_2=-I_2$ which, of course, implies $I_2=0$ as expected!

  • $\begingroup$ Looks that the other answer given by @xpaul also agrees with yours, so in fact the OP has an incorrect answer. $$ \frac{b-a}{2(a+b)} = \frac12 - \frac{a}{a+b} $$ $\endgroup$ – Leo Sera Jul 17 '15 at 21:05
  • $\begingroup$ @achaire You're welcome. My pleasure. $\endgroup$ – Mark Viola Jul 18 '15 at 19:27
  • $\begingroup$ @achaire Please let me know how I can improve my answer. I just want to give you the best answer I can. $\endgroup$ – Mark Viola Jul 22 '15 at 15:07
  • $\begingroup$ @Dr.MV I am very glad for your interest, you can improve your answer but I don't understand your last comment... $\endgroup$ – Achaire Jul 23 '15 at 1:11
  • $\begingroup$ @Achaire Which comment? I was asking how I can improve my answer so that it would be more useful for you. $\endgroup$ – Mark Viola Jul 23 '15 at 2:21

For $I_2$, one way is to use $$\frac{d}{dt}\left(a^2\sin^2(t)+b^2\cos^2(t)\right)=2a^2\sin t\cos t-2b^2\cos t\sin t=(a^2-b^2)\sin(2t).$$


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