For \(n\in\mathbb Z\), define \[ f_n(x)=\frac{(x-i)^n}{\sqrt\pi\,(x+i)^{n+1}}, \qquad x\in\mathbb R. \] Prove that the functions \(f_n\) are orthonormal: \[ \int_{-\infty}^{\infty}f_m(x)\overline{f_n(x)}\,dx = \begin{cases} 1,&m=n,\\ 0,&m\ne n. \end{cases} \]
Solution
1. For real \(x\) and every \(n\in\ZZ\),\[ \overline{f_n(x)} = \frac{(x+i)^n}{\sqrt\pi\,(x-i)^{n+1}}. \]
For real \(x\), \[ \overline{x-i}=x+i, \qquad \overline{x+i}=x-i. \] Complex conjugation commutes with every integer power because \(x\pm i\neq0\). Applying conjugation to the definition of \(f_n\) gives the formula.
2. For all \(m,n\in\ZZ\),\[ f_m(x)\overline{f_n(x)} = \frac{1}{\pi(1+x^2)} \left(\frac{x-i}{x+i}\right)^{m-n}. \]
Using step 1, \[ \begin{aligned} f_m(x)\overline{f_n(x)} &= \frac1\pi \frac{(x-i)^m(x+i)^n} {(x+i)^{m+1}(x-i)^{n+1}}\\ &= \frac1\pi \left(\frac{x-i}{x+i}\right)^{m-n} \frac1{(x-i)(x+i)}\\ &= \frac{1}{\pi(1+x^2)} \left(\frac{x-i}{x+i}\right)^{m-n}. \end{aligned} \]
3. Under the substitution\[ x=\cot\frac{\theta}{2}, \qquad 0<\theta<2\pi, \]one has\[ \frac{x-i}{x+i}=e^{-i\theta} \]and\[ \frac{dx}{1+x^2}=-\frac12\,d\theta. \]
Writing \(t=\theta/2\), \[ \frac{\cot t-i}{\cot t+i} = \frac{\cos t-i\sin t}{\cos t+i\sin t} = \frac{e^{-it}}{e^{it}} = e^{-2it} = e^{-i\theta}. \] Also, \[ dx = -\frac12\csc^2\frac{\theta}{2}\,d\theta, \] while \[ 1+x^2 = 1+\cot^2\frac{\theta}{2} = \csc^2\frac{\theta}{2}. \] Dividing gives the second formula.
4. For every \(m,n\in\ZZ\),\[ \int_{-\infty}^{\infty} f_m(x)\overline{f_n(x)}\,dx = \frac1{2\pi} \int_0^{2\pi}e^{-i(m-n)\theta}\,d\theta. \]
By step 2, the absolute value of the integrand is \[ \frac1{\pi(1+x^2)}, \] because \[ \abs{\frac{x-i}{x+i}}=1 \] for real \(x\). Thus the integral is absolutely convergent.
As \(\theta\) increases from \(0\) to \(2\pi\), \[ \cot\frac{\theta}{2} \] decreases from \(+\infty\) to \(-\infty\). Applying step 3 and reversing the limits gives \[ \begin{aligned} \int_{-\infty}^{\infty} f_m(x)\overline{f_n(x)}\,dx &= \frac1\pi \int_{2\pi}^{0} e^{-i(m-n)\theta} \left(-\frac12\right)d\theta\\ &= \frac1{2\pi} \int_0^{2\pi}e^{-i(m-n)\theta}\,d\theta. \end{aligned} \]
5. The integral in step 4 equals\[ \boxed{ \begin{cases} 1,&m=n,\\ 0,&m\neq n. \end{cases} } \]
If \(m=n\), the integrand in step 4 is \(1\), so the normalized integral equals \(1\).
If \(m\neq n\), set \(r=m-n\in\ZZ\setminus\{0\}\). Then \[ \int_0^{2\pi}e^{-ir\theta}\,d\theta = \left[ \frac{e^{-ir\theta}}{-ir} \right]_0^{2\pi} = 0 \] because \(e^{-2\pi ir}=1\).
6. Therefore the family \((f_n)_{n\in\ZZ}\) is orthonormal.
Step 5 is exactly the stated orthonormality relation.
7. Q.E.D.
Step 6 proves the claim.