Integral of the following function:

Question

$$I=\int_{0}^{\dfrac\pi4}\log(\cos(x))\mathop{\mathrm{d}x}$$ I can solve it if the limit is from $0$ to $\frac\pi2$. How to do it? I have done like this, tried not to use any knowledge of series but just simple integration rules. $I_1=\int_{0}^{\frac{\pi}{4}} \ln(\sin x)+\ln (\cos x) dx \\ =\int_{0}^{\frac{\pi}{4}}\ln\left(\frac{\sin 2x}{2}\right) dx \\ =-\frac{\pi}{4}\ln 2 -\frac{\pi}{4}\ln 2 \\ =-\frac{\pi}{2}\ln 2$

Now ,

$I_2=\int_{0}^{\frac{\pi}{4}} \ln(\cos x)-\ln(\sin x) dx \\ =\int _{0}^{\frac{\pi}{4}} \ln(\cot x) dx \\ =-\int_{-\infty}^{0} \frac{(e^z)^2}{1+(e^z)^2} dz$

Say $$[\cot x=e^z \implies dx=-\frac{e^z}{1+(e^z)^2} dz]$$ Now from above, $$=-\int_{0}^{\infty} \frac{1}{1+(e^z)^2} dz \\ =-\int_{0}^{\infty} \frac{e^{-z}}{e^z+e^{-z}} dz \\ =-\frac{1}{2} \int_{0}^{\infty} 1-\frac{e^z-e^{-z}}{e^z+e^{-z}}dz \\ =-\frac{1}{2}\left[z-\ln (e^z+e^{-z})\right]^\infty_0 \\ =-\frac{1}{2}\ln 2$$

This site solves it and shows the steps, along with some other interesting properties quite well: http://www.integral-calculator.com/ — Irregular User, Mar 30 '16 at 06:34
See here: Convergence of the integral $\int_0^{\frac{\pi}{2}}\ln(\cos(x))dx$ — hamid kamali, Mar 30 '16 at 06:42

score 6 · Answer 1 · answered Mar 30 '16 at 06:45

$$I(t)=\int_0^{t}\log\big(\cos(x)\big)\,dx$$ is a very difficult integral involving the polylogarithm function. For $t=\frac \pi 4$, the explicit result is $$I(\frac \pi 4)=\frac{C}{2}-\frac{\pi}{4} \log (2)$$ where appears Catalan number.

As Archis Welankar commented, it could be a good idea to compose Taylor series for the integrand. This would lead to $$\log\big(\cos(x)\big)=-\frac{x^2}{2}-\frac{x^4}{12}-\frac{x^6}{45}-\frac{17 x^8}{2520}-\frac{31 x^{10}}{14175}+O\left(x^{11}\right)$$ Integrating and using bounds, this will give $\approx -0.0864107$ for an exact solution $\approx -0.0864137$

score 4 · Answer 2 · edited Apr 13 '17 at 12:20

As shown in this answer, $$ \int_0^{\pi/2}\log(1+\cos(x))\,\mathrm{d}x=2\mathrm{G}-\frac\pi2\log(2) $$ where $\mathrm{G}$ is Catalan's Constant.

Therefore, $$ \begin{align} \int_0^{\pi/4}\log(\cos(x))\,\mathrm{d}x &=\frac12\int_0^{\pi/2}\log(\cos(x/2))\,\mathrm{d}x\\ &=\frac14\int_0^{\pi/2}\log\left(\frac{1+\cos(x)}2\right)\,\mathrm{d}x\\ &=\frac14\int_0^{\pi/2}\log(1+\cos(x))\,\mathrm{d}x-\frac\pi8\log(2)\\[3pt] &=\frac{\mathrm{G}}2-\frac\pi8\log(2)-\frac\pi8\log(2)\\[6pt] &=\frac{\mathrm{G}}2-\frac\pi4\log(2) \end{align} $$

Integral of the following function:

2 Answers2