I decided to look through Tristan Needham's Complex Analysis book since it's usually mentioned with great praise. Just doing some exercises, I got stuck on #4 of Chapter 9).

Here $P_n(z)$ denotes the $n$-th Legendre polynomial. I've been able to derive that $$ P_n(z)=\frac{1}{2\pi i}\int_K\frac{(Z^2-1)^n}{2^n(Z-z)^{n+1}}dZ $$ for $K$ any simple loop around $z$.

Then the book says by taking $K$ to be a circle of radius $\sqrt{|z^2-1|}$ centered at $z$, $$ P_n(z)=\frac{1}{\pi}\int_0^\pi(z+\sqrt{z^2-1}\cos t)^n dt. $$

I tried to rewrite the RHS of the original equation by reparametrizing $Z=z+\sqrt{|z^2-1|}e^{it}$. However, upon rewriting in terms of the standard substitutions, the integral becomes unmanageable. I have $dZ=i\sqrt{|z^2-1|}e^{it}dt$, $Z^2-1=z^2+2z\sqrt{|z^2-1|}e^{it}+|z^2-1|e^{2it}-1$, $(Z-z)=\sqrt{|z^2-1|}e^{it}$.

Substituting in, $$ \frac{1}{2^{n+1}\pi i}\int_0^{2\pi}\left(\frac{Z^2-1}{Z-z}\right)^2\frac{i\sqrt{|z^2-1|}e^{it}dt}{\sqrt{|z^2-1|}e^{it}} $$ which simplifies to $$ \frac{1}{2^{n+1}\pi}\int_0^{2\pi}\left(\frac{z^2-1}{\sqrt{|z^2-1|}e^{it}}+2z+\sqrt{|z^2-1|}e^{it}\right)^n dt. $$ Is there a way to put this into the final desired form? Thanks.

  • $\begingroup$ +1 for reading Needham's Visual Complex Analysis. $\endgroup$
    – lhf
    Mar 19, 2012 at 22:18
  • $\begingroup$ well if $z^2-1\ge 0$ $\frac{z^2-1}{\sqrt{|z^2-1|}e^{it}}=\sqrt{|z^2-1|}e^{-it}$ and you'll get $\left(2\sqrt{|z^2-1|}\cos(t)+2z\right)^n$ $\endgroup$ Mar 20, 2012 at 1:11
  • $\begingroup$ Thanks @Raymond, but what if $z^2-1$ isn't real, let alone nonnegative? $\endgroup$
    – Yong Pan
    Mar 20, 2012 at 1:18
  • $\begingroup$ hmmm... looks like $\left(2i\sqrt{|z^2-1|}\sin(t)+2z\right)^n$ seems not good but... the integral is over a period and $\left(i\sqrt{|z^2-1|}\right)=\left(\sqrt{z^2-1}\right)$ so perhaps good after all! $\endgroup$ Mar 20, 2012 at 1:24
  • $\begingroup$ @RaymondManzoni Sorry, how did you get a $\sin$ in the integrand, and that last equality $\left(i\sqrt{|z^2-1|}\right)=\left(\sqrt{z^2-1}\right)$? If you want to put it as an answer, I'd accept, since I think this will clear up my trouble. $\endgroup$
    – Yong Pan
    Mar 20, 2012 at 1:35

1 Answer 1


Let's start with : $$ I(z)=\frac{1}{2^{n+1}\pi}\int_0^{2\pi}\left(\frac{z^2-1}{\sqrt{|z^2-1|}e^{it}}+2z+\sqrt{|z^2-1|}e^{it}\right)^n dt $$

If $z^2-1\ge 0$ then : $$ I(z)=\frac{1}{2^{n+1}\pi}\int_0^{2\pi}\left(\sqrt{|z^2-1|}e^{-it}+2z+\sqrt{|z^2-1|}e^{it}\right)^n dt $$ $$ =\frac{1}{2\pi}\int_0^{2\pi}\left(\sqrt{z^2-1}\cos(t)+z\right)^n dt $$

If $z^2-1< 0$ then : $$ I(z)=\frac{1}{2^{n+1}\pi}\int_0^{2\pi}\left(\frac{-(1-z^2)e^{-it}}{\sqrt{1-z^2}}+2z+\sqrt{1-z^2}e^{it}\right)^n dt $$ $$ =\frac{1}{2^{n+1}\pi}\int_0^{2\pi}\left(-\sqrt{1-z^2}e^{-it}+2z+\sqrt{1-z^2}e^{it}\right)^n dt $$

$$ =\frac{1}{2^{n+1}\pi}\int_0^{2\pi}\left(\sqrt{1-z^2}(-e^{-it}+e^{it})+2z\right)^n dt $$ $$ =\frac{1}{2^{n+1}\pi}\int_0^{2\pi}\left(\sqrt{z^2-1}(e^{it}-e^{-it})i+2z\right)^n dt $$ (using $\sqrt{-1}=i$, $\sqrt{-1}=-i$ would give the same result as we will see) (see too this) $$ =\frac{1}{2\pi}\int_0^{2\pi}\left(\sqrt{z^2-1}(-\sin(t))+z\right)^n dt $$ use the change of variable $t=x+3\frac{\pi}2$ to get (cutting the integral in two parts $(0, 2\pi-3\pi/2)$ and $(-3\pi/2,0)$ with the second interval replaced by $2\pi-3\pi/2,2\pi$) : $$ =\frac{1}{2\pi}\int_0^{2\pi}\left(\sqrt{z^2-1}\cos(x)+z\right)^n dx $$

so that for any real $z$ we have : $$ I(z)=\frac{1}{2\pi}\int_0^{2\pi}\left(\sqrt{z^2-1}\cos(t)+z\right)^n dt $$ Analytic continuation should allow you to extend this in the complex plane (we are searching polynomials after all!) but this is probably not the simple answer you hoped...

  • $\begingroup$ Strange, I didn't expect the computation to be so involved. Thanks for writing this. $\endgroup$
    – Yong Pan
    Mar 20, 2012 at 2:28
  • $\begingroup$ yes there should be a shorter path but I don't see it so late... $\endgroup$ Mar 20, 2012 at 2:33

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