Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
+1 for reading Needham's Visual Complex Analysis. – lhf Mar 19 '12 at 22:18
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 $\left(2\sqrt{|z^2-1|}\cos(t)+2z\right)^n$ – Raymond Manzoni Mar 20 '12 at 1:11
Thanks @Raymond, but what if $z^2-1$ isn't real, let alone nonnegative? – Yong Pan Mar 20 '12 at 1:18
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! – Raymond Manzoni Mar 20 '12 at 1:24
@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. – Yong Pan Mar 20 '12 at 1:35
up vote 2 down vote accepted

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...

share|cite|improve this answer
Strange, I didn't expect the computation to be so involved. Thanks for writing this. – Yong Pan Mar 20 '12 at 2:28
yes there should be a shorter path but I don't see it so late... – Raymond Manzoni Mar 20 '12 at 2:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.