2

For tackling the integral $$I=\int_0^{\frac{\pi}{2}} \frac{\ln (\cos x)}{1+\sin ^2 x} d x,$$ Feynman’s trick can’t be directly applied. I started to express the integrand in terms of $\tan^2x$ as: $$ \begin{aligned} I & =-\frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x \ln \left(1+\tan ^2 x\right)}{\sec ^2 x+\tan ^2 x} d x \\ & =-\frac{1}{2} \int_0^{\infty} \frac{\ln \left(1+t^2\right)}{1+2 t^2} d t, \quad \textrm{ where } t=\tan x\\&= -\frac{1}{2} I(1), \end{aligned} $$ where $$ I(a)=\int_0^{\infty} \frac{\ln \left(1+a t^2\right)}{1+2 t^2} d t $$ whose derivative w.r.t. $a$ can be found using partial fractions as: $$ \begin{aligned} I^{\prime}(a) & =\int_0^{\infty} \frac{t^2}{\left(1+a t^2\right)\left(1+2 t^2\right)} d t \\ & =\frac{\pi}{2 \sqrt{2} \sqrt{a}(\sqrt{a}+\sqrt{2})} \end{aligned} $$ Integrating back yields $$ \begin{aligned} I&=-\frac{1}{2} \int_0^1 \frac{\pi}{2 \sqrt{2} \sqrt{a}(\sqrt{a}+\sqrt{2})} d a \\ & =-\left.\frac{\pi}{2 \sqrt{2}} \ln (\sqrt{a}+\sqrt{2})\right|_0 ^1\\&= \frac{\pi}{2 \sqrt{2}} \ln (2-\sqrt{2}) \end{aligned} $$

My Question

Are there any alternatives to find the exact value of the integral?

Your comments and alternatives are highly appreciated.

Bowei Tang
  • 3,763
Lai
  • 31,615
  • 1
    Possible generalization: at $a\geqslant0$ and $b>0$ $$\int_0^\frac\pi2\frac{\ln(a^2+\cos^2x)}{b^2+\sin^2x}dx=\frac\pi{\sqrt{b^2+1}}\ln\frac{\sqrt{(a^2+1)(b^2+1)}+ba}{\sqrt{b^2+1}+b}$$ (by means of complex integration) – Svyatoslav May 20 '25 at 17:05
  • 1
    $$\int_0^\infty\frac{\ln(x^2+a^2)}{x^2+b^2}\mathrm dx=\frac\pi{b}\ln(a+b)$$ – Integreek May 20 '25 at 17:15

0 Answers0