# Deriving the integral form of the Legendre polynomials from the derivative form

I have the Legendre polynomials $$P_n(x)= \frac{1}{n!2^n}\frac{d^n}{dx^n} (x^2−1)^n$$ and have to show/derive the integral form $$P_n(x)=\frac{1}{\pi} \int_0^{\pi}(x+\sqrt{1−x^2}\cos t)^n dt$$ for $$|x|<1$$ by expressing the derivative form as a complex integral

So as I understand it, I get $$P_n (x) = \frac{1}{n!2^n} \int \frac{d^n}{dz^n} (z^2 - 1)^n dz$$. I'm not sure if this is correct and how to proceed further? Any tip would be much appreciated

• note that your formula is incorrect as you need $P_n(x)=\frac{1}{\pi} \int_0^{\pi}(x+\sqrt{x^2-1}\cos t)^n dt$ since as it stands you get $P_n(0) >0$ for $n$ even and that is not true since $P_{2n}(0)=(-1)^n\frac{{2n \choose n}}{4^n}$ Oct 17, 2022 at 19:09
• @Conrad which formula is incorrect? $P_n (x) = \frac{1}{n!2^n} \int \frac{d^n}{dz^n} (z^2 - 1)^n dz$ ? Oct 17, 2022 at 19:15
• no -the integral which needs to have $\sqrt{x^2-1}\cos t$ not $\sqrt{1-x^2}\cos t$ - for $x=0$ the first gives you for $P_n(0)$ the integral of $(\pm i \cos t)^n$ which is correct, the second gives you only the integral of $(\cos t)^n$ which is incorrect; see proof below which is correct though misses some details which are fairly straightforward Oct 17, 2022 at 19:22
• @Conrad I wrote it as $\sqrt{1 - x^2}$ because that is what my professor wrote on the exercise. So the correct integral expression is $P_n(x)=\frac{1}{\pi} \int_0^{\pi}(x+\sqrt{x^2 - 1}\cos t)^n dt$ for $|x|<1$ ? Oct 17, 2022 at 19:31
• yes that is correct (note that you have a complex square root which as noted is needed); the result holds for all $x$ btw Oct 17, 2022 at 19:34

We start with Rodrigue's Formula $$\begin{eqnarray*} P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} (x^2-1)^n. \end{eqnarray*}$$ Using Cauchy's $$n^{th}$$ derivative formula we have Schlafli's formula $$\begin{eqnarray*} P_n(x) = \frac{1}{2 \pi i} \int_C \frac{(t^2-1)^n}{2^n(t-x)^{n+1}} dt \end{eqnarray*}$$ where $$C$$ is any contour going around $$t=x$$.
Now choose this contour to be $$\begin{eqnarray*} t=x + (x^2-1)^{1/2} e^{i \phi} \end{eqnarray*}$$ where $$\phi$$ varies from $$- \pi$$ to $$\pi$$ ... giving $$\begin{eqnarray*} P_n(x) = \frac{ 1}{ 2 \pi} \int_{- \pi }^{\pi} ( x+ (x^2-1)^{1/2} \cos( \phi ) )^{n} d \phi. \end{eqnarray*}$$ Finally note that the integrand is an even function of $$\phi$$, so the interval can be split in half and you have the desired formula.
Edit: Note that $$\begin{eqnarray*} t^2-1 &=& x^2-1 + 2x(x^2-1)^{1/2}e^{i \phi} + (x^2-1)e^{2 i \phi} \\ &=& (x^2-1)^{1/2}e^{i \phi}(2x + (x^2-1)^{1/2}(e^{ i \phi}+ e^{ -i \phi}))\\ &=& 2 (x^2-1)^{1/2}e^{i \phi}(x + (x^2-1)^{1/2}\cos \phi ).\\ \end{eqnarray*}$$
• Thanks for the reply!! If I understand correctly, I have $\frac{d^n}{dx^n} (z^2 - 1)^n$ and use the Cauchy integral formula $f^{(n)} (a) = \frac{n!}{2 \pi i} \oint_\gamma \frac{f(z)}{(z-a)^n+1} dz$ and my $f(z)$ is $(z^2 - 1)^n$ ? Oct 17, 2022 at 19:22
• sorry, I meant my $f(z)$ is $\frac{(z^2 - 1)^n}{2^n}$ Oct 17, 2022 at 20:05
• and shouldn't it be $\frac{d^n}{dz^n} (z^2 - 1)^n$ ? Oct 17, 2022 at 20:17