Evaluate or justify the following:
- For \(x>0\), the infinite nested radical \[ R(x)=\sqrt{x\sqrt[3]{x\sqrt[4]{x\sqrt[5]{x\cdots}}}} \] and an antiderivative of \(R(x)\).
- \(\displaystyle \int_0^1\frac{\sin(\log x)}{\log x}\,dx\).
- \(\displaystyle \int_{-\infty}^{\infty}\frac{\sin^2x}{x^2}\,dx\).
- \(\displaystyle \lim_{n\to\infty}\int_0^1\frac{x^n}{1+x}\,dx\).
- \(\displaystyle \int_0^\infty \frac{\sin x}{x e^{\sqrt3x}}\,dx\).
- \(\displaystyle \int_{-\infty}^{\infty}\frac{\cos x}{1+x^2}\,dx\).
- \(\displaystyle \int\sqrt{\frac{1+x}{1-x}}\,dx\) for \(-1<x<1\).
Solution
1. Evaluate the infinite nested radical.
Truncate after the \(N\)-th root and write the result as \(R_N(x)=x^{a_{2,N}}\). If the exponent at the \(k\)-th level is \(a_{k,N}\), then \[ a_{N,N}=\frac1N, \qquad a_{k,N}=\frac{1+a_{k+1,N}}k. \] Unrolling gives \[ a_{2,N}=\frac1{2!}+\frac1{3!}+\cdots+\frac1{N!}. \] Hence \[ a_{2,N}\longrightarrow e-2, \] so for \(x>0\), \[ R(x)=x^{e-2}. \] Therefore \[ \boxed{\int R(x)\,dx=\frac{x^{e-1}}{e-1}+C.} \]
2. Evaluate the logarithmic sine integral.
Put \(t=-\log x\). Then \(x=e^{-t}\), \(dx=-e^{-t}dt\), and \[ \int_0^1\frac{\sin(\log x)}{\log x}\,dx =\int_0^\infty e^{-t}\frac{\sin t}{t}\,dt. \] For \(a>0\), define \[ F(b)=\int_0^\infty e^{-at}\frac{\sin(bt)}t\,dt. \] Differentiation under the integral gives \[ F'(b)=\int_0^\infty e^{-at}\cos(bt)\,dt=\frac{a}{a^2+b^2}, \] and \(F(0)=0\). Thus \(F(b)=\arctan(b/a)\). Taking \(a=b=1\), \[ \boxed{\int_0^1\frac{\sin(\log x)}{\log x}\,dx=\frac\pi4.} \]
3. Evaluate the squared-sinc integral.
Use the Fourier transform \(\widehat f(\xi)=\int_{\mathbb R}f(t)e^{-it\xi}\,dt\). For \(f=\mathbf1_{[-1,1]}\), \[ \widehat f(\xi)=\frac{2\sin\xi}{\xi}. \] Plancherel’s theorem in this normalization gives \[ \int_{\mathbb R}|\widehat f(\xi)|^2\,d\xi =2\pi\int_{\mathbb R}|f(t)|^2\,dt=4\pi. \] Therefore \[ 4\int_{-\infty}^{\infty}\frac{\sin^2\xi}{\xi^2}\,d\xi=4\pi, \] so \[ \boxed{\int_{-\infty}^{\infty}\frac{\sin^2x}{x^2}\,dx=\pi.} \]
4. Pass the limit through the integral.
For \(x\in[0,1)\), \(x^n/(1+x)\to0\), while \[ 0\le\frac{x^n}{1+x}\le1. \] Dominated convergence therefore gives \[ \boxed{\lim_{n\to\infty}\int_0^1\frac{x^n}{1+x}\,dx=0.} \]
5. Evaluate the exponentially damped sine integral.
The calculation in step 2 gives, for \(a>0\), \[ \int_0^\infty e^{-ax}\frac{\sin x}{x}\,dx=\arctan\frac1a. \] With \(a=\sqrt3\), \[ \boxed{\int_0^\infty\frac{\sin x}{xe^{\sqrt3x}}\,dx =\arctan\frac1{\sqrt3}=\frac\pi6.} \]
6. Evaluate the Cauchy-kernel cosine transform.
Integrate \[ \frac{e^{iz}}{1+z^2} \] over the upper semicircle of radius \(R>1\). The only enclosed pole is \(z=i\), with residue \[ \operatorname*{Res}_{z=i}\frac{e^{iz}}{1+z^2} =\frac{e^{-1}}{2i}. \] The residue theorem gives a contour integral tending to \(\pi/e\). The semicircular contribution tends to \(0\): away from the endpoints the factor \(e^{-\operatorname{Im}z}\) gives exponential decay, and the two endpoint pieces are \(O(R^{-1})\). Hence \[ \int_{-\infty}^{\infty}\frac{e^{ix}}{1+x^2}\,dx=\frac\pi e. \] Taking real parts yields \[ \boxed{\int_{-\infty}^{\infty}\frac{\cos x}{1+x^2}\,dx=\frac\pi e.} \]
7. Integrate the square-root quotient.
Put \(x=\cos\theta\), where \(0<\theta<\pi\). Then \[ \sqrt{\frac{1+x}{1-x}} =\sqrt{\frac{1+\cos\theta}{1-\cos\theta}} =\cot\frac\theta2, \qquad dx=-\sin\theta\,d\theta. \] Thus \[ \sqrt{\frac{1+x}{1-x}}\,dx =-(1+\cos\theta)\,d\theta. \] Consequently \[ \int\sqrt{\frac{1+x}{1-x}}\,dx =-\theta-\sin\theta+C. \] Since \(-\arccos x=\arcsin x-\pi/2\) and \(\sin\theta=\sqrt{1-x^2}\), the constant absorbs \(-\pi/2\), giving \[ \boxed{\int\sqrt{\frac{1+x}{1-x}}\,dx =\arcsin x-\sqrt{1-x^2}+C.} \]