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It is known (can be deduced in one turn) from this post that $$ I=\int^{\pi/2}_{0}\frac{\theta \cos(\theta)}{1+\sin^{2}(\theta)}d\theta=\frac{1}{2}\ln^2(1+\sqrt{2}) $$ I'm trying to use different approach of evaluation:

  1. Generalize with parametrization: $\theta\to \arccos(b\cos(\theta))$. $b\in[-1,1]$

  2. Differentiate over $b$: $$\frac{\partial I}{\partial b}=-\int^{\pi/2}_{0}\frac{\cos^2(\theta)}{(1+\sin^{2}(\theta))\sqrt{1-b^2\cos^2(\theta)}}d\theta$$ Note: argument is continuous at all point exept $\theta=\pi k, k\in\mathbb{Z}, b=\pm1$.

  3. Use symmetry of reflection: $$-\frac{1}{2}\int^{\pi/2}_{-\pi/2}\frac{\cos^2(\theta)}{(1+\sin^{2}(\theta))\sqrt{1-b^2\cos^2(\theta)}}d\theta$$

  4. Shift $\theta\to\theta +\pi/2$: $$-\frac{1}{2}\int^{\pi}_{0}\frac{\sin^2(\theta)}{(1+\cos^{2}(\theta))\sqrt{1-b^2\sin^2(\theta)}}d\theta$$

  5. Use identities: $\cos^2(\theta)=\frac{1+\cos(2\theta)}{2},\sin^2(\theta)=\frac{1-\cos(2\theta)}{2}$ and substitute $t=2\theta$: $$-\frac{1}{4}\int^{2\pi}_{0}\frac{1-\cos(t)}{(3+\cos(t))\sqrt{1-\frac{b^2}{2}(1-\cos(t))}}dt$$

  6. Use substitution from complex analysis: $\cos(t)=\frac{z^2+1}{2z},dt=\frac{dz}{iz}$, which after simplification gives: $$\frac{1}{2|b|i}\int_{|z|=1}\frac{z^2-2z+1}{(z^2+6z+1)\sqrt{z^2+\left(\frac{4}{b^2}-2\right)z+1}}\frac{dz}{\sqrt{z}}$$

  7. Nonzero residue only at $z_1=-3+2\sqrt{2}$, which after all simplifications leads to $$\frac{\partial I}{\partial b}=\frac{\pi}{2i}\frac{1}{\sqrt{b^2-\frac{1}{2}}}$$

Which definitely is not right, because for $b>\frac{\sqrt{2}}{2}$ result appears to be complex, while it should be real. So, I wonder where I made a mistake.

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Kostiantyn Lapchevskyi
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    write $2 \int_0^{\pi/2}\frac{x \cos(x)}{1+\sin^2(x)}= \int_0^{\pi}\frac{x \cos(x)}{1+\sin^2(x)}+\pi \int_0^{\pi/2}\frac{ \cos(x)}{1+\sin^2(x)}$. The latter integral is standard whereas the first one can be tackled by complex analysis as nicely explained here: http://residuetheorem.com/2014/12/27/a-tale-of-two-integrals/ – tired Sep 15 '16 at 23:36
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    @tired: thank you, it is exactly what I looked for – Kostiantyn Lapchevskyi Sep 15 '16 at 23:42
  • Don't think the proposed duplicate targets apply. The question here is "I wonder where I made a mistake" or at least asking for a derivation that's very close to his. – Lee David Chung Lin Feb 02 '20 at 11:05

1 Answers1

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A real-analytic approach. Through integration by parts we have $$ I = \int_{0}^{\pi/2}\frac{\theta\cos\theta}{1+\sin^2\theta}\,d\theta=\frac{\pi^2}{8}-\int_{0}^{\pi/2}\arctan(\sin\theta)\,d\theta \tag{1}$$ but since $$ \int_{0}^{\pi/2}\sin(\theta)^{2n+1}\,d\theta = \frac{\sqrt{\pi}\,\Gamma\left(n+1\right)}{2\,\Gamma(n+\frac{3}{2})}\tag{2} $$ by exploiting the Taylor series of the arctangent function and performing termwise integration, we have, after some simplification: $$ I = \frac{\pi^2}{8}-\sum_{n\geq 0}\frac{(-1)^n 4^n}{(2n+1)^2\binom{2n}{n}} \tag{3}$$ and $I$ can be written in terms of a squared (hyperbolic) arcsine: $$ I = \frac{1}{2}\text{arcsinh}^2(1) = \color{red}{\frac{1}{2}\log^2(1+\sqrt{2})}\tag{4}$$ as wanted.

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Jack D'Aurizio
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  • Thank you for attention, but this answer overlaps with linked one and says nothing about why complex analysis aproach fails, which is the main question – Kostiantyn Lapchevskyi Sep 15 '16 at 17:23
  • @KostiantynLapchevskyi: it does not fail! In the linked answer there are multiple approaches to prove the key identity about the squared arcsine, and one of them exploits complex analysis. – Jack D'Aurizio Sep 15 '16 at 17:24
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    We may directly exploit the residue theorem, but we have to be careful in dealing with the square root branch point. Or we may remove the branch point through a suitable substitution and deal with a simpler integral. – Jack D'Aurizio Sep 15 '16 at 17:26
  • Maybe it just me, but not seeing complex analysis among those approaches. However, you're right regarding beeing careful with branch points. Most likely that I was wrong thinking that small (almost) circle around $z=0$ retuns $0$. – Kostiantyn Lapchevskyi Sep 15 '16 at 19:41