3
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

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

Then

\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

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

$\endgroup$

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.