Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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 am trying to evaluate the following integral:

$$I_n=\int\limits_{-1}^1 f(x)P_n(x)dx$$ where $f(x)=1$ for $x\in[-1,0)$ and $f(x)=-1$ for $x\in(0,1]$ and $P_n(x)$ is the Legendre polynomial of degree n.

What I did (which is most probably wrong):

Argue that $I_n=0$ for all even n since $P_{even \,\,\, n}$ is symmetric along $x=0$. But then when I tried to evaluate the odd $n$'s , I get $I_n=2\int\limits_{-1}^0 P_n(x) dx=2\int\limits_{-1}^0 P'_{n+1}(x) -P'_{n-1}(x)dx=0$??

Thanks for helping!

share|cite|improve this question
up vote 0 down vote accepted

The Legendre polynomials satisfy $(2n+1)P_n(x)=P'_{n+1}(x)-P'_{n-1}(x)$, and $P_n(-1)=(-1)^n$, so for odd $n$ we have:

$$ \begin{eqnarray} I_n&=&2\int_{-1}^0 P_n(x)dx\\ &=&\frac{2}{2n+1}\left[ (P_{n+1}(0)-P_{n-1}(0)) - (P_{n+1}(-1)-P_{n-1}(-1)) \right] \\ &=&\frac{2}{2n+1}(P_{n+1}(0)-P_{n-1}(0))\end{eqnarray} $$ But for odd $n$, $P_{n+1}(0)\ne P_{n-1}(0)$, so $I_n \ne 0$.

share|cite|improve this answer
Also, for the only even $n = 0$, it yields $2$. – Felix Marin Jan 28 '14 at 1:57

We have \begin{align*}\int_0^1P_n(x)dx&=-\frac 1{n(n+1)}\int_0^1\frac{d }{dx}\left((1-x^2)\frac d{dx}P_n(x)\right)dx\\ &=-\frac 1{n(n+1)}\left[(1-x^2)\frac d{dx}P_n(x)\right]_{x=0}^{x=1}\\ &=\frac 1{n(n+1)}P_n'(0). \end{align*} Thanks to the formula $P_n(x)=\frac 1{2^nn!}\frac{d^n}{dx^n}(1-x^2)^n$, we get that $P_n'(0)=P_{n-1}(0)$ for all $n$. Since $P_n(0)=\frac 1{2^n}\sum_{k=0}^n(-1)^k\binom nk^2$, we finally find that $I_{2p}=0$ for all $p$ and $$I_{2p+1}=\frac 1{(2p+1)(2p+2)2^{2p}}\sum_{k=0}^{2p}(-1)^k\binom{2p}k^2.$$

share|cite|improve this answer

Well, your $f(x)$ is odd, and Legendre polynomials are even for even $n$ and odd for odd $n$. The product of an even and an odd function is odd (so your integral should indeed be zero in that case), but the product of two odd functions is even. So...

share|cite|improve this answer

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \sum_{\ell = 0}^{\infty}h^{\ell}\int_{-1}^{0}{\rm P}_{\ell}\pars{x} &=\int_{-1}^{0}{\dd x \over \root{1 - 2xh + h^{2}}} =\left.{\root{1 - 2xh + h^{2}} \over -h}\right\vert_{x\ =\ 0}^{x\ =\ 1} ={1 \over h}\bracks{\root{1 + h^{2}} - 1 + h} \\[3mm]&={1 \over h}\bracks{\sum_{\ell = 0}^{\infty} {1/2 \choose \ell}h^{2\ell} - 1 + h}= 1 + \sum_{\ell = 1}^{\infty} {1/2 \choose \ell}h^{2\ell - 1} \end{align}

\begin{align} {1/2 \choose \ell} &= {\Gamma\pars{3/2} \over \ell!\,\Gamma\pars{3/2 - \ell}} = {\root{\pi} \over 2}\,{1 \over \ell!\,\Gamma\pars{3/2 - \ell}} \\[3mm]&= {\root{\pi} \over 2}\, {1 \over \ell!\braces{\pi/\bracks{\Gamma\pars{\ell - 1/2}} \sin\pars{\pi\bracks{\ell - 1/2}}}} = {1 \over 2\root{\pi}}\, {\pars{-1}^{\ell + 1}\Gamma\pars{\ell - 1/2} \over \ell!} \\[2mm]&= {1 \over \root{\pi}}\, {\pars{-1}^{\ell + 1}\Gamma\pars{\ell + 1/2} \over \pars{2\ell - 1}\ell!} = {1 \over \root{\pi}}\, {\pars{-1}^{\ell + 1} \over \pars{2\ell - 1}\ell!}\, {\Gamma\pars{2\ell}\root{\pi} \over 2^{2\ell - 1}\Gamma\pars{\ell}} \\[3mm]&= {\pars{-1}^{\ell + 1} \over \pars{2\ell - 1}\ell!}\, {\pars{2\ell - 1}! \over 2^{2\ell - 1}\pars{\ell - 1}!} \end{align}

$$\color{#00f}{\large\left\lbrace% \begin{array}{rcl} \int_{-1}^{0}{\rm P}_{0}\pars{x}\,\dd x & = & 1 \\[2mm] \int_{-1}^{0}{\rm P}_{2\ell - 1}\pars{x}\,\dd x & = & \pars{-1}^{\ell + 1}\, {2^{-\pars{2\ell - 1}} \over 2\ell - 1}\,{2\ell - 1 \choose \ell}\,,\quad \ell = 1,2,3,\ldots \\[2mm] && 0\ \mbox{otherwise} \end{array}\right.} $$

share|cite|improve this answer

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.