How to find $\int_{0}^{\pi/2} \log ({1+\cos x}) dx$ using real-variable methods?

Question

How do you find the value of this integral, using real methods?

$$I=\displaystyle\int_{0}^{\pi/2} \log ({1+\cos x}) dx$$

The answer is $2C-\dfrac{\pi}{2}\log {2}$ where $C$ is Catalan's constant.

Answer 1

By using $\cos(x)=\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$ we have: $$\begin{eqnarray*} I = 2\int_{0}^{1}\frac{\log 2-\log(1+t^2)}{1+t^2}dt&=&2\int_{0}^{1}\frac{\log 2-\log t-\log(1/t+t)}{1+t^2}dt\\&=&2I_1-2I_2-2I_3,\end{eqnarray*}$$ where: $$I_1=\int_{0}^{1}\frac{\log 2}{1+t^2}\,dt = \frac{\pi}{4}\log 2,$$ $$I_2=\int_{0}^{1}\frac{\log t}{1+t^2}\,dt = \sum_{k=0}^{+\infty}(-1)^k\int_{0}^{1}t^{2k}\log t\,dt = -C,$$ $$I_3=\int_{0}^{1}\frac{\log(t+1/t)}{t^2+1}\,dt=\int_{1}^{+\infty}\frac{\log u}{u\sqrt{u^2-1}}\,du=\int_{0}^{1}\frac{-\log\nu}{\sqrt{1-\nu^2}}\,d\nu,$$ $$ I_3=-\int_{0}^{\pi/2}\log\cos t\,dt=\frac{\pi}{2}\log 2,$$ hence:

$$ I = 2C-\frac{\pi}{2}\log 2.$$

Answer 2

If you don't mind, this is another solution that is more simplistic. Integrating by parts gives \begin{align} I &=\int^{\pi/2}_0\ln(1+\cos{x})dx\\ &=\int^{\pi/2}_0\frac{x\sin{x}}{1+\cos{x}}dx\\ &=\int^{\pi/2}_0\frac{x\tan{x}}{\sec{x}+1}\frac{\sec{x}-1}{\sec{x}-1}dx\\ &=\underbrace{\int^{\pi/2}_0\frac{x}{\sin{x}}dx}_{2G}-\int^{\pi/2}_0\frac{x}{\tan{x}}dx\\ &=2G-\int^{\pi/2}_0x\cot{x}dx\\ &=2G+\int^{\pi/2}_0\ln\sin{x}dx\\ &=2G-\frac{\pi}{2}\ln{2} \end{align}

Answer 3

Using Weierstrass substitution $$ t=\tan\frac x2\qquad;\qquad\cos x=\frac{1-t^2}{1+t^2}\qquad;\qquad dx=\frac{2}{1+t^2}\ dt $$ we obtain \begin{align} \int_0^{\Large\frac\pi4}\ln(1+\cos x)\ dx&=2\underbrace{\int_0^1\frac{\ln2}{1+t^2}\ dt}_{\color{blue}{\text{set}\ t=\tan\theta}}-2\color{red}{\int_0^1\frac{\ln\left(1+t^2\right)}{1+t^2}\ dt}\\ &=\frac{\pi}{2}\ln2-2\color{red}{\int_0^1\frac{\ln\left(1+t^2\right)}{1+t^2}\ dt}.\tag1 \end{align} Consider \begin{align} \int_0^\infty\frac{\ln\left(1+t^2\right)}{1+t^2}\ dt&=\int_0^1\frac{\ln\left(1+t^2\right)}{1+t^2}\ dt+\underbrace{\int_1^\infty\frac{\ln\left(1+t^2\right)}{1+t^2}\ dt}_{\large\color{blue}{t\ \mapsto\ \frac1t}}\\ &=2\int_0^1\frac{\ln\left(1+t^2\right)}{1+t^2}\ dt-2\int_0^1\frac{\ln t}{1+t^2}\ dt\\ \color{red}{\int_0^1\frac{\ln\left(1+t^2\right)}{1+t^2}\ dt}&=\frac12\underbrace{\int_0^\infty\frac{\ln\left(1+t^2\right)}{1+t^2}\ dt}_{\color{blue}{\text{set}\ t=\tan\theta}}+\int_0^1\frac{\ln t}{1+t^2}\ dt\\ &=-\underbrace{\int_0^{\Large\frac\pi2}\ln\cos\theta\ d\theta}_{\color{blue}{\Large\text{*}}}+\sum_{k=0}^\infty(-1)^k\underbrace{\int_0^1 t^{2k}\ln t\ dt}_{\color{blue}{\Large\text{**}}}\\ &=\frac\pi2\ln2-\text{G},\tag2 \end{align} where $\text{G}$ is Catalan's constant.

