Integration 02

[Papapetros Vaggelis]

Compute the integral \(\displaystyle{I=\int_{0}^{1}\ln\,\left(\sqrt{1-x}+\sqrt{1+x}\right)\,dx}\) .

[Papapetros Vaggelis]

The function \(\displaystyle{f:\left[0,1\right]\longrightarrow \mathbb{R}\,\,,f(x)=\ln\,\left(\sqrt{1-x}+\sqrt{1+x}\right)}\) is continuous, so

\(\displaystyle{\int_{0}^{1}f(x)\,dx=\int_{0}^{1}\ln\,\left(\sqrt{1-x}+\sqrt{1+x}\right)\,dx=I\in\mathbb{R}}\) .

We apply the substitution \(\displaystyle{x=\cos\,t}\) and we have \(\displaystyle{1-x=1-\cos\,t=2\,\sin^2\,\frac{t}{2}\,\,,1+x=1+\cos\,t=2\,\cos^2\,\frac{t}{2}}\),

\(\displaystyle{dx=-\sin\,t\,dt}\) and \(\displaystyle{t\in\left[0,\frac{\pi}{2}\right]}\) . Then,

\(\displaystyle{\sqrt{1-x}=\sqrt{2}\,\sin\,\frac{t}{2}\,\,,\sqrt{1+x}=\sqrt{2}\,\cos\,\frac{t}{2}}\) and

\(\displaystyle{I=-\int_{\frac{\pi}{2}}^{0}\sin\,t\cdot \ln\,\left(\sqrt{2}\,\sin\,\frac{t}{2}+\sqrt{2}\,\cos\,\frac{t}{2}\right)\,dt=\int_{0}^{\frac{\pi}{2}}\sin\,t\cdot \ln\,\left(\sqrt{2}\,\sin\,\frac{t}{2}+\sqrt{2}\,\cos\,\frac{t}{2}\right)\,dt}\) .

By the substitution \(\displaystyle{t=2\,u}\), we have that \(\displaystyle{dt=2\,du\,,u\in\left[0,\frac{\pi}{4}\right]}\) and then

\(\displaystyle{I=\int_{0}^{\frac{\pi}{4}}2\,\sin\,\left(2\,u\right)\cdot \ln\,\left[\sqrt{2}\,\left(\sin\,u+\cos\,u\right)\right]\,du\,\,(I)}\) .

\(\displaystyle{I_1=\int_{0}^{\frac{\pi}{4}}2\,\ln\,\sqrt{2}\,\sin\,\left(2\,u\right)\,du=\left[-\ln\,\sqrt{2}\,\cos\,\left(2\,u\right)\right]_{0}^{\frac{\pi}{4}}=\frac{\ln\,2}{2}}\)

Let \(\displaystyle{I_2=\int_{0}^{\frac{\pi}{4}}2\,\sin\,\left(2\,u\right)\cdot \ln\,\left[\sin\,u+cos\,u\right]\,du}\) .

\(\displaystyle{\sin\,u+\cos\,u=\frac{1}{\cos\,\frac{\pi}{4}}\,\left[\cos\,u\cdot \cos\,\frac{\pi}{4}+\sin\,u\cdot \sin\,\frac{\pi}{4}\right]=\sqrt{2}\,\cos\,\left(u-\frac{\pi}{4}\right)}\), so

\(\displaystyle{I_2=\int_{0}^{\frac{\pi}{2}}2\,\sin\,\left(2\,u\right)\cdot \ln\,\left(\sqrt{2}\,\cos\,\left(u-\frac{\pi}{4}\right)\right)\,du}\) .

By substituting \(\displaystyle{u-\frac{\pi}{4}=y}\) , we have that \(\displaystyle{du=dy\,,y\in\left[-\frac{\pi}{4},0\right]\,,\sin\,\left(2\,u\right)=\cos\,\left(2\,y\right)}\) .

Therefore,

\(\displaystyle{\begin{aligned} I_2&=\int_{-\frac{\pi}{4}}^{0}2\,\cos\,\left(2\,y\right)\cdot \ln\,\left(\sqrt{2}\,\cos\,y\right)\,dy\\&=\int_{-\frac{\pi}{4}}^{0}\ln\,\left(\sqrt{2}\,\cos\,y\right)\,d\,\left(\sin\,\left(2\,y\right)\right)\\&=\left[\sin\,\left(2\,y\right)\cdot \ln\,\left(\sqrt{2}\,\cos\,y\right)\right]_{-\frac{\pi}{4}}^{0}-\int_{-\frac{\pi}{4}}^{0}\sin\,\left(2\,y\right)\,d\,\left[\ln\,\left(\sqrt{2}\,\cos\,y\right)\right]\\&=0-\int_{-\frac{\pi}{4}}^{0}\sin\,\left(2\,y\right)\cdot \frac{-\sqrt{2}\,\sin\,y}{\sqrt{2}\,\cos\,y}\,dy\\&=\int_{-\frac{\pi}{4}}^{0}\frac{2\,\sin^2\,y\,\cos\,y}{\cos\,y}\,dy\\&=\int_{-\frac{\pi}{4}}^{0}2\,\sin^2\,y\,dy\\&=\int_{-\frac{\pi}{4}}^{0}\left(1-\cos\,\left(2\,y\right)\right)\,dy\\&=\left[y-\frac{\sin\,\left(2\,y\right)}{2}\right]_{-\frac{\pi}{4}}^{0}\\&=\frac{\pi-2}{4}\end{aligned}}\)

and according to the relation \(\displaystyle{(I)}\) we have that

\(\displaystyle{I=\int_{0}^{1}\ln\,\left(\sqrt{1-x}+\sqrt{1+x}\right)\,dx=I_1+I_2=\frac{2\,\ln\,2+\pi-2}{4}}\) .