The question is to evaluate this definite integral: $$\int_{0}^{\pi/2}\frac{x}{(a^2\cos^2x+b^2\sin^2x)}dx$$

I know the definite integration with limits from $0$ to $\pi$, where we use the identity $\int_{a}^{b}f(x)dx = \int_{a}^{b}f(a+b-x)dx$, but it obviously cannot be used here. I also tried breaking the $0$ to $\pi$ one into two parts to no avail. Since the other one had such a simple solution, I can't wrap my head around this one, and would love some help!

  • $\begingroup$ Integration by parts? $\endgroup$ – lab bhattacharjee Mar 19 '18 at 18:04
  • $\begingroup$ Do you know uv integration? If you know try using it here. $\endgroup$ – Netravat Pendsey Mar 19 '18 at 18:04
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    $\begingroup$ We can adopt the Feynman’s trick. The answer is $$\frac{1}{ab}\left( \frac{\pi^2}{8} + \chi_2\left(\frac{a-b}{a+b}\right)\right)$$ where $\chi_2$ is the Legendre chi function. $\endgroup$ – Sangchul Lee Mar 19 '18 at 20:38
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    $\begingroup$ Set $z=e^{i \phi}$ $$I=\oint_{QC} dz \frac{4z\log(z)}{(z^4+1)(a^2+b^2)+2(a^2-b^2)z^2}$$ Cauchy integral formula (no singularities) $$ (a^2+b^2)I=4\int_0^1dx\frac{\log(x)}{x^2+2cx+1}+4 i\int_0^1dx\frac{\log(x)+i\pi/2}{-x^2+2icx+1} $$ Partial fractions reduces this to standard integrals of the form $$ \int_0^1\frac{\log(x)}{\zeta+x} $$ now the boring part begins.. $\endgroup$ – tired Mar 19 '18 at 20:54

By enforcing the substitution $x=\arctan u$ we are left with

$$\mathcal{J}(a,b)=\int_{0}^{\pi/2}\frac{x}{a^2\cos^2 x+b^2\sin^2 x}=\int_{0}^{+\infty}\frac{\arctan u}{a^2+b^2 u^2}\,du $$ and for any $a,b,c>0$ $$ \frac{d}{dc}\int_{0}^{+\infty}\frac{\arctan(c u)}{a^2+b^2 u^2}\,du = \int_{0}^{+\infty}\frac{u\,du}{(a^2+b^2 u^2)(1+c^2 u^2)}\stackrel{\text{PFD}}{=}\frac{\log(c)-\log(b)+\log(a)}{a^2 c^2-b^2}$$ such that $$\mathcal{J}(a,b)=\int_{0}^{1}\frac{\log(c)-\log(b)+\log(a)}{a^2 c^2-b^2}\,dc $$ is related to $\log$ and $\text{Li}_2$ as pointed out by Sangchul Lee and tired in the comments.

  • $\begingroup$ this polylogarithmic integrals are really annoying :-( $\endgroup$ – tired Mar 19 '18 at 20:55
  • $\begingroup$ @tired: I agree, but as you already remarked, everything boils down to $$ \int_{0}^{1}\frac{\log x}{x+\xi}=\text{Li}_2\left(-\frac{1}{\xi}\right)$$ for $\text{Re}(\xi)>0$. $\endgroup$ – Jack D'Aurizio Mar 19 '18 at 21:07

Write $\alpha = a/b$ and substitute $\tan x = \alpha t$ to obtain

$$ \int_{0}^{\frac{\pi}{2}} \frac{x}{a^2\cos^2 x + b^2\sin^2 x} \, dx = \frac{1}{ab} \int_{0}^{\infty} \frac{\arctan(\alpha t)}{1 + t^2} \, dt =: \frac{1}{ab}I(\alpha). $$


\begin{align*} I'(\alpha) &= \int_{0}^{\infty} \frac{t}{(1+t^2)(1+\alpha^2 t^2)} \, dt \\ &\hspace{1.5em} = \left[ -\frac{1}{2(1-\alpha^2)} \log\left( \frac{1+\alpha^2 t^2}{1+t^2} \right) \right]_{0}^{\infty} = -\frac{\log \alpha}{1 - \alpha^2}. \end{align*}

Now it is easy to check that, if we write $\chi_2(z) = \int_{0}^{z} \frac{\operatorname{artanh} \xi}{\xi} \, d\xi$ for the Legendre chi function, then

$$ \frac{d}{d\alpha} \chi_2\left(\frac{1-\alpha}{1+\alpha}\right) = \frac{\log \alpha}{1-\alpha^2}. $$

So it follows that

$$ I(\alpha) = \chi_2(1) - \chi_2\left(\frac{1-\alpha}{1+\alpha}\right) = \frac{\pi^2}{8} + \chi_2\left(\frac{\alpha-1}{\alpha+1}\right). $$

Plugging this back and manipulating a bit, we obtain

$$ \int_{0}^{\frac{\pi}{2}} \frac{x}{a^2\cos^2 x + b^2\sin^2 x} \, dx = \frac{1}{ab} \left( \frac{\pi^2}{8} + \chi_2\left(\frac{a-b}{a+b}\right) \right). $$


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