$(*)$ can be proven by using the symmetry of $\ln\cos\theta$ and $\ln\sin\theta$ in the interval $\left[0,\frac\pi2\right]$ and $(**)$ can be proven by using formula $$ \int_0^1 x^\alpha \ln^n x\ dx=\frac{(-1)^n n!}{(\alpha+1)^{n+1}}, \qquad\text{for }\ n=0,1,2,\ldots $$ Thus, plugging in $(2)$ to $(1)$ yields \begin{align} \int_0^{\Large\frac\pi4}\ln(1+\cos x)\ dx =\large\color{blue}{2\text{G}-\frac{\pi}{2}\ln2}.\tag{Q.E.D.} \end{align}

Answer 4

If I may give my two cents for what it's worth.

Another way would be to consider the handy Fourier series for $\log(\cos(x/2))$

Starting with the identity mentioned up top, that $1+\cos(x)=2\cos^{2}(x/2)$

write $\displaystyle \log(2\cos^{2}(x/2))=\log(2)+2\log(\cos(x/2))$

Using the series $\displaystyle \log(\cos(x/2))=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}\cos(kx)}{k}-\log(2)$

$=\displaystyle 2\sum_{k=1}^{\infty}\int_{0}^{\frac{\pi}{2}}\frac{(-1)^{k+1}\cos(kx)}{k}-\frac{\pi}{2}\log(2)\int_{0}^{\frac{\pi}{2}}dx$

$=\displaystyle 2\sum_{k=1}^{\infty}\frac{(-1)^{k+1}\sin(\pi k/2)}{k^{2}}-\frac{\pi}{2}\log(2)\tag{1}$

Notice the Clausen series now obtained. It has period:

$\displaystyle \begin{array}{rcl}k=1&k=2&k=3& k=4& k=5& k=6&k=7&k=8 \\ 1&0&-1&0&1&0&-1&0\end{array}$

See how it repeats with period 4 from 1 to -1?.

What we now have is the series $2\displaystyle \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{(2k-1)^{2}}=2G$

Put this together with $\frac{-\pi}{2}\log$ in (1) and get:

$$2G-\frac{\pi}{2}\log(2)$$

Answer 5

$$ \begin{aligned} \int_{0}^{\frac{\pi}{2}} \ln (1+\cos x) d x &=\int_{0}^{\frac{\pi}{2}} \ln \left(2 \cos ^{2} \frac{x}{2}\right) d x \\ &=\int_{0}^{\frac{\pi}{2}} \ln 2 d x+2 \int_{0}^{\frac{\pi}{2}} \ln \left(\cos \frac{x}{2}\right) d x \\ &=\frac{\pi}{2} \ln 2+4 \int_{0}^{\frac{\pi}{4}} \ln (\cos x) d x \\ &=\frac{\pi}{2} \ln 2+4\left(-\frac{\pi}{2} \ln 2+\frac{G}{2}\right) \end{aligned} $$ where the last integral comes from my post: $\int_{0}^{\frac{\pi}{4}} \ln (\cos x) d x=-\frac{\pi}{4} \ln 2+\frac{G}{2}$. Hence we can conclude that $$ \boxed{\int_{0}^{\frac{\pi}{2}} \ln (1+\cos x) d x= \frac{G}{2}-\frac{\pi}{2} \ln 2,} $$ where G is the Catalan’s constant.

Answer 6

\begin{align}J&=\int_0^{\frac{\pi}{2}}\ln(1+\cos x)dx\\ K&=\int_0^{\frac{\pi}{2}}\ln(1-\cos x)dx\\ J+K&=2\int_0^{\frac{\pi}{2}}\ln(\sin x)dx\\ &=\int_0^{\frac{\pi}{2}}\ln(\sin x)dx+\underbrace{\int_0^{\frac{\pi}{2}}\ln(\sin x)dx}_{u=\frac{\pi}{2}-x}\\ &=\int_0^{\frac{\pi}{2}}\ln(\sin u\cos u)du\\ &=\int_0^{\frac{\pi}{2}}\ln\left(\frac{\sin(2u)}{2}\right)du\\ &\overset{z=2u}=\frac{1}{2}\int_0^\pi \ln\left(\frac{\sin z}{2}\right)dz \\ &=\frac{1}{2}\int_0^\pi \ln\left(\sin z\right)dz-\frac{1}{2}\pi\ln 2 \\ &=\frac{1}{2}\int_0^{\frac{\pi}{2}} \ln\left(\sin z\right)dz+\frac{1}{2}\underbrace{\int_{\frac{\pi}{2}}^\pi \ln\left(\sin z\right)dz}_{t=\pi-z}-\frac{1}{2}\pi\ln 2\\ &=\frac{1}{2}(J+K)-\frac{1}{2}\pi\ln 2\\ J+K&=\boxed{-\pi\ln 2}\\ K-J&=\int_0^{\frac{\pi}{2}}\ln\left(\frac{1-\cos x}{1+\cos x}\right)dx\\ &\overset{u=\sqrt{\frac{1-\cos x}{1+\cos x}}}=4\int_0^1 \frac{\ln u}{1+u^2}du=-4\text{G}\\ J&=\frac{1}{2}\Big((J+K)-(K-J)\Big)=\frac{1}{2}(-\pi\ln 2+4\text{G})=\boxed{2\text{G}-\frac{1}{2}\pi\ln 2} \end{align}