definite Integral

[jacks]

Evaluation of \(\displaystyle \int_{0}^{1}\ln \left(\frac{1+\sqrt{1-x^2}}{x}\right)dx\)

[Papapetros Vaggelis]

Let \(\displaystyle{t\in\left(0,1\right]}\) and \(\displaystyle{I(t)=\int_{t}^{1}\ln\,\left(\frac{1+\sqrt{1-x^2}}{x}\right)\,dx}\) . Then,

\(\displaystyle{I(t)=\int_{t}^{1}\left(\ln\,\left(1+\sqrt{1-x^2}\right)-\ln\,x\right)\,dx}\).

\(\displaystyle{\int_{t}^{1}\ln\,x\,dx=\left[x\,\ln\,x\right]_{t}^{1}-\int_{t}^{1}x\cdot \frac{1}{x}\,dx=t\,\ln\,t+t-1}\)

In order to calculate the integral \(\displaystyle{\int_{t}^{1}\ln\,\left(1+\sqrt{1-x^2}\right)\,dx}\) , we apply the substitution

\(\displaystyle{x=\sin\,y}\) and we have \(\displaystyle{dx=\cos\,y\,dy\,\,,y\in\left[\arcsin\,t,\frac{\pi}{2}\right]}\) , so

\(\displaystyle{\begin{aligned} \int_{t}^{1}\ln\,\left(1+\sqrt{1-x^2}\right)\,dx&=\int_{\arcsin\,t}^{\frac{\pi}{2}}\ln\,\left(1+\cos\,y\right)\,\cos\,y\,dy\\&=\int_{\arcsin\,t}^{\frac{\pi}{2}}\ln\,\left(1+\cos\,y\right)\,\left(\sin\,y\right)'\,dy\\&=\left[\sin\,y\,\ln\,\left(1+\cos\,y\right)\right]_{\arcsin\,t}^{\frac{\pi}{2}}-\int_{\arcsin\,t}^{\frac{\pi}{2}}\sin\,y\,\frac{\left(-\sin\,y\right)}{1+\cos\,y}\,dy\\&=-t\,\ln\,\left(1+\cos\,\left(\arcsin\,t\right)\right)+\int_{\arcsin\,t}^{\frac{\pi}{2}}\frac{\sin^2\,y}{1+\cos\,y}\,dy\\&=-t\,\ln\,\left(1+\cos\,\left(\arcsin\,t\right)\right)+\int_{\arcsin\,t}^{\frac{\pi}{2}}\frac{1-\cos^2\,y}{1+\cos\,y}\,dy\\&=-t\,\ln\,\left(1+\cos\,\left(\arcsin\,t\right)\right)+\int_{\arcsin\,t}^{\frac{\pi}{2}}\left(1-\cos\,y\right)\,dy\\&=-t\,\ln\,\left(1+\cos\,\left(\arcsin\,t\right)\right)+\left[y-\sin\,y\right]_{\arcsin\,t}^{\frac{\pi}{2}}\\&=-t\,\ln\,\left(1+\cos\,\left(\arcsin\,t\right)\right)+\frac{\pi}{2}-1-\arcsin\,t+t\\&=\frac{\pi}{2}-t\,\ln\,\left(1+\cos\,\left(\arcsin\,t\right)\right)-\arcsin\,t+t\end{aligned}}\)

Thus,

\(\displaystyle{\int_{t}^{1}\ln\,\left(\frac{1+\sqrt{1-x^2}}{x}\right)\,dx=\frac{\pi}{2}-1-t\,\ln\,\left(1+\cos\,\left(\arcsin\,t\right)\right)-\arcsin\,t-t\,\ln\,t+1-t\,,t\in\left(0,1\right]}\)

\(\displaystyle{\lim_{t\to 0^{+}}\left(\frac{\pi}{2}-t\,\,\cos\,\left(\arcsin\,t\right)-\arcsin\,t-t\,\ln\,t+t-1\right)=\frac{\pi}{2}}\) because

\(\displaystyle{\lim_{t\to 0^{+}}t\,\cos\,\left(\arcsin\,t\right)=0\cdot \cos\,0=0\,,\lim_{t\to 0^{+}}\arcsin\,t=0}\) and

\(\displaystyle{\lim_{t\to 0^{+}}t\,\ln\,t=\lim_{t\to 0^{+}}\frac{\ln\,t}{1/t}\stackrel{D.L.H}{=}\lim_{t\to 0^{+}}\frac{1/t}{-1/t^2}=\lim_{t\to 0^{+}}\left(-t\right)=0}\)

So, \(\displaystyle{\int_{0}^{1}\ln\,\left(\frac{1+\sqrt{1-x^2}}{x}\right)\,dx=\lim_{t\to 0^{+}}I(t)=\frac{\pi}{2}}\).