For \(0<a<b\), evaluate \[ I=\frac1{2\pi}\int_0^{2\pi}\frac{d\theta}{|ae^{i\theta}-b|^4}. \]
Solution
Set \[ r=\frac ab, \] so \(0<r<1\).
1.\[ I = \frac1{b^4}\frac1{2\pi} \int_0^{2\pi} \frac{d\theta}{\abs{1-re^{i\theta}}^4}. \]
Since \[ ae^{i\theta}-b=-b(1-re^{i\theta}), \] taking fourth powers of absolute values gives the identity.
2. For every \(\theta\),\[ \frac1{(1-re^{i\theta})^2} = \sum_{m=0}^{\infty}(m+1)r^m e^{im\theta}. \]
For \(\abs z<1\), \[ \frac1{1-z}=\sum_{m=0}^{\infty}z^m. \] Differentiating gives \[ \frac1{(1-z)^2} = \sum_{m=0}^{\infty}(m+1)z^m. \] Substitute \(z=re^{i\theta}\). Since \(r<1\), the series converges absolutely and uniformly in \(\theta\).
3.\[ \frac1{2\pi} \int_0^{2\pi} \frac{d\theta}{\abs{1-re^{i\theta}}^4} = \sum_{m=0}^{\infty}(m+1)^2r^{2m}. \]
2. Expandingthe absolutely convergent series and using\[ \frac1{2\pi}\int_0^{2\pi}e^{i(m-n)\theta}\,d\theta = \begin{cases} 1,&m=n,\\ 0,&m\neq n, \end{cases} \]leaves exactly the displayed sum.
4. For \(\abs x<1\),\[ \sum_{m=0}^{\infty}(m+1)^2x^m = \frac{1+x}{(1-x)^3}. \]
Starting from \[ \sum_{m=0}^{\infty}(m+1)x^m=\frac1{(1-x)^2}, \] differentiate after multiplying by \(x\): \[ \sum_{m=0}^{\infty}(m+1)^2x^m = \frac{d}{dx}\left(\frac{x}{(1-x)^2}\right) = \frac{1+x}{(1-x)^3}. \]
5.\[ I = \boxed{\frac{a^2+b^2}{(b^2-a^2)^3}}. \]
By steps 1, 3, and 4 with \(x=r^2\), \[ I = \frac1{b^4}\frac{1+r^2}{(1-r^2)^3}. \] Substituting \(r=a/b\) and simplifying gives \[ I = \frac1{b^4} \frac{1+a^2/b^2}{(1-a^2/b^2)^3} = \frac{a^2+b^2}{(b^2-a^2)^3}. \]
6. Q.E.D.
Step 5 gives the requested value